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  • If the value of a is -1 and b is 1 then matrix A, <img alt="A=\left[\begin{array}{lll}{a-b} & {a^2-b^2} & {\;a^3-b^3} \\ {a\times b} & {a^2\times b} & {(b-a)^2} \\ {a+b} &

If the value of a is -1 and b is 1 then matrix A, A=\left[\begin{array}{lll}{a-b} & {a^2-b^2} & {\;a^3-b^3} \\ {a\times b} & {a^2\times b} & {(b-a)^2} \\ {a+b} & {(a-b^2)} & {\;a^2-b}\end{array}\right] is equal matrix to

Option: 1

B=\left[\begin{array}{lll}{-2} & {\;\;0} & {-2} \\ {-1} & {\;\;1} & {\;\;\;4} \\ {\;\;\;0} & \;\;{4} & {\;\;\;0}\end{array}\right]


Option: 2

B=\left[\begin{array}{lll}{\;\;\;0} & {\;\;0} & {-2} \\ {-1} & {\;\;1} & {\;\;\;4} \\ {\;\;\;0} & \;\;{4} & {\;\;\;0}\end{array}\right]


Option: 3

B=\left[\begin{array}{lll}{\;\;\;0} & {\;\;0} & {-2} \\ {-1} & {-1} & {\;\;\;4} \\ {\;\;\;0} & \;\;{4} & {\;\;\;0}\end{array}\right]


Option: 4

B=\left[\begin{array}{lll}{\;\;\;0} & {\;\;0} & {-2} \\ {-1} & {\;\;0} & {\;\;\;4} \\ {\;\;\;0} & \;\;{4} & {\;\;\;0}\end{array}\right]


Answers (1)

best_answer

 

 

Types of Matrices - Part 1 -

Equal Matrices: Two matrices are said to be equal if they have the same order and each element of one matrix is equal to the corresponding elements of another matrix or we can say a_{ij}=b_{ij} where a is the element of one matrix and b is the element of another matrix.

 

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A=\left[\begin{array}{lll}{a-b} & {a^2-b^2} & {\;a^3-b^3} \\ {a\times b} & {a^2\times b} & {(b-a)^2} \\ {a+b} & {(a-b^2)} & {\;a^2-b}\end{array}\right]

on putting a=-1 and b=1,

A=\left[\begin{array}{lll}{-1-1} & {(-1)^2-1^2} & {\;(-1)^3-1^3} \\ {-1\times 1} & {(-1)^2\times 1} & {(1-(-1))^2} \\ {-1+1} & {(-1-(1)^2)} & {\;(-1)^2-1}\end{array}\right]

hence,

B=\left[\begin{array}{lll}{-2} & {\;\;0} & {-2} \\ {-1} & {\;\;1} & {\;\;\;4} \\ {\;\;\;0} & \;\;{4} & {\;\;\;0}\end{array}\right]

correct option is (a)

Posted by

Pankaj

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