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Prime and Composite Numbers

2021-12-27T19:59:00.008+05:45

Identify if a number is prime or composite

Prime and composite number calculator between two numbers.

From: To:

What are prime numbers?

Those numbers greater than 1 which upon division with any natural number except, one(1) and itself, results a decimal quotient are called prime numbers.
Examples of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 97, 197, 199, 991, 997, 9949, 9967, 9973 and so on.

What are composite numbers?

Those numbers which can be expressed as product of atleast two or more prime numbers are called composite numbers.
Examples of composite numbers are 4, 6, 8, 9, 10, 20, 100, 111, 1000, 1010, 1122, 9975 and so on.

Composite numbers Product of prime numbers

4 2×2

64 2×2×2×2×2×2

121 11×11

1024 2×2×2×2×2×2×2×2×2×2

Composite numbers	Product of prime numbers
4	2×2
64	2×2×2×2×2×2
121	11×11
1024	2×2×2×2×2×2×2×2×2×2

Prime Numbers from 1 to 50

Prime Numbers from 1 to 100

Prime Numbers from 1 to 1000

ENIAC

2021-12-16T17:45:00.006+05:45

What does ENIAC stand for?

ENIAC stands for "Electronic Numerical Integrator and Computer".

Which was the first general purpose digital computer?

ENIAC was the world's first general purpose digital computer.

What was some salient features of ENIAC?

Memory size : 20×10 digits
word length: 10 digits
Weight: 30 tons (30,000 Kg)
Dimension: 1,500 square feet.
Power consumption: 140 kilowatts ≈ 300 Macbooks power consumption
Cost: $400,000

Source:-Britannica

Was ENIAC a binary machine?

No, ENIAC was not a binary machine.
It was a decimal machine.

Which number system was used by ENIAC?

Decimal number system was used by ENIAC.

How was a decimal number represented in ENIAC?

To represent a digit, 10 vaccum tubes were arranged in a ring fashion. Each tubes were assigned a unique digit(0-9). At a time only one of these tubes can be in "ON" state. This tube in "ON" state represents the digit.

Where were instructions and data stored in ENIAC?

Data were stored in the 20 accumulators of the ENIAC. These accumulators work as the memory for ENIAC.
There was no facility for storage of instructions. Instructions were provided to the ENIAC manually. This is done through switching and cabling.

Structure and function of a computer system

2021-12-14T19:55:00.001+05:45

Structure

In general:-
Structure of a system or a sub-system is defined as the interconnection of its components.

Defining structure in terms of a hierarchical system:-
When a complex systems like computer is depicted as an hierarchical system then the interconnection of the components at a particular level of the hierarchy is defined as the structure at that level.

Structure of a computer

Structure of a CPU

Structure of a control unit

Topmost level components of computer structure are as follows:-

CPU(Central processing unit)/Processor
Memory
System bus
I/O
Communication lines
Remote devices

All these above components are explained below:-

CPU(Central processing unit)/Processor

A computer system can be seen as a centralized system where CPU lies at the center. It controls all other components of the computer system. It processes the data and all data transfer operations are routed via CPU(except DMA transfer technique).

Memory

Memory is a storage component of the computer system. All the instructions and data are stored here.

System bus

System bus is a combination of parallel wires. All the data movement from one address to other address among the componets of computer structure except remote devices is done via system bus.

I/O devices

These I/O devices moves data between the computer and external environment. Input devices send data to the computer system from the external environment.
For examples:- Mouse, Keyboard, Microphone, Camera, Scanner, etc.
While, output devices receive data from the computer system.
For example:- Monitor, Speaker, Printer etc

Communication lines

To communicate devices which are at remote places or can't be connected to system bus directly are connected to the computer via communication lines. These communication lines can be wired or wireless.
For example:- Ethernet cable, Bluetooth, WiFi, Wireless internet connection etc.

Remote devices

Those devices which communicates with the computer over communication lines are called remote devices.
Foe example a keyboard connected to the computer via bluetooth.

Computer structure

The structural components of cpu are as follows:-

Control unit
Register array
ALU(Arithmetic logic unit)
Internal bus

All these above components are explained below:-

Control unit

Contol unit manages all the resources of the computer system. All the instruction of the computer are achieved when sequence of some microoperations are are executed. This sequence of microoperations is controlled by control unit.

Registers array

Registers array has a number of arrays. It stores data temporarily before data is fed to ALU and after data is received from ALU. Registers are faster than memory but expensive.

ALU(Arithmetic logic unit)

All the data processing tasks are carried out in ALU.

Internal bus

It provides the connection between ALU and registers array for data movement inside CPU.

Structural components of CPU

The structural components of control unit are as follows:-

Address sequencer
CAR(Control address register)
Control memory
CDR(Control data register)

All these above components are explained below:-

Address sequencer

Address sequences generates the address of next microoperation in control memory.

CAR(Control address register)

Contol address register holds the address generated by address sequencer.

Control memory

Every computer has finite number of microoperations. Every microoperation is represented by string of 0's and 1's. This is called control word. These control words are stored in the control memory.

CDR(Control data register)

Control data register holds the control word provided by the control memory.

Structural components of control unit

Function

The operations performed by a system or a sub-system or a component is called its function.

All the functions of computer can be classified into following four functions:-

Control
Data processing
Data storage
Data transfer

All these functions are explained below:-

Control

A computer controls all its other functions like data transfer, storage and processing. A computer is connected to various peripherals. All these devices use resources like memory, ALU, registers, etc. Computer also make these resources available as per the instructions. In case of multiple request for a particular resource a computer decides how to handle requests coming for these resources.

Data transfer

Data has to be moved from one one location to another location in computer world for retrieval, processing or storing at some different location. Most frequent data tranfer routes are:-
memory-memory,
memory-peripherals,
memory-remote devices over communication lines,
peripherals-peripherals,
peripherals-remote devices over communication lines and
remote devices-remote devices over communication lines.

If the data transfer is between the computer and a device that is directly connected to it then it is called I/O communication or peripheral communication.
Here devices receive and send data over system bus.

If the data transfer is between the computer and devices which is connected to the computer with communication lines then it is called data communication.
Here devices receive and send data over communication lines.

Data storage

A computer stores data in its memory or registers. On the fly data (data which come for processing and are send) are stored temporarily. These data are stored in processor registers. While data which are to be retrieved or updated further are stored in its memory.

Data processing

Data are processed to obtain new data. All the data processing operation like arithmetic and logical operation are performed inside computer.

The differences between computer function and computer structure are as follows:-

Computer function	Computer structure
Definition
The operation perfomed by a computer is called computer function.	The interconnection of top-level componets of the computer system is called computer structure.
Control, data processing, data storage and data transfer are computer functions.	CPU, memory, system bus, IOP(Input-output processor), input devices, output devics, communication lines and remote devices are the components the computer structure.
Relation
Computer functions are realized by the computer structure.	Computer structure develops a system to provide computer functions.

Citation:-
Stallings, W. (2010). Computer Organization and Architecture: Designing for performance. Prentice Hall.

Multiprocessors and Multiprocessing

2021-12-13T08:21:00.003+05:45

Multiprocessors and multiprocessing

Definition of multiprocessors

A single chip having two or more interconnected processors incorporated in it is called a multiprocesor.
For example:-
Dual core processors like
Quad core processors like
Octa core processors like
Hexa core processors like

Definition of multiprocessors system

A computer system having two or more interconnected CPUs with its memory and peripheral devices is called a multiprocessor system.
All the commercial PCs and smartphones are widely used multiprocessors system. For example:-

Definition of multiprocessing

The parallel programming technique of executing multiple instructions stream and multiple data stream concurrently in a environment of two or more processors controlled by one operating system is called multiprocessing.

Advantage of mutiprocessing:-

Multiprocessing improves the system performance.

How does multiprocessing improves the system performance?

Multiprocessing improves the system performance.

Classification of multiprocessors

On the basis of the organization of memory, multiprocessors are classified into tightly coupled multiprocesor and loosely coupled multiprocesor.

The multiprocesor system having a global common memory is called a shared memory multiprocesor or tightly coupled multiprocessor.

> The multiprocessor system not having a global common memory but each processor has their own local memory is called distributed-memory multiprocesor or loosely coupled multiprocesor.

The differences between tightly coupled multiprocessors and loosely coupled multiprocesors are as follows:-

Tightly coupled multiprocessors	Loosely coupled multiprocesors
It has a global common memory.	It has no global memory. Each processor has its own local memory.
Efficient when data dependency among task are very frequent.	Efficient when data dependency among task are rare.

NTC: Nepal Telecom Vaccency, Syllabus and Study Materials

2021-12-13T01:14:00.003+05:45

Nepal Telecom

Vaccencies

Promotion Based Competition
Internal Competition
Open Competition
Published date	2078-08-03(2021-Nov-19)
Deadline	2078-08-23(2021-Dec-09)
Deadline(Double Fee Charge)	2078-09-01(2021-Dec-16)

Notice for online application

Syllabus

Post:- Account Manager

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Level	10
Service	Administration
Group	Accoutant
Qualification
Competition	Internal

Post:- Telecom Manager

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Service	Technical
Group	Telecommunication
Qualification
Competition	Internal

Post:- Deputy Account Manager

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Level	9
Service	Administration
Group	Accountant
Qualification	Master in Management + 5 years experience as an Officer Or, Master in Commerce + 5 years experience as an Officer Or, Master in Business Administration + 5 years experience as an Officer Or, PhD + 3 years experience as an Officer
Competition	Internal & Open

Post:- Deputy Administrative Manager

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Service	Administration
Group	Administration
Qualification	Masters + 5 Years experience as an Officer Or, PhD + 3 Years experience as an Officer
Competition	Internal & Open

Post:- Deputy Telecom Manager

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Service	Technical
Group	Telecom
Qualification	Master in Engineering + Registered in NEC + 5 Years work Experience as An Officer Or, PhD + 3 Years work Experience as An Officer
Competition	Internal & Open

Post:- Senior Account Officer

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Level	8
Service	Administration
Group	Account
Qualification	Master in Management + 5 years experience as an Officer Or, Master in Commerce + 5 years experience as an Officer Or, Master in Business Administration + 5 years experience as an Officer Or, PhD + 3 years experience as an Officer
Competition	Internal & Open

Post:- Chartered Accountant

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Service	Administration
Group	Account(C.A.)
Qualification	CA/ACCA/CPA
Competition	Open

Post:- Senior Administrative Officer

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Service	Administration
Group	Administration
Qualification	Masters + 5 Years experience as an Officer Or, PhD + 3 Years experience as an Officer
Competition	Internal & Open

Post:- Senior Business Officer

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Level	8
Service	Administration
Group	Administration
Qualification	Master in Management + 5 years experience as an Officer Or, Master in Commerce + 5 years experience as an Officer Or, Master in Economics + 5 years experience as an Officer Or, Master in Mathematics + 5 years experience as an Officer Or, Master in Statistics + 5 years experience as an Officer Or, Master in Business Administration + 5 years experience as an Officer Or, PhD + 3 years experience as an Officer
Competition	Internal & Open

Post:- Senior Technical Officer

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Service	Technical
Group	Technical
Qualification
Competition	Internal

Post:- Senior Engineer

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Service	Technical
Group	Telecommunication
Qualification	Master in Engineering + Registered in NEC + 5 Years work Experience as An Officer Or, PhD + 3 Years work Experience as An Officer
Competition	Internal & Open

Post:- Account Officer

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Level	7
Service	Administration
Group	Administration/Account
Qualification	-
Competition	Internal

Post:- Business Officer

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Service	Administration
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Post:- Admin officer

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Service	Administration
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Qualification	-
Competition	Internal

Post:- Technical officer

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Service	Technical
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Post:- Electronics and Communication Engineer

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Level	7
Service	Technical
Group	Telecommunication
Qualification	B.E. In Electronics & Communication
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Post:- Computer Engineer

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Service	Technical
Group	Telecommunication(Computer)
Qualification	B.E.
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Post:- Electrical Engineer

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Group	Telecommunication(Electrical)
Qualification	B.E.
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Post:- Civil Engineer

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Service	Technical
Group	Telecommunication(Civil)
Qualification	B.E.
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Post:- Assistant Account Officer

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Level	6
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Qualification	Bachelor in management, Commerce or Business.
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Post:- Assistant Law Officer

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Service	Administration
Group	Administration
Qualification	Bachelor in Law.
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Post:- Assistant Business Officer

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Level	6
Service	Administration
Group	Administration
Qualification	Bachelor in management; Bachelor in commerce; Bachelor in business; Bachelor in mathematics;
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Post:- Assistant Admin Officer

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Post:- Assistant Technical Officer

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Post:- Overseer

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Level	5
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Qualification	Diploma in Architect/Civil/Survey engineering.
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Post:- Senior Assistant Admin

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Post:- Senior Technician

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Post:- Assistant Admin

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Post:- Technician

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Qualification	SEE + (Transmission Training/ Switching Training/ Power Training)
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Post:- Junior Technician

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Service	Technical
Group	Technical
Qualification	SEE + (Transmission Training/ Switching Training/ Power Training)
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Differences between computer architecture and computer organization

2021-12-12T20:25:00.012+05:45

Computer architecture

The portrayal of a computer system from perspective of machine language programming is called computer architecture.
Examples of computer architecture:-
IBM 370 architecture, PDP-11 architecture, Intel 8080 and 8085 architecture, Intel x86 architecture, ARM(Advanced RISC machines) architecture, etc
Images of some computer architectures:-
Computer architecture covers the software aspects of a computer system. Characteristics of a computer system covered by computer architecture:-

Word length :- Word length is defined as the number of bits that a internal data register can hold and process at a time.A programmer should have idea about the word length before coding in order to write instructions of bit-length equals to the word length.
Instruction set :- A programmer knows what a computer can do from its instruction set. S/he may not have idea about the hardware details. That is why a instruction set is an architectural aspect.
Number of registers and their functions :- There are various registers in a processor. They have their own functions. Some registers are used to store data, while some stores addresses. Register called accumulator is used to store results by default. These aspects should be exposed to the programmer. That is why number of registers and their functions is an architectural aspect.
Addressing modes :- Instructions fetch and stores data from registers and memory addresses defined by various addressing modes. Addressing modes direct impacts the logical execution of the program that is why, addresing modes are an architectural aspect.
Bus structure,
I/O mechanism,
Memory addressing techniques,

Computer organization

The portrayal of hardwares and their interconnections in order to realize the computer system architecture is called computer organization.
For a given computer architecture there can be various computer organizations.
Computer organization covers the electronic aspects of a computer system. Characteristics of a computer system covered by computer organization:-

Details about internal structure of memories,
Interfacing circuitary for bus, memory and peripherals,
Circuitary details for timing and control signals,

Computer organization vs computer architecture

Computer architecture	Computer organization
Definition
The portrayal of a computer system from perspective of machine language programming is called computer architecture.	The portrayal of hardwares and their interconnections in order to realize the computer system architecture is called computer organization
Objectives
Computer architecture depicts the entire computer system as a programming model so that machine-language programmer can develop softwares.	The objective of computer organization is to realize computer architecture.
Relation A computer architecture can be realized by numerous forms of computer organizations.
Choosing an instruction is an architectural concern.	How the instruction will be executed is an organizational concern.
Number of registers, types of registers and word length is an architectural concern.	The electronic aspect that how a register will be selected for a particular instruction is an organizational concern.
Aspects included
Computer architecture covers the software aspects of a computer system. Characteristucs of a computer system covered by computer architecture:- Word length, Instruction set, Number of registers, Registers bit length and functions Addressing modes, Bus structure, I/O mechanism, Memory addressing techniques etc	Computer organization covers the electonic aspects of the compter system. Characteristucs of a computer system covered by computer organization are as follows:- Details about internal structure of memories, Interfacing circuitary for bus, memory and peripherals, Circuitary details for timing and control signals,
Hardware exposure
Hardware exposure at architectural level is superficial since only those hardware details are included which are important from view point of machine language programming.	Detailed circuitary of internal structure and interfacing provided.
Applications
For developing softwares.	For realising the computer architecture.
A computer architecture lasts longer than a computer organizational.	A computer organization is replaced by another organization belonging to same architecture with change in technology.

Computer design

Computer design is defined as the process of designing circuitary for the hardware components of the computer system.

RBB: Rastriya Banijya Bank Vaccencies and Study Materials

2021-12-11T22:22:00.016+05:45

Rastriya Banijya Bank

Vaccencies

Published date :- 2078-07-14(2021-Oct-31)
Start of application:- 2078-07-15(2021-Nov-01)
Deadline:- 2078-08-05(2021-Nov-21)

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Syllabus(Open Competition), Administration
Level	Post	Group	Syllabus
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9	Senior manager	Administration	Download
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NRB: Nepal Rastra Bank Vaccencies and Study Materials

2021-12-11T20:00:00.010+05:45

Vacancy Notice(Open Competition)

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Syllabus

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Online Application Portal(Perform Based Promotion & Internal Competition)

Post:- Assistant Director

Rank:- Officer Third
Service:- Administration
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Qualification:- Master in Economics Or Master in Management or Master in Public Administration.
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Post:- Assistant Director(CA)

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Qualification:- CA/ACCA/CPA
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Post:- Assistant Director(Commercial Law)

Rank:- Officer Third
Service:- Administration
Group:- Commercial Law
Qualification:- Masters in Law Or Master in Economics + Bachelor in Law Or Master in Management + Bachelor in Law Or Master in Public Administration + Bachelor in Law.
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Post:- Assistant Director(International Law)

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Service:- Administration
Group:- International Law
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Post:- Assistant Director(IT)

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Group:- IT(Information & Technology)
Qualification:- Bachelor in Computer Science or Engineering Or Bachelor in IT Or Bachelor in Electronics & Communication
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Post:- Assistant

Rank:- Assistant 2nd
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Qualification:- Diploma(3 Years) Or 10+2
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Post:- Assistant(Law)

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Post:- Assistant(IT)

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Service:- Technical
Group:- IT
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Machine language vs assembly language

2021-12-08T23:08:00.004+05:45

Differences between machine and assembly language
Machine Language	Assembly Language
Definition :
The programming language in which both the operations and operands are expressed in numeric form (usually hexadecimal or binary form) is called a machine language.	The machine dependent programming language in which operations are expressed in mnemonics (symbolic word from human language) and operands are expressed in alpha-numeric form is called an assembly language.
Generation of language
1G : Machine language is defined as the first generation of language.	2G : Machine language is defined as the second generation of language.
Execution speed
Its execution spped is the fastet among both lower-level and higher-level programming language.	Execution speed of assembly language is slower than machine language but faster than higher-level programming language.
Memory requirement
It requires comparatively less memory than assembly language.	It requires more memory than machine language.
Translator
As machine understand binary codes so translator is not required.	An additional software called complier is required to translate assembly into machine language.
Human understandability
Machine language is hard to understand and code.	As assemnbly language uses mnemonics which are like words from human language, so it is comparatively easy to understand and code.
Error fixing
Error fixing in machine language is a very tough task as entire code is only string of 0s and 1s.	Error fixing is comparatively easier.
Example of instructions
01001101 0101000011111111 etc	MOV A,B ADD A,FF etc

Measurement

2021-12-04T22:02:00.005+05:45

What is measurement?

Measurement is the process of comparing an unknown quantity with a known quantity.
Or,
Measurement is expressing characteristics or properties of an object in terms of number and unit. Like expressing length of a table as 1.5 meter, mass of a rice pack as 20 kg etc.

Importance of measurement

Standardization:- Measurement has made standardization of physical quantities in terms of predefined known quantities possible. This led to a revolution in field of science, technology, medicine, trade and commerce. In the absence of standard quantities the ways of expressing quantities were not uniform. Conficts used to arise while exchanging goods.
Quantification:- Measurement helps to quantify unknown quantities. The measured values can be used for further calculation while estimitating, planning or replacing with similar quantity.

Applications of measurement

Every field uses measurement. Some sectors where measurement is used are as follows:-

Physics
Engineering and technology
Medicine
Agriculture
Trade and commerce
Sports and Entertainment

The use of knowledge of measurement and measurement techniques is very frequent in our daily life. Some usages of measurement in our daily life are as follows:-

Selling or purchasing of goods,
Offices, school and TV shows schedules,
Expressing distance between two places,
Taking size of apparel,
Adjusting volume and brightness of electronic gadgets,
Expressing age,
To determine expiry date of groceries, medicine and documents,
and so on.

Quantity

Quantities are characteristics or properties we are trying to measure, such as the length of an object.
Quantities can be classified into following categories:-

Physical quantities
Non-physical quantities

Physical quantities

Those quantities which can be measured are called physical quantities. Examples of physical quantities are time, length, mass, temperature, weight, area, height, speed, angle of elevation, angle of depression, etc.

Non-physical quantities

Those quantities which can't be measured are called non-physical quantities. Examples of non-physical quantities are love, anger, desire, lust, pain, happiness, etc.

Physical quantities vs non-physical quantities

Physical quantity	Non-physical quantity
It can be measured.	It can't be measured.
It is expressed in terms of number and units.	It can not be expressed in terms of number and units. It is expressed in an abstract way using words like slight, severe, true etc.
Physical quantities of the same kind can be compared.	Non-physical quantities can’t be compared.
Examples:- Mass, Weight, Length, Luminous intensity, et cetera.	Examples:- Love, Anger, Lust, Anxiety, Pain, Happiness, Sadness, et cetera.

Classification of physical quantities

Fundamental quantity and
Derived quantity

In any measurement system those physical quantities which are independent of other physical quantities are called base or fundamental or independent quantities.
There are 7 fundamental quantities in SI system and they are distance, mass, duration, luminous intensity, temperature, amount of substance and electric current.

Those physical quantities which dependen upon fundamental quantities and are expressed as algebric combinations of one or more fundamental quantities are called derived or dependent quantities.
Examples of derived quantities are area, volume, force, speed, weight, acceleration, power, energy, work, pressure, resistance, etc

Fundamental vs derived quantities :-

Fundamental quantities	Derived quantities
Those physical quantities which are independent of other quantities are called fundamental or independent quantities.	Those physical quantities which dependen upon fundamental quantities for their expression in mathematical form are called derived or dependent quantities.
There are 7 fundamental quantities.	There are numerous derived quantities. Its hard to figure out exact number of derived quantities.
Fundamental quantities are elementary quantities so they can not be further broken down to other quantities.	Derived quantities are composite quantities so it can be broken down in terms of fundamental quantities.
All fundamental quantities have a unit associated with it.	Some derived quantities are unitless as they are ratio of same physical quantities. For example angle between two lines, efficiency, etc.
Examples Length, mass, time, amount of substance, temperature, luminous intensity and electric current.	Examples Speed, Frequency, plain angle, solid angle, force, pressure, energy, work, power, electric voltage, magnetic flux density, etc
Examples of SI units associated with fundamental quantities :- meter, kilogram, second, mole, kelvin, candela and current.	Examples of SI units associated with derived quantities :- meter per second(m/s), hertx(Hz), degree, newton(N), pascal(Pa), joule(J), watt(W), voltage(V), tesla(T), etc.

Unit

A standard measuring value associated with a particular physical quantity which is used to express or quantify physical quantities of same type is called a unit.
For example 'meter' is unit associated with length so it can be used to quantify other similar quantities like distance, breadth, height, etc but not other dissimilar quantities like area, volume, weight, duration, etc.
Physical quantities are expressed in term of some standard quantity of same kind termed as unit. Without unit, a physical quantity is just a number which gives no sense about the standard quantity in terms of which the physical quantity is expressed. A physical quantity only has sense when expressed in some terms of standard units.

Types of units

Fundamental unit and
Derived unit

Fundamental unit

Units associated with fundamental quantities are called fundamental units.
Or,
In any measurement system those units which are independent of other units are called fundamental units.
Examples of fundamental units are meter(m), kilometer(km), gram(g), kilogram(kg), millisecond(ms), degree centigrade(°C), kelvin(K), mole(mol), milliampere(mA), etc.

Derived unit

Units associated with derived quantities are called derived units.
Or,
Those units which are expressed as algebric cominations of fundamental units are called derived units.
Examples of derived units are meter-squared(m²), hertz(Hz), newton(N), pascal(Pa), joule(J), calorie(cal), watt(W), horsepower(Hp), etc.

Fundamental units vs Derived units

Fundamental units	Derived units
In any measurement system those units which are independent of other units are called fundamental units.	Those units which are expressed as algebric cominations of fundamental units are called derived units.
Fundamental units are associated with fundamental quantities.	Derived units are associated with derived quantities.
There are 7 fundamental units in SI unit.	There are numerous derived units. Its hard to figure out exact number of derived units.
Fundamental units are elementary units so they can not be further broken down to other units.	Derived units are composite units so it can be broken down in terms of fundamental units.
Examples :- meter, kilogram, second, mole, kelvin, candela and current.	Examples :- meter per second(m/s), hertx(Hz), degree, newton(N), pascal(Pa), joule(J), watt(W), voltage(V), tesla(T), etc.

Systems of measurement

Systems of measurement are standard which declares fundamental quantities to be considered for measurement.
Or,
Systems of measurement are defined as system which declares base physical quantities and units of measurement and all other physical quantities are expressed in some algebric expression of these base quantities.
SI, MKS, MKSA, CGS, FPS are some widely used measurement system.
Measurement systems are necessary as they draw a general consensus by declaring base quantities and units. In absence of such system conflicts would arise among different parties. Also measurement systems lead to formation of standard derived quantities.

Following types of systems of measurements are widely used:-

SI (système international)
MKS system
MKSA system
CGS system
FPS system

SI units

To bring uniformity in measurement system and make trade and commerce convenient across the borders some global metric system was needed. To answer this SI was intoduced in 1960. The system of measurement approved by the General Conference on Weights and Measures in 1960 AD is called SI system.
Or,
SI system is a metric system having 7 fundamental units viz meter(m) for length, kilogram(kg) for mass, second(s) for time, ampere(A) for electric current, Candela(cd) for luminous intensity, kelvin(K) for temperature and mole(mol) for amount of substance.
SI stans for système international (System of international).
The base units approved by the General Conference on Weights and Measures in 1960 AD and all other units derived from these base units are called SI units.

Base quantities	Base units
Length or distance	Meter(m)
Mass	Kilogram(kg)
Time	Second(s)
Electric current	Ampere(A)
Temperature	Kelvin(K)
Luminous intensity	Candela(cd)
Amount of substance	Mole(mol)

MKS

MKS stands for meter, kilogram and second.
MKS system is a metric system having 3 fundamental units viz meter(m) for distance, kilogram(kg) for mass and second(s) for time.

MKSA

MKSA stands for meter, kilogram, second and ampere.
MKSA system is a metric system having 4 fundamental units viz meter(m) for distance, kilogram(kg) for mass, second(s) for time and ampere for electric current.

CGS

CGS stands for centimeter, gram and second.
CGS system is a metric system having 3 fundamental units viz centimeter(cm) for distance, gram(g) for mass and second(s) for time.

FPS

FPS stands for foot, pound and second.
FPS system is a metric system having 3 fundamental units viz foot for distance, pound for mass and second for time. It is also called British or English system of measurement.

MKS vs SI

MKS system	SI system
MKS stands for meter kilogram and second.	SI stands for "Système International" in French which means International system.
MKS system has 3 fundamental units.	S.I. has 7 fundamental units.
MKS units are:- Meter, Kilogram and Second	SI units are:- Meter, Kilogram, Second, Ampere, Candela, Kelvin and Mole
It is the basis for the development of the S.I. system.	SI system is an extended version of MKS system. Some base units were added to SI system time to time. The last added unit is mole.
It has only units for mechanical quantities.	It has units for mechanical quantities along with electromagnetics, themodynamics and optical quantities. It has the Avogadro constant as mole unit.
It was adopted as international metric system in May 20, 1875.	It was adopted as international metric system in 1960.

Definition of SI units

1 second :-
One second is defined as the reciprocal of the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom, which is 9,192,631,770 Hz.
In short:-
One second is defined as the reciprocal caesium frequency.

1 meter :-
Before November 2018, one meter is defined as the length of the international prototype of platinum-iridium bar kept at International Bureau of Weights and Measures (BIPM) in Sèvres, France.
The metre, symbol m, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299 792 458 when expressed in the unit m/s, where the second is defined in terms of ΔνCs.

1 kg :-
Before November 2018 1 kg mass was defined as the mass of the platinum-iridium cylinder kept at International Bureau of Weights and Measures (BIPM) in Sèvres, France.
One kilogram is defined as the taking the fixed numerical value of the Planck constant h to be 6.626 070 15 × 10–34 when expressed in the unit J s, which is equal to kg m2 s–1, where the metre and the second are defined in terms of c and ΔνCs.

1 ampere :- One ampere is defined as time rate of flow charge.

1 Kelvin :- One kelvin(K) is defined as the product of one kg and one meter squared divided by the product of one second squared and Boltzmann constant(k) which is 1.380649 × 10⁻²³ Kg.m².s⁻².K⁻¹.
Mathematically, 1K = 1kg . 1m² . 1s⁻² . k

1 Candela :- The candela (cd) is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency 540 × 10¹² Hz, K_cd, to be 683 when expressed in the unit lm W^-1, which is equal to cd sr W^-1, or cd sr kg^-1 m^-2 s³, where the kilogram, meter and second are defined in terms of h, c and ∆ν_Cs.

One mole (mol) is defined as exactly 6.02214076 × 10²³ numbers of elementary entities.

Mass

The quantity of matter contained in a body is called mass. It is measured using beam balance or physical balance.
A beam balance which has two pans one at each end. At the middle there is a pivot on which a needle is hinged. The inclination of the needle toward any pan indicates the heavier side than the other. When it is not inclined to any side this means both the sides has equal masses. One of its side contains standard mass whose mass is known while on the other side the object which mass is to be known is kept. Using hit and trial with various standard masses when the inclination is zero the unkown mass is calculated.

Weight

Weight is defined as the force by which the earth attracts an object toward it center. It is measured using spring balance or top-pan balance.
Mathematically,
Weight(W) = mass(M) * acceleration due to gravity(g)
As weight of body depends upon acceleration due to gravity(g) it varies as value of "g" varies.

Mass vs weight

Mass	Weight
The amount of matter contained in a body is called its mass.	The force with which a body is attracted toward the centre of a celestial body is called weight of the body at that place.
Mass of a body remains constant.	Weight of a body varies with place.
Mass is a fundamental quantity.	Weight is a derived quantity.
Mass is a scalar quantity.	Weight is a vector quantity.
Mass of a body can’t be zero.	Weight of a body can be zero.
Mass is measured using beam balance.	Weight is measured using spring balance or top balance.
S.I. unit of mass is Kg.	S.I. Unit of weight is Newton (N).
CGS unit of mass is gram.	CGS unit of weight is dyne.

Distance

Distance is defined as measure of apartness between two points either along a straight path or combination of straight paths or curved path.
Rular and measuring tape are used to measure distance. Rular are convenient only for measuring short distance along straight path while measuring tapes can be used even to measure distance in kilometer range and along crved path too. Beside these in physics laboratory vernier calipers and micrometer screw gauge are used to measure distance in range smaller than millimeter.

Time

Time is defined as duration between two events.
Clock and watches are used to measure time. Atomic watches are used by scientist and physicist. They uses atomic watches because atomic watches use resonance frequency for to track time. These resonance frequency are in range of gigahertz. So such atomic watches can be used to measure hundreds of thousand fraction of a second and also there time fluctuation may not vary a single second in a 100 million years.
The time taken by the earth to complete one rotation about its axis is called one solar day.

Differentiate between fundamental quantities and derived quantities

2021-11-16T19:11:00.006+05:45

Base quantities, fundamental quantities and independent quantities terms are used interchangeably.
While
Derived quantities and dependent quantities terms are used interchangeably.

The differences between fundamental and derived quantities are as follows:-

Fundamental quantities	Derived quantities
Those physical quantities which are independent of other quantities are called fundamental or independent quantities.	Those physical quantities which dependen upon fundamental quantities for their expression in mathematical form are called derived or dependent quantities.
There are 7 fundamental quantities.	There are numerous derived quantities. Its hard to figure out exact number of derived quantities.
Fundamental quantities are elementary quantities so they can not be further broken down to other quantities.	Derived quantities are composite quantities so it can be broken down in terms of fundamental quantities.
All fundamental quantities have a unit associated with it.	Some derived quantities are unitless as they are ratio of same physical quantities. For example angle between two lines, efficiency, etc.
Examples Length, mass, time, amount of substance, temperature, luminous intensity and electric current.	Examples Speed, Frequency, plain angle, solid angle, force, pressure, energy, work, power, electric voltage, magnetic flux density, etc
Examples of SI units associated with fundamental quantities :- meter, kilogram, second, mole, kelvin, candela and current.	Examples of SI units associated with derived quantities :- meter per second(m/s), hertx(Hz), degree, newton(N), pascal(Pa), joule(J), watt(W), voltage(V), tesla(T), etc.

Physics : definition and applications

2021-11-16T12:04:00.006+05:45

Physics is defined as the scientific study which is concerned with describing the interactions of energy, matter, space, and time, and it is especially interested in what fundamental mechanisms underlie every phenomenon.

Applications of physics
Knowledge of physics is useful in everyday situations as well as in nonscientific professions. Physics is the foundation of many important disciplines and contributes directly to others. Some applications of physics are as follows:-

Engineering :- Most branches of engineering are applied physics. In architecture, physics is at the heart of structural stability, and is involved in the acoustics, heating, lighting, and cooling of buildings.
Chemistry :- In chemistry knowledge of physics is required to understand the interactions of atoms and molecules.
Biology :- Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cell walls and cell membranes (Figure 1.6 and Figure 1.7). On the macroscopic level, it can explain the heat, work, and power associated with the human body. Physics is involved in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements. Medical therapy sometimes directly involves physics; for example, cancer radiotherapy uses ionizing radiation. Physics can also explain sensory phenomena, such as how musical instruments make sound, how the eye detects color, and how lasers can transmit information.
Geology :- Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and heat transfer in the Earth.

Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.
https://openstax.org/details/books/college-physics-ap-courses

If xy = yx, where x≠y, then find the value of x and y.

2021-11-13T21:56:00.000+05:45

Solution
Given
x^y = y^x      .....(i)
x≠y      .....(ii)
Let x = ky      .....(iii)
where k is a constant
Equation .....(i), .....(ii) and .....(iii) implies that k≠1.
Inserting x = ky in equation .....(i)
(ky)^y = y^ky
Or, k^y . y^y = y^ky
Or, k^y = y^ky . y^-y
Or, k^y = y^{ky - y}
Or, k^y = y^y(k-1)
Or, k^y = (y^(k-1))^y
Here power of both the sides are y so.
Or, k = y^(k-1)

Raising both sides by power of
1
k-1 we get

Or, k^1
k-1 = y

y = k^1
k-1 .....(iv)

So, x = ky is given by

x = k (k^1
k-1)

x = k^{1
k-1 +1}

x = k^{1 + k - 1
k-1}

x = k^k
k-1 .....(v)

Equation .....(v) and .....(iv) gives the values of x and y respectively for k ≠ 1

∴x = k^k
k-1
∴y = k^1
k-1

Computer : An Introduction

2021-09-09T08:29:00.013+05:45

Computer

COMPUTER stands for Common Operating Machine Purposely Used for Technological and Educational Research.

A computer is a device either electronic or mechanical or electro-mechanical which takes input, has internal computaion mechanism to process the input and provides output.

These days mecanical computers are obsolete while electronic computers are ubiquitous, so computer usually means an electronic computer.
A computer is an electronic device either analog, digital or both which takes input, stores and processes it in its processing unit having a hard-wired or microprogrammed control unit and provides output.

For example:-
Analog clock, Superheterodyne radio receiver, Video cassette Recorder(VCR), Magnetic tape recorder, Analog television receiver, TV Remote, Calculator, Videogames, Wi-fi Routers, Mobilephones, Smartphones, Personal Computers, ATM(Automatic Teller Machine), Web Hosting Server, Supercomputers like Fugaku, Summit, Sierra, PARAM Series.

Charles Babbage's difference engine.

Cray supercomputer

Analog clock

Calculator

Smartphone

Laptop

Desktop

Mainframe(Web hosting server)

Summit supercomputer

Amisha Shahi

2021-08-26T12:31:00.012+05:45

Message us your homework @ E-cikai FB page

2078-Bhadra-13-Sunday

2021-August-29

Mathematics

Multiply a) 0.8 and b) 0.16 by 10 and 100.

a) 0.8
0.8 * 10 = 8
0.8 * 100 = 80

b) 0.16
0.16 * 10 = 1.6
0.16 * 100 = 16

Simplify
a)6√5 + 2√5 - 3√5
b)2√8 - √18 + 3√50

a)6√5 + 2√5 - 3√5
= 8√5 - 3√5
= 5√5

b)2√8 - √18 + 3√50
= 2√(2*2*2) - √(3*3*2) + 3√(5*5*2)
= 2√(2² * 2) - √(3² * 2) + 3√(5² * 2)
= 2*2√2 - 3√2 + 3*5√2
= 4√2 - 3√2 + 15√2
= 19√2 - 3√2
= 16√2

Simplify : 100 ÷ 2[500 ÷ { 6 + (25 + 7 - 24 )}]

= 100 ÷ 2[500 ÷ { 6 + (21 + 7 - 24 )}]
= 100 ÷ 2[500 ÷ { 6 + (21 + (-17) )}]
= 100 ÷ 2[500 ÷ { 6 + (21 - 17)}]
= 100 ÷ 2[500 ÷ { 6 + 4 }]
= 100 ÷ 2[500 ÷ { 10 }]
= 100 ÷ 2[500 ÷ 10 ]
= 100 ÷ 2[50]
= 100 ÷ 100
= 1

11.

a) Convert 491 into binary number.

Solution 
24911
22451
21220
2611
2300
2151
271
231
211
 0 

Arranging all these remainder from buttom to top in left to right order we get the binary equivalent of given decimal number.
111101011₂
    ∴ 491₁₀ = 111101011₂ .

b) Convert 142 into quinary number.

Solution 
51411
5283
550
511
 0 

Arranging all these remainder from buttom to top in left to right order we get the binary equivalent of given decimal number.
1031₅
    ∴ 141₁₀ = 1031₅ .

12) Convert 111101₂ into quinary number.

Solution 
Given binary number is 111101₂
Converting binary number into decimal number : 
111101₂
 = 1 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 1 × 2⁰
= 32 + 16 + 8 + 4 + 1
 = 61
   ∴111101₂ = 61₁₀  
5611
5122
522
 0 

Arranging all these remainder from buttom to top in left to right order we get the binary equivalent of given decimal number.
221₅
    ∴ 61₁₀ = 221₅ .

2078-Bhadra-10-Thrusday

2021-August-26

First Term Exam 2078

Health & Physical Education

Group A 'Health Education'

1. What is circulatory system? [2]

2. What are the organs of circulatory system? [2]

3. What is the main function of heart? [2]

4. What are the components of blood? [2]

5. What is the importance of blood circulation? [2]

6. Describe blood circulation process of human body briefly. [2]

7. Why pancrease is called mixed gland? [2]

8. What are the parts of central nervous system? [2]

9. What is the function of cerebellum? [2]

10. Where is midbrain located? [2]

11. Why pituitary gland is called a master gland? [2]

12. Differentiate between Exocrine gland and Endocrine gland. [3]

13. What is a spinal cord? [2]

What is circulatory system?

[2]

Circulatory system is defined as the system of organ that supplies essential nutrients, oxygen, hormone to all the body parts by pumping blood.

What are the organs of circulatory system?

[2]

Heart, blood and blood vessels are the organs of circulatory system.

What is the main function of heart?

[2]

The main function of heart is to pump pure blood to the body parts.

What are the components of blood?

[2]

Plasma, Red blood Cells, white blood cells and platelets are the components of blood.

What is the importance of blood circulation?

[2]

Describe blood circulation process of human body briefly.

[2]

Why pancrease is called mixed gland?

[2]

Pancreas is called mixed gland as one of its secretion is transported by a separate duct while other secretion is transported by blood.

What are the parts of central nervous system?

[2]

Brain and spinal cord are the parts of central nervous system.

What is the function of cerebellum?

[2]

The main functions of the cerebellum are as follows :

It senses equilibrium.

Coordinates eye movement.

Coordinates and maintains body balance and posture during walking, running, riding, swimming.

The cerebellum is also essential for making fine adjustments to motor actions.

10.

Where is midbrain located?

[2]

The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain.

11.

Why pituitary gland is called a master gland?

[2]

Pituitary gland is called as a master gland because it controls the activities of other endocrine glands and many other body functions by producing appropriate hormones including growth hormones.

12.

Differentiate between Exocrine gland and Endocrine gland.

[3]

Exocrine glands	Endocrine glands
Exocrine glands are ductful glands. They have ducts.	Endocrine glands are ductless glands as they do not have ducts.
Exocrine glands secrets zymes.	Endocrine glands secrets hormones.
Examples:- Salivary glands, liver, Brunner’s glands, oesophagal glands	Examples:- Pituitary gland, Thyroid glands, Parathyroid glands, Adrenal glands.

13.

What is a spinal cord?

[2]

The spinal cord is a part of the central nervous system which looks like a long pipe structure arising from the medulla oblongata, part of the brain consisting of a collection of nerve fibres, running through the vertebral column of the backbone.

2078-Bhadra-10-Thrusday

2021-August-26

First Term Exam 2078

Science

Group A 'Physics'

1.a What is measurement? Differentiate between mass and weight. (1+2)

The process of measuring an unknown quantity with a known quantity is called measurement.

The differences between mass and weight are as follows :-

Mass	Weight
The amount of matter contained in a body is called its mass.	The force with which a body is attracted toward the centre of a celestial body is called weight of the body at that place.
Mass of a body remains constant.	Weight of a body varies with place.
Mass is a fundamental quantity.	Weight is a derived quantity.
Mass is a scalar quantity.	Weight is a vector quantity.
Mass of a body can’t be zero.	Weight of a body can be zero.
Mass is measured using beam balance.	Weight is measured using spring balance or top balance.
S.I. unit of mass is Kg.	S.I. Unit of weight is Newton (N).
CGS unit of mass is gram.	CGS unit of weight is dyne.

1.b The mass of a girl is 40 kg. Calculate her weight.(2)

Solution
Given
Mass, m = 40 kg
Acceleration due to gravity, g = 9.8m/s²
So, weight of the girl, w = m.g = 40 * 9.8 = 392 Newton.

2.a What does it mean by the fact that the acceleration of a car is 5m/s² ? (2)

This means that the velocity of the car is increasing 5m/s every second. If at t = 0 s velocity was 8m/s then at t = 1s velocity becomes (8+5)m/s = 13 m/s. Similarly at t = 2s velocity becomes (13+5)m/s = 18 m/s and so on.

2.b What is acceleration ? Differentiate between speed and velocity. (1+2)

The time rate of change of velocity is called acceleration.

Differences between speed and velocity are as follows :-

Speed	Velocity
Speed is defined as the distance covered per unit of time.	Velocity is defined as the displacement covered per unit of time.
Speed has magnitude but no direction.	Velocity has both magnitude and direction.
Speed is a scalar quantity.	Velocity is a vector quantity.
Value of speed can't be negative.	Value of velocity can be positive, zero and negative.

3.a If a train starts from rest and attains velocity of 100m/s in 25 seconds. Calculate the acceleration produced by the train. (2)

Solution
Given
Initial velocity, u = 0 m/s
Final velocity, v = 100 m/s
Time taken, t = 25 s
Acceleration, a =
(v - u)/t
= (100 m/s - 0 m/s) / 25s
= 100/25 m/s²
= 4 m/s²
The acceleration produced by the train is 4 m/s².

3.b A body starts moving from rest and attains the acceleration of 0.5 m/s². Calculate the velocity at the end of 3 minutes. Also find the distance travelled by it during that time. (3)

Solution
Given
Initial velocity, u = 0 m/s
Acceleration, a = 0.5 m/s²
Time taken, t = 3 minutes = 3 * 60 seconds = 180 seconds
Final velocity, v = u + at = 0 + 0.5 * 180 = 90 m/s.
Distance travelled, d = v*t = 90*180 = 16200 m.
Hence the velocity of the body at the end of 3 minutes is 90 m/s and distance travelled by it is 16200 m.

Quantitive Analysis Assignment Solution For 2021 July

2021-07-30T10:38:00.003+05:45

Sri Dev Suman Uttrakhand Vishwavidhalaya

2. Calculate the Mean from the folowing data (07)

Marks	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
No. of Students	5	17	31	41	49

Solution

Marks	0-10	10-20	20-30	30-40	40-50
No. of Students	5	17	31	41	49


  Observations(x)    Frequency(f)    Class mark        x × f     
  0 - 10   5    5    5×5 = 25  
  10 - 20   17    15    15×17 = 255  
  20 - 30   31    25    25×31 = 775  
  30 - 40   41    35    35×41 = 1435  
  40 - 50   49    45    45×49 = 2205  
    Σf = 143      Σfx = 4695  



Arithmetic mean (x) =  Σfx 
Arithmetic mean (x) =      
Arithmetic mean (x) =  Σf  


Arithmetic mean (x) =   4695  
Arithmetic mean (x) =         
Arithmetic mean (x) =   143   


 ∴ Arithmetic mean (x) = 32.832

Three different chemiclas are mixed in a specific proportion to produce 100 units of a product. The company has to meet a constraint in the production of the product that chemical X should be one-third of the difference between chemicals Y and Z used. The cost of chemicals X, Y and Z are respecvtively Rs 50 Rs 20 and Rs 40 and the total cost of all the chemicals used is Rs 2900. Determine each unit of chemical X, Y and Z used by matrix method.

2021-07-29T21:18:00.001+05:45

Solution
Let,
x units of X
y units of Y and
z units of Z are used to produce 100 units of the product.
So, x + y + z = 100 ......(i)
Second condition is :
Chemical X should be one-third of the difference between chemicals Y and Z used
x =
1
3(y - z)

Or, 3x = y-z
So, 3x - y + z = 0 ......(ii)
Third condition is :
Total cost = 2900 which is sum of cost of X, Y and Z
So, 50x + 20y + 40z = 2900 ......(iii)

Writing the equation in the matrix form

AX = B

A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
1

3

50

1

-1

20

1

1

40

Om
Om
Om
Om
Om
Om
Om
Om
Om

X =
Om
Om
Om
Om
Om
Om
Om
Om
Om
x

y

z

Om
Om
Om
Om
Om
Om
Om
Om
Om

B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
100

0

2900

Om
Om
Om
Om
Om
Om
Om
Om
Om

Om
Om
Om
Om
Om
Om
Om
Om
Om
1

3

50

1

-1

20

1

1

40

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
x

y

z

Om
Om
Om
Om
Om
Om
Om
Om
Om =
Om
Om
Om
Om
Om
Om
Om
Om
Om
100

0

2900

Om
Om
Om
Om
Om
Om
Om
Om
Om

X = A⁻¹ B.
So, now we find A⁻¹.

M₁₁ =
Om
Om
Om
Om
Om
Om
-1

20

1

40

Om
Om
Om
Om
Om
Om

  =  -1 × 40 - 1 × 20   =  -60
Hi!
  A₁₁ = (-1)¹⁺¹  M₁₁ = -60

M₁₂ =
Om
Om
Om
Om
Om
Om
3

50

1

40

Om
Om
Om
Om
Om
Om

  =  3 × 40 - 1 × 50   =  70
Hi!
  A₁₂ = (-1)¹⁺²  M₁₂ = -70

M₁₃ =
Om
Om
Om
Om
Om
Om
3

50

-1

20

Om
Om
Om
Om
Om
Om

  =  3 × 20 - -1 × 50   =  110
Hi!
  A₁₃ = (-1)¹⁺³  M₁₃ = 110

M₂₁ =
Om
Om
Om
Om
Om
Om
1

20

1

40

Om
Om
Om
Om
Om
Om

  =  1 × 40 - 1 × 20   =  20
Hi!
  A₂₁ = (-1)²⁺¹  M₂₁ = -20

M₂₂ =
Om
Om
Om
Om
Om
Om
1

50

1

40

Om
Om
Om
Om
Om
Om

  =  1 × 40 - 1 × 50   =  -10
Hi!
  A₂₂ = (-1)²⁺²  M₂₂ = -10

M₂₃ =
Om
Om
Om
Om
Om
Om
1

50

1

20

Om
Om
Om
Om
Om
Om

  =  1 × 20 - 1 × 50   =  -30
Hi!
  A₂₃ = (-1)²⁺³  M₂₃ = 30

M₃₁ =
Om
Om
Om
Om
Om
Om
1

-1

1

1

Om
Om
Om
Om
Om
Om

  =  1 × 1 - 1 × -1   =  2
Hi!
  A₃₁ = (-1)³⁺¹  M₃₁ = 2

M₃₂ =
Om
Om
Om
Om
Om
Om
1

3

1

1

Om
Om
Om
Om
Om
Om

  =  1 × 1 - 1 × 3   =  -2
Hi!
  A₃₂ = (-1)³⁺²  M₃₂ = 2

M₃₃ =
Om
Om
Om
Om
Om
Om
1

3

1

-1

Om
Om
Om
Om
Om
Om

  =  1 × -1 - 1 × 3   =  -4
Hi!
  A₃₃ = (-1)³⁺³  M₃₃ = -4

The matrix of cofactors of A is C_A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-60

-20

2

-70

-10

2

110

30

-4

Om
Om
Om
Om
Om
Om
Om
Om
Om

As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."

∴ Adjoint of matrix A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-60

-70

110

-20

-10

30

2

2

-4

Om
Om
Om
Om
Om
Om
Om
Om
Om

Now we find the determinant of the matrix A

|A| =
Om
Om
Om
Om
Om
Om
Om
Om
Om
1

3

50

1

-1

20

1

1

40

Om
Om
Om
Om
Om
Om
Om
Om
Om

= 1 × (-1) × 40 + 1 × 1 × 50 + 1 × 3 × 20 - 1 × 1 × 20 - 1 × 3 × 40 - 1 × (-1) × 50

= -40 + 50 + 60-20-120+50

= -20

As determinant of the matrix ≠ 0. Hence its inverse exists.

Inverse of a matrix is scalar multiplication of its adjugate matrix with the reciprocal of its determinant.

Inverse of matrix A = A^-¹

=
1
|A|

Om
Om
Om
Om
Om
Om
Om
Om
Om
-60

-70

110

-20

-10

30

2

2

-4

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
1
-20

Om
Om
Om
Om
Om
Om
Om
Om
Om
-60

-70

110

-20

-10

30

2

2

-4

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴ Inverse matrix of A = A^-¹ =
Om
Om
Om
Om
Om
Om
Om
Om
Om
3

3.5

-5.5

1

0.5

-1.5

-0.1

-0.1

0.2

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
3

3.5

-5.5

1

0.5

-1.5

-0.1

-0.1

0.2

Om
Om
Om
Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
100

0

2900

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
3×100+1×0+(-0.1)×2900

3.5×100+0.5×0+(-0.1)×2900

(-5.5)×100+(-1.5)×0+0.2×2900

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
300+0-290

350+0-290

-550+0+580

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴ A^-¹ B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
10

60

30

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴x = 10

∴y = 60

∴z = 30

Diary Corporation is planning to produce 1200 kgs of baby food mixing 3 nutrients A, B and C. The 3 nutrient components per kg cost Rs 12, Rs 16 and Rs 14 respectively. The final output of 1 kg of baby food pack must contain nutrient C to be twice that of nutrient A . The budget available to buy component nutrients are to the corporation is Rs 18,480. Find amount of each nutrient which should be mixed in the final output by using matrix method.

2021-07-29T21:14:00.002+05:45

Solution
Let,
a kgs of A
b kgs of B and
c kgs of C are used to produce 1200 kgs of the baby food.
So, a + b + c = 1200 ......(i)
Second condition is :
Total cost of the baby food is Rs. 18,480
So, 12a + 16b + 14c = 18480 ......(ii)
Third condition is :
Nutrient C should be used twice nutrient A.
So, c = 2a
2a - c = 0 ......(iii)

Writing the equation in the matrix form

AX = B

A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
1

12

2

1

16

0

1

14

-1

Om
Om
Om
Om
Om
Om
Om
Om
Om

X =
Om
Om
Om
Om
Om
Om
Om
Om
Om
a

b

c

Om
Om
Om
Om
Om
Om
Om
Om
Om

B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
1200

18480

0

Om
Om
Om
Om
Om
Om
Om
Om
Om

Om
Om
Om
Om
Om
Om
Om
Om
Om
1

12

2

1

16

0

1

14

-1

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
a

b

c

Om
Om
Om
Om
Om
Om
Om
Om
Om =
Om
Om
Om
Om
Om
Om
Om
Om
Om
1200

18480

0

Om
Om
Om
Om
Om
Om
Om
Om
Om

X = A⁻¹ B.
So, now we find A⁻¹.

M₁₁ =
Om
Om
Om
Om
Om
Om
16

0

14

-1

Om
Om
Om
Om
Om
Om

  =  16 × -1 - 14 × 0   =  -16
Hi!
  A₁₁ = (-1)¹⁺¹  M₁₁ = -16

M₁₂ =
Om
Om
Om
Om
Om
Om
12

2

14

-1

Om
Om
Om
Om
Om
Om

  =  12 × -1 - 14 × 2   =  -40
Hi!
  A₁₂ = (-1)¹⁺²  M₁₂ = 40

M₁₃ =
Om
Om
Om
Om
Om
Om
12

2

16

0

Om
Om
Om
Om
Om
Om

  =  12 × 0 - 16 × 2   =  -32
Hi!
  A₁₃ = (-1)¹⁺³  M₁₃ = -32

M₂₁ =
Om
Om
Om
Om
Om
Om
1

0

1

-1

Om
Om
Om
Om
Om
Om

  =  1 × -1 - 1 × 0   =  -1
Hi!
  A₂₁ = (-1)²⁺¹  M₂₁ = 1

M₂₂ =
Om
Om
Om
Om
Om
Om
1

2

1

-1

Om
Om
Om
Om
Om
Om

  =  1 × -1 - 1 × 2   =  -3
Hi!
  A₂₂ = (-1)²⁺²  M₂₂ = -3

M₂₃ =
Om
Om
Om
Om
Om
Om
1

2

1

0

Om
Om
Om
Om
Om
Om

  =  1 × 0 - 1 × 2   =  -2
Hi!
  A₂₃ = (-1)²⁺³  M₂₃ = 2

M₃₁ =
Om
Om
Om
Om
Om
Om
1

16

1

14

Om
Om
Om
Om
Om
Om

  =  1 × 14 - 1 × 16   =  -2
Hi!
  A₃₁ = (-1)³⁺¹  M₃₁ = -2

M₃₂ =
Om
Om
Om
Om
Om
Om
1

12

1

14

Om
Om
Om
Om
Om
Om

  =  1 × 14 - 1 × 12   =  2
Hi!
  A₃₂ = (-1)³⁺²  M₃₂ = -2

M₃₃ =
Om
Om
Om
Om
Om
Om
1

12

1

16

Om
Om
Om
Om
Om
Om

  =  1 × 16 - 1 × 12   =  4
Hi!
  A₃₃ = (-1)³⁺³  M₃₃ = 4

The matrix of cofactors of A is C_A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-16

1

-2

40

-3

-2

-32

2

4

Om
Om
Om
Om
Om
Om
Om
Om
Om

As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."

∴ Adjoint of matrix A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-16

40

-32

1

-3

2

-2

-2

4

Om
Om
Om
Om
Om
Om
Om
Om
Om

Now we find the determinant of the matrix A

|A| =
Om
Om
Om
Om
Om
Om
Om
Om
Om
1

12

2

1

16

0

1

14

-1

Om
Om
Om
Om
Om
Om
Om
Om
Om

= 1 × 16 × (-1) + 1 × 14 × 2 + 1 × 12 × 0 - 1 × 14 × 0 - 1 × 12 × (-1) - 1 × 16 × 2

= -16 + 28 + 0-0+12-32

= -8

As determinant of the matrix ≠ 0. Hence its inverse exists.

Inverse of a matrix is scalar multiplication of its adjugate matrix with the reciprocal of its determinant.

Inverse of matrix A = A^-¹

=
1
|A|

Om
Om
Om
Om
Om
Om
Om
Om
Om
-16

40

-32

1

-3

2

-2

-2

4

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
1
-8

Om
Om
Om
Om
Om
Om
Om
Om
Om
-16

40

-32

1

-3

2

-2

-2

4

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴ Inverse matrix of A = A^-¹ =
Om
Om
Om
Om
Om
Om
Om
Om
Om
2

-5

4

-0.125

0.375

-0.25

0.25

0.25

-0.5

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
2

-5

4

-0.125

0.375

-0.25

0.25

0.25

-0.5

Om
Om
Om
Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
1200

18480

0

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
2×1200+(-0.125)×18480+0.25×0

(-5)×1200+0.375×18480+0.25×0

4×1200+(-0.25)×18480+(-0.5)×0

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
2400-2310+0

-6000+6930+0

4800-4620+0

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴ A^-¹ B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
90

930

180

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴a = 90

∴b = 930

∴c = 180

The three commodities C1, C2 , C3 are purchased and sold by the three persons A, B and C. Mr. A purchases 4 units of C3 and sells 3 units of C1 and 5 units of C2 . Mr B purchases 3 units of C1 and sells 2 units of C2 and 1 units of C3. Mr C purchases 1 units of C2 and sells 4 units of C1 and 6 units of C3. In the purchase – sell process Mr. A suffered a loss of Rs 1,000; Mr B earns no profit and Mr C earns Rs 40,000. Find the prices per unit of these commodities by using matrix method.

2021-07-29T21:09:00.007+05:45

Solution
Let,
Price of commodity C1 per unit be x
Price of commodity C2 per unit be y
Price of commodity C3 per unit be z
Amount asociated with sells and profit is considered positive while amount considered with purchase and loss is considered negative.
Equation for Mr.A
3x + 5y - 4z = -1000 ......(i)

Equation for Mr.B
-3x + 2y + z = 0 ......(ii)
Equation for Mr.C

4x - y + 6z = 40000 ......(iii)

Writing the equation in the matrix form

AX = B

A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
3

-3

4

5

2

-1

-4

1

6

Om
Om
Om
Om
Om
Om
Om
Om
Om

X =
Om
Om
Om
Om
Om
Om
Om
Om
Om
x

y

z

Om
Om
Om
Om
Om
Om
Om
Om
Om

B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-1000

0

40000

Om
Om
Om
Om
Om
Om
Om
Om
Om

Om
Om
Om
Om
Om
Om
Om
Om
Om
3

-3

4

5

2

-1

-4

1

6

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
x

y

z

Om
Om
Om
Om
Om
Om
Om
Om
Om =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-1000

0

40000

Om
Om
Om
Om
Om
Om
Om
Om
Om

X = A⁻¹ B.
So, now we find A⁻¹.

M₁₁ =
Om
Om
Om
Om
Om
Om
2

-1

1

6

Om
Om
Om
Om
Om
Om

  =  2 × 6 - 1 × -1   =  13
Hi!
  A₁₁ = (-1)¹⁺¹  M₁₁ = 13

M₁₂ =
Om
Om
Om
Om
Om
Om
-3

4

1

6

Om
Om
Om
Om
Om
Om

  =  -3 × 6 - 1 × 4   =  -22
Hi!
  A₁₂ = (-1)¹⁺²  M₁₂ = 22

M₁₃ =
Om
Om
Om
Om
Om
Om
-3

4

2

-1

Om
Om
Om
Om
Om
Om

  =  -3 × -1 - 2 × 4   =  -5
Hi!
  A₁₃ = (-1)¹⁺³  M₁₃ = -5

M₂₁ =
Om
Om
Om
Om
Om
Om
5

-1

-4

6

Om
Om
Om
Om
Om
Om

  =  5 × 6 - -4 × -1   =  26
Hi!
  A₂₁ = (-1)²⁺¹  M₂₁ = -26

M₂₂ =
Om
Om
Om
Om
Om
Om
3

4

-4

6

Om
Om
Om
Om
Om
Om

  =  3 × 6 - -4 × 4   =  34
Hi!
  A₂₂ = (-1)²⁺²  M₂₂ = 34

M₂₃ =
Om
Om
Om
Om
Om
Om
3

4

5

-1

Om
Om
Om
Om
Om
Om

  =  3 × -1 - 5 × 4   =  -23
Hi!
  A₂₃ = (-1)²⁺³  M₂₃ = 23

M₃₁ =
Om
Om
Om
Om
Om
Om
5

2

-4

1

Om
Om
Om
Om
Om
Om

  =  5 × 1 - -4 × 2   =  13
Hi!
  A₃₁ = (-1)³⁺¹  M₃₁ = 13

M₃₂ =
Om
Om
Om
Om
Om
Om
3

-3

-4

1

Om
Om
Om
Om
Om
Om

  =  3 × 1 - -4 × -3   =  -9
Hi!
  A₃₂ = (-1)³⁺²  M₃₂ = 9

M₃₃ =
Om
Om
Om
Om
Om
Om
3

-3

5

2

Om
Om
Om
Om
Om
Om

  =  3 × 2 - 5 × -3   =  21
Hi!
  A₃₃ = (-1)³⁺³  M₃₃ = 21

The matrix of cofactors of A is C_A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
13

-26

13

22

34

9

-5

23

21

Om
Om
Om
Om
Om
Om
Om
Om
Om

As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."

∴ Adjoint of matrix A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
13

22

-5

-26

34

23

13

9

21

Om
Om
Om
Om
Om
Om
Om
Om
Om

Now we find the determinant of the matrix A

|A| =
Om
Om
Om
Om
Om
Om
Om
Om
Om
3

-3

4

5

2

-1

-4

1

6

Om
Om
Om
Om
Om
Om
Om
Om
Om

= 3 × 2 × 6 + 5 × 1 × 4 + (-4) × (-3) × (-1) - 3 × 1 × (-1) - 5 × (-3) × 6 - (-4) × 2 × 4

= 36 + 20-12+3+90+32

= 169

As determinant of the matrix ≠ 0. Hence its inverse exists.

Inverse of a matrix is scalar multiplication of its adjugate matrix with the reciprocal of its determinant.

Inverse of matrix A = A^-¹

=
1
|A|

Om
Om
Om
Om
Om
Om
Om
Om
Om
13

22

-5

-26

34

23

13

9

21

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
1
169

Om
Om
Om
Om
Om
Om
Om
Om
Om
13

22

-5

-26

34

23

13

9

21

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴ Inverse matrix of A = A^-¹ =
Om
Om
Om
Om
Om
Om
Om
Om
Om
0.077

0.13

-0.03

-0.154

0.201

0.136

0.077

0.053

0.124

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
0.077

0.13

-0.03

-0.154

0.201

0.136

0.077

0.053

0.124

Om
Om
Om
Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
-1000

0

40000

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
0.077×(-1000)+(-0.154)×0+0.077×40000

0.13×(-1000)+0.201×0+0.053×40000

(-0.03)×(-1000)+0.136×0+0.124×40000

Om
Om
Om
Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
-77+0+3080

-130+0+2120

30+0+4960

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴ A^-¹ B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
3003

1990

4990

Om
Om
Om
Om
Om
Om
Om
Om
Om

∴x = 3003

∴y = 1990

∴z = 4990

Hence the price of C1 per unit = 3003
The price of C2 per unit = 1990
The price of C3 per unit = 4990

Application of matrix algebra (Numerical)

2021-07-29T21:04:00.002+05:45

A biscuit manufacturing company has two plants located at two different places. Each plant produces three different types of biscuits Tiger, Glucose and Salty. Matrix A shows the number of packets of each type of biscuit produced per day and matrix B shows their cost and selling price.

A =
Om
Om
Om
Om
Om
Om
Om
Om
Om


Plant I

Plant II

Tiger

300

200

Glucose

200

300

Salty

100

200

Om
Om
Om
Om
Om
Om
Om
Om
Om

B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
C.P.

10

11

12

S.P.

13

12

14

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om

If plant I and Plant II operate for 20 and 25 days respectively in a month. Using matrix algebra, find the total :
i) Production of biscuits from each plant per day.
ii) Production of each type of biscuits in the month,
iii) Profit in the month.

Solution

A =
Om
Om
Om
Om
Om
Om
300

200

200

300

100

200

Om
Om
Om
Om
Om
Om

Column matrix of cost price, C =
Om
Om
Om
Om
Om
Om
Om
Om
Om
10

11

12

Om
Om
Om
Om
Om
Om
Om
Om
Om

Column matrix of selling price, S =
Om
Om
Om
Om
Om
Om
Om
Om
Om
13

12

14

Om
Om
Om
Om
Om
Om
Om
Om
Om

Row matrix of operating days, D = [ 20 25 ]

i)Production of biscuits from each plant per day :

=
Om
Om
Om
Om
Om
Om
300

200

200

300

100

200

Om
Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
1

1

1

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
Om
Om
Om
Om
Om
Om
300×1+200×1+100×1

200×1+300×1+200×1

Om
Om
Om
Om
Om
Om

=
Om
Om
Om
Om
Om
Om
300+200+100

200+300+200

Om
Om
Om
Om
Om
Om

=
Om
Om
Om
Om
Om
Om
600

700

Om
Om
Om
Om
Om
Om

Production of biscuits from each plant per day =
Om
Om
Om
Om
Om
Om
Plant I

Plant II

600

700

Om
Om
Om
Om
Om
Om

ii)Production of each type of biscuits in the month :

= [ 20     25 ]

Om ×
Om
Om
Om
Om
Om
Om
Om
300

200

200

300

100

200

Om
Om
Om
Om
Om
Om

= [ 20×300+25×200     20×200+25×300     20×100+25×200 ]

= [ 6000+5000     4000+7500     2000+5000 ]

= [ 11000     11500     7000 ]

Production of each type of biscuits in the month, P =
Om
Om
Om
Om
Om
Om
Tiger

11000

Glucose

11500

Salty

7000

Om
Om
Om
Om
Om
Om

ii)Profit in the month :

Calculating total selling price in a month :

= [ 11000     11500     7000 ]

Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
13

12

14

Om
Om
Om
Om
Om
Om
Om
Om
Om

= [ 11000×13+11500×12+7000×14 ]

= [ 143000+138000+98000 ]

= [ 379000 ]

Total selling price in a month = Rs. 379,000

Calculating total cost price in a month :

= [ 11000     11500     7000 ]

Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
10

11

12

Om
Om
Om
Om
Om
Om
Om
Om
Om

= [ 11000×10+11500×11+7000×12 ]

= [ 110000+126500+84000 ]

= [ 320500 ]

Total cost price in a month = Rs. 320,500

Profit in the month = Total selling price in a month - Total cost price in a month = Rs. 379,000 - Rs. 320,500 = Rs. 58500

Alternatively,

i)
Production of biscuits from Plant I = Sum of elements of Row 1 = 300 + 200 + 100 = 600
Production of biscuits from Plant II = Sum of elements of Row 2 = 200 + 300 + 200 = 700

ii)
Plant I:
Operating days in a month = 20 days
Production of Tiger biscuits from Plant I in a month = (Production of Tiger biscuits from Plant I in a day × Operating days in a month) = 300 × 20 = 6000
Production of Glucose biscuits from Plant I in a month = (Production of Glucose biscuits from Plant I in a day × Operating days in a month) = 200 × 20 = 4000
Production of Salty biscuits from Plant I in a month = (Production of Salty biscuits from Plant I in a day × Operating days in a month) = 100 × 20 = 2000

Plant II:
Operating days in a month = 25 days
Production of Tiger biscuits from Plant II in a month = (Production of Tiger biscuits from Plant II in a day × Operating days in a month) = 200 × 25 = 5000
Production of Glucose biscuits from Plant II in a month = (Production of Glucose biscuits from Plant II in a day × Operating days in a month) = 300 × 25 = 7500
Production of Salty biscuits from Plant II in a month = (Production of Salty biscuits from Plant II in a day × Operating days in a month) = 200 × 25 = 5000
Production of Tiger biscuits in a month = 6000 + 5000 = 11000
Production of Glucose biscuits in a month = 4000 + 7500 = 11500
Production of Salty biscuits in a month = 2000 + 5000 = 7000

iii)
Profit in a month = Production of Tiger biscuits in a month(S.P. -C.P.) + Production of Glucose biscuits in a month(S.P. -C.P.) + Production of Salty biscuits in a month(S.P. -C.P.)
Profit in a month = 11000(13 - 10) + 11500(12 - 11) + 7000(14 - 12)
= 11000 × 3 + 11500 × 1 + 7000 × 2
33000 + 11500 + 14000 = 58500
Profit in a month is found to be Rs. 58,500.

Solve using techniques of matrix algebra

2021-07-29T20:59:00.001+05:45

A manufacturing company Ltd. Produces two types of drugs D1 and D2 with the help of three chemicals C1 , C2 and C3 . The quantity of chemical requirements per kg of each D1 and D2 are given below in suitable units.

	D₁	D₂
C₁	10	12
C₂	16	15
C₃	8	16

The price of chemicals in three different markets M1, M2 and M3 are as follows:

	C₁	C₂	C₃
M₁	20	19	16
M₂	8	9	7
M₃	6	7	8

Find the price per kg of each drug in each market by using matrix algebra.

Solution

Let matrix A represents prices of chemical in different market.

A =
Om
Om
Om
Om
Om
Om
Om
Om
Om
20

8

6

19

9

7

16

7

8

Om
Om
Om
Om
Om
Om
Om
Om
Om

Let matrix B represents quantity of chemical required for 1 kd of each drugs.

B =
Om
Om
Om
Om
Om
Om
Om
Om
Om
10

16

8

12

15

16

Om
Om
Om
Om
Om
Om
Om
Om
Om

Multiplication of A and B gives price per kg of each drugs in each market.

=
Om
Om
Om
Om
Om
Om
Om
Om
Om
20

8

6

19

9

7

16

7

8

Om
Om
Om
Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
10

16

8

12

15

16

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
Om
Om
Om
Om
Om
Om
Om
Om
Om
20×10+19×16+16×8

8×10+9×16+7×8

6×10+7×16+8×8

20×12+19×15+16×16

8×12+9×15+7×16

6×12+7×15+8×16

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
Om
Om
Om
Om
Om
Om
Om
Om
Om
200+304+128

80+144+56

60+112+64

240+285+256

96+135+112

72+105+128

Om
Om
Om
Om
Om
Om
Om
Om
Om

=
Om
Om
Om
Om
Om
Om
Om
Om
Om
632

280

236

781

343

305

Om
Om
Om
Om
Om
Om
Om
Om
Om

	D₁	D₂
M₁	632	781
M₂	280	343
M₃	236	305

The per kg price of drug D₁ in Market M₁, M₂ and M₃ are 632, 280 and 236 respectively.
The per kg price of drug D₂ in Market M₁, M₂ and M₃ are 781, 343 and 305 respectively.

If A= OmOm OmOmOmOm1 3 2 1 OmOmOmOm OmOmThen show that A2-2A-5I = 0 where 0 is the 2×2 null matrix and I be the identity matrix of order 2.

2021-07-29T20:53:00.002+05:45

Solution

A × A =
Om
Om
Om
Om
Om
Om
1

3

2

1

Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
1

3

2

1

Om
Om
Om
Om
Om
Om

A² =
Om
Om
Om
Om
Om
Om
1×1+2×3

3×1+1×3

1×2+2×1

3×2+1×1

Om
Om
Om
Om
Om
Om

A² =
Om
Om
Om
Om
Om
Om
1+6

3+3

2+2

6+1

Om
Om
Om
Om
Om
Om

∴ A² =
Om
Om
Om
Om
Om
Om
7

6

4

7

Om
Om
Om
Om
Om
Om

2A = 2
Om
Om
Om
Om
Om
Om
1

3

2

1

Om
Om
Om
Om
Om
Om

2A =
Om
Om
Om
Om
Om
Om
2 × 1

2 × 3

2 × 2

2 × 1

Om
Om
Om
Om
Om
Om

2A =
Om
Om
Om
Om
Om
Om
2

6

4

2

Om
Om
Om
Om
Om
Om

5I = 5
Om
Om
Om
Om
Om
Om
1

0

0

1

Om
Om
Om
Om
Om
Om

5I =
Om
Om
Om
Om
Om
Om
5 × 1

5 × 0

5 × 0

5 × 1

Om
Om
Om
Om
Om
Om

5I =
Om
Om
Om
Om
Om
Om
5

0

0

5

Om
Om
Om
Om
Om
Om

A² -2A-5I =
Om
Om
Om
Om
Om
Om
Om
Om
7

6

4

7

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
2

6

4

2

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
5

0

0

5

Om
Om
Om
Om
Om
Om

A² -2A-5I =
Om
Om
Om
Om
Om
Om
7-2-5

6-6-0

4-4-0

7-2-5

Om
Om
Om
Om
Om
Om

A² -2A-5I =
Om
Om
Om
Om
Om
Om
0

0

0

0

Om
Om
Om
Om
Om
Om

A² – 2A – 5I = 0 proved.

Solve by using matrix method 6x+8y = 30 9x+5y =42

2021-07-29T20:48:00.004+05:45

Solution

Writing the equation in the matrix form

AX = B

A =
Om
Om
Om
Om
Om
Om
6

9

8

5

Om
Om
Om
Om
Om
Om

X =
Om
Om
Om
Om
Om
Om
x

y

Om
Om
Om
Om
Om
Om

B =
Om
Om
Om
Om
Om
Om
30

42

Om
Om
Om
Om
Om
Om

Om
Om
Om
Om
Om
Om
6

9

8

5

Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
Om
x

y

Om
Om
Om
Om
Om
Om =
Om
Om
Om
Om
Om
Om
30

42

Om
Om
Om
Om
Om
Om

X = A⁻¹ B.
So, now we find A⁻¹.

|A| =
Om
Om
Om
Om
Om
Om
6

9

8

5

Om
Om
Om
Om = 6×5 - 9×8 = 30 - 72 = -42
Om
Om

As determinant of the matrix ≠ 0. Hence its inverse exists.

Inverse of matrix A = A^-¹

=
1
|A|

Om
Om
Om
Om
Om
Om
5

-9

-8

6

Om
Om
Om
Om
Om
Om

=
1
-42

Om
Om
Om
Om
Om
Om
5

-9

-8

6

Om
Om
Om
Om
Om
Om

∴ Inverse matrix of A = A^-¹ =
Om
Om
Om
Om
Om
Om
-0.119

0.214

0.19

-0.143

Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
-0.119

0.214

0.19

-0.143

Om
Om
Om
Om
Om
Om ×
Om
Om
Om
Om
Om
Om
Om
Om
30

42

Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
(-0.119)×30+0.19×42

0.214×30+(-0.143)×42

Om
Om
Om
Om
Om
Om

A^-¹ × B =
Om
Om
Om
Om
Om
Om
-3.57+7.98

6.42-6.006

Om
Om
Om
Om
Om
Om

∴ A^-¹ B =
Om
Om
Om
Om
Om
Om
4.41

0.41

Om
Om
Om
Om
Om
Om

∴x = 4.41

∴y = 0.41

Find the adjoint matrix of the matrix given below:A = OmOm OmOmOmOm1 3 2 -5 OmOmOmOmOmOm

2021-07-29T20:43:00.004+05:45

Solution

A₁₁ = -5CikaiA₁₂ = -3CikaiA₂₁ = -2CikaiA₂₂ = 1Cikai

The matrix of cofactors of A is C_A =
Om
Om
Om
Om
Om
Om
-5

-2

-3

1

Om
Om
Om
Om
Om
Om

As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."

∴ Adjoint of matrix A =
Om
Om
Om
Om
Om
Om
-5

-3

-2

1

Om
Om
Om
Om
Om
Om

Observations(x)	Frequency(f)	Class mark	x × f
0 - 10	5	5	5×5 = 25
10 - 20	17	15	15×17 = 255
20 - 30	31	25	25×31 = 775
30 - 40	41	35	35×41 = 1435
40 - 50	49	45	45×49 = 2205
	Σf = 143		Σfx = 4695

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Quantitive Analysis Assignment Solution For 2021 July

Application of matrix algebra (Numerical)

Solve using techniques of matrix algebra

If A= OmOm OmOmOmOm1 3 2 1 OmOmOmOm OmOmThen show that A2-2A-5I = 0 where 0 is the 2×2 null matrix and I be the identity matrix of order 2.

Solve by using matrix method 6x+8y = 30 9x+5y =42

Find the adjoint matrix of the matrix given below:A = OmOm OmOmOmOm1 3 2 -5 OmOmOmOmOmOm

Those numbers greater than 1 which upon division with any natural number except, one(1) and itself, results a decimal quotient are called prime numbers.
Examples of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 97, 197, 199, 991, 997, 9949, 9967, 9973 and so on.

Simplify
a)6√5 + 2√5 - 3√5
b)2√8 - √18 + 3√50

12) Convert 111101₂ into quinary number.

2.a What does it mean by the fact that the acceleration of a car is 5m/s² ? (2)

3.b A body starts moving from rest and attains the acceleration of 0.5 m/s². Calculate the velocity at the end of 3 minutes. Also find the distance travelled by it during that time. (3)