Question

Given X + Y = $\left[\begin{array}{ccc}1&2\\1&1\\\end{array}\right]$ and X - Y = $\left[\begin{array}{ccc}3&0\\-1&1\\\end{array}\right]$. Find X and Y ?

Answer 1

Answer:-

Given:-

$\sf \: X + Y= \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} \: \: \: - \: equation \: (1)$

And,

$\sf \: X - Y = \begin{bmatrix} \sf \: 3& \sf \: 0 \\ \sf \: - 1& \sf \: 1 \end{bmatrix} \: \: - \: \: equation \: (2)$

Add equations (1) & (2)

$\implies \sf \: X + Y + X - Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} + \begin{bmatrix} \sf \: 3& \sf \: 0 \\ \sf \: - 1& \sf \: 1 \end{bmatrix} \\ \\ \\ \implies \sf \:2X = \begin{bmatrix} \sf \: 1 + 3& \sf \: 2 + 0\\ \sf \: 1 - 1& \sf \: 1 + 1 \end{bmatrix} \\ \\ \\ \implies \sf \:2X = \begin{bmatrix} \sf \: 4& \sf \: 2 \\ \sf \: 0& \sf \: 2\end{bmatrix} \\ \\ \\ \implies \sf \: X = \begin{bmatrix} \sf \: \frac{4}{2} & \sf \: \frac{2}{2} \\ \\ \sf \: \frac{0}{2} & \sf \: \frac{2}{2} \end{bmatrix} \\ \\ \\ \sf \implies \red{X = \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix}}$

Substitute the value of X in equation (1).

$\implies \sf \: \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix} + Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} \\ \\ \\ \implies \sf \:Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} - \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix} \\ \\ \\ \implies \sf \:Y =\begin{bmatrix} \sf \: 1 - 2& \sf \: 2 - 1 \\ \sf \: 1 - 0& \sf \: 1 - 1 \end{bmatrix} \\ \\ \\ \implies \sf \red{ \:Y =\begin{bmatrix} \sf \: - 1 & \sf \: 1 \\ \sf \: 1 & \sf \: 0 \end{bmatrix}}$