If X={bmatrix}3 & -4 \\ 1 & -1 {bmatrix}, B={bmatrix}5 & 2 \\ -2 & 1 {bmatrix} and A={bmatrix}p & q \\ r & s {bmatrix} satisfy the equation AX = B then matrix A is equal to [NDA I 2015]

Correct Answer is : [−7261−5]{bmatrix}-7 & 26 \\ 1 & -5 {bmatrix}[−7126−5]

{bmatrix}-7 & 26 \\ 1 & -5 {bmatrix}

{bmatrix}7 & 26 \\ 4 & 17 {bmatrix}

{bmatrix}-7 & -4 \\ 26 & 13 {bmatrix}

{bmatrix}-7 & 26 \\ -6 & 23 {bmatrix}

Correct Answer is : [−7261−5]{bmatrix}-7 & 26 \\ 1 & -5 {bmatrix}[−7126−5]

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NDA Math Matrices

Question : 59 of 69

Marks: +1, -0

X=\begin{bmatrix}3 & -4 \\ 1 & -1 \end{bmatrix}, B=\begin{bmatrix}5 & 2 \\ -2 & 1 \end{bmatrix}

A=\begin{bmatrix}p & q \\ r & s \end{bmatrix}

Solution:

Consider the given matrices

A=\begin{pmatrix}p & q \\ r & s \end{pmatrix}

X=\begin{pmatrix}3 & -4 \\ 1 & -1 \end{pmatrix}

Ax=\begin{pmatrix}p & q \\ r & s \end{pmatrix}\begin{pmatrix}3 & -4 \\ 1 & -1 \end{pmatrix}

=\begin{pmatrix}3p+q & -4p-q \\ 3r+s & -4r-s \end{pmatrix}

=B

=\begin{pmatrix}5 & 2 \\ -2 & 1 \end{pmatrix}

Hence by comparing the elements of the matrices, we get

3p+q=5

-4p-q=2

Solving for p and q, yields

p=-7

q=26

Similarly

3r+s=-2

-4r-s=1

r=1

s=-5

Hence matrix A is equal to

A=\begin{pmatrix}p & q \\ r & s \end{pmatrix}=\begin{pmatrix}-7 & 26 \\ 1 & -5 \end{pmatrix}

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