Question

Find matrices A and B such that, can u plz help me
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Answer 1

Question:-

Find matrices A and B such that ,

$\sf \: A + B = \begin{bmatrix} \sf \: 5& \sf \: 4 \\ \\ \sf \:7 & \sf \: 3 \end{bmatrix} \: \& \: \: \: A - B = \begin{bmatrix} \sf \: 11& \sf \: 2 \\ \\ \sf \: - 1 & \sf \: 7 \end{bmatrix}$

Answer:-

Given:-

$\: \sf \: A + B = \begin{bmatrix} \sf \: 5& \sf \: 4 \\ \\ \sf \:7 & \sf \: 3 \end{bmatrix} \: \: \: \: - - \: equation \: (1). \\ \\ \\ \sf \: A - B = \begin{bmatrix} \sf \: 11& \sf \: 2 \\ \\ \sf \: - 1 & \sf \: 7 \end{bmatrix} \: \: - - \: equation \: (2).$

Adding equations (1) & (2) we get;

$\implies \sf \: A + B + A - B = \begin{bmatrix} \sf \: 5& \sf \: 4 \\ \\ \sf \: 7 & \sf \: 3 \end{bmatrix} + \begin{bmatrix} \sf \: 11& \sf \: 2 \\ \\ \sf \: - 1 & \sf \: 7 \end{bmatrix} \\ \\ \\ \implies \sf \:2A = \begin{bmatrix} \sf \: 5 + 11& \sf \: 4 + 2 \\ \\ \sf \: 7- 1 & \sf \: 3 + 7 \end{bmatrix} \\ \\ \\ \implies \sf \:2A = \begin{bmatrix} \sf \: 16& \sf \: 6 \\ \\ \sf \: 6 & \sf \: 10 \end{bmatrix} \\ \\ \\ \implies \sf \:A = \begin{bmatrix} \sf \: \dfrac{16}{2} & \sf \: \dfrac{6}{2} \\ \\ \sf \: \dfrac{6}{2} & \sf \: \dfrac{10}{2} \end{bmatrix} \\ \\ \\ \implies \sf \red{\:A = \begin{bmatrix} \sf \: 8 & \sf \: 3 \\ \\ \sf \: 3 & \sf \: 5 \end{bmatrix}}$

Substitute the value of A in equation (1).

$\implies \sf \: \begin{bmatrix} \sf \: 8& \sf \: 3 \\ \\ \sf \: 3& \sf \: 5 \end{bmatrix} + B = \begin{bmatrix} \sf \: 5& \sf \: 4 \\ \\ \sf \: 7& \sf \: 3 \end{bmatrix} \\ \\ \\ \implies \sf \: B = \begin{bmatrix} \sf \: 5& \sf \: 4 \\ \\ \sf \: 7& \sf \: 3 \end{bmatrix} - \begin{bmatrix} \sf \: 8& \sf \: 3 \\ \\ \sf \: 3& \sf \: 5 \end{bmatrix} \\ \\ \\ \implies \sf \: B = \begin{bmatrix} \sf \: 5 - 8& \sf \: 4 - 3 \\ \\ \sf \: 7 - 3& \sf \: 3 - 5 \end{bmatrix} \\ \\ \\ \implies \sf \red{B = \begin{bmatrix} \sf \: - 3& \sf \: 1 \\ \\ \sf \:4 & \sf \: - 2 \end{bmatrix}}$

Answer 2

Question:-

Find matrices A and B such that ,

$\rm \: A + B \: = \begin{bmatrix} 5 & 4\\ 7 & 3\end{bmatrix}$

$\rm \: A - B \: = \begin{bmatrix} 11 & 2\\ - 1 & 7\end{bmatrix}$

Answer:-

Given:-

$\rm \: A + B \: = \begin{bmatrix} 5 & 4\\ 7 & 3\end{bmatrix} - - - (i)$

$\rm \: A - B \: = \begin{bmatrix} 11 & 2\\ - 1 & 7\end{bmatrix} - - - (ii)$

On adding (i) and (ii), we get

$\rm :\implies\:2A \: = \begin{bmatrix} 5 + 11 & 4 + 2\\ 7 - 1 & 3 + 7\end{bmatrix}$

$\rm :\implies\: \rm \: 2A \: = \begin{bmatrix} 16 & 6\\ 6 & 10\end{bmatrix}$

$\rm :\implies\: \boxed{ \pink{ \bf \: A \: = \tt \: \begin{bmatrix} 8 & 3\\ 3 & 5\end{bmatrix}}}$

Now,

Again,

We have,

$\rm \: A + B \: = \begin{bmatrix} 5 & 4\\ 7 & 3\end{bmatrix} - - - (i)$

$\rm \: A - B \: = \begin{bmatrix} 11 & 2\\ - 1 & 7\end{bmatrix} - - - (ii)$

On Subtracting (ii) from (i), we get

$\rm :\implies\: \rm \: 2B \: = \begin{bmatrix} 5 - 11 & 4 - 2\\ 7 + 1 & 3 - 7\end{bmatrix}$

$\rm :\implies\: \rm \: 2 B \: = \begin{bmatrix} - 6 & 2\\ 8 & - 4\end{bmatrix}$

$\rm :\implies\: \boxed{ \pink{ \bf \: B\: = \tt \: \begin{bmatrix} - 3 & 1\\ 4 & - 2\end{bmatrix} }}$