Determine the matrices A and B when

A + 2B = $\begin{bmatrix$}[r] 1 & 2 \ 6 & -3 \end{bmatrix} \text{ and 2A - B} = \begin{bmatrix}[r] 2 & -1 \ 2 & -1 \end{bmatrix}.

Given,

$\text{A + 2B} = \begin{bmatrix$}[r] 1 & 2 \ 6 & -3 \end{bmatrix} \qquad \text{….[Eq 1] } \\[1em] \text{2A - B} = \begin{bmatrix}[r] 2 & -1 \ 2 & -1 \end{bmatrix} \qquad \text{….[Eq 2] } \\[1em]

Multiplying Eq 1 by 2,

$\Rightarrow 2A + 4B = \begin{bmatrix$}[r] 2 & 4 \ 12 & -6 \end{bmatrix} \\[1em]

Subtracting Eq 2 from above equation we get,

$\Rightarrow 2A + 4B - (2A - B) = \begin{bmatrix$}[r] 2 & 4 \ 12 & -6 \end{bmatrix} - \begin{bmatrix}[r] 2 & -1 \ 2 & -1 \end{bmatrix} \\[1em] \Rightarrow 2A - 2A + 4B - (-B) = \begin{bmatrix}[r] 2 - 2 & 4 - (-1) \ 12 - 2 & -6 - (-1) \end{bmatrix} \\[1em] \Rightarrow 5B = \begin{bmatrix}[r] 0 & 5 \ 10 & -5 \end{bmatrix} \\[1em] \Rightarrow B = \dfrac{1}{5}\begin{bmatrix}[r] 0 & 5 \ 10 & -5 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix}[r] 0 & 1 \ 2 & -1 \end{bmatrix}. \\[1em]

Putting value of matrix B in Eq 1,

$\Rightarrow A + 2 \begin{bmatrix$}[r] 0 & 1 \ 2 & -1 \end{bmatrix} = \begin{bmatrix}[r] 1 & 2 \ 6 & -3 \end{bmatrix} \\[1em] \Rightarrow A + \begin{bmatrix}[r] 0 & 2 \ 4 & -2 \end{bmatrix} = \begin{bmatrix}[r] 1 & 2 \ 6 & -3 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix}[r] 1 & 2 \ 6 & -3 \end{bmatrix} - \begin{bmatrix}[r] 0 & 2 \ 4 & -2 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix}[r] 1 - 0 & 2 - 2 \ 6 - 4 & -3 - (-2) \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix}[r] 1 & 0 \ 2 & -1 \end{bmatrix}. \\[1em] \therefore A = \begin{bmatrix}[r] 1 & 0 \ 2 & -1 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] 0 & 1 \ 2 & -1 \end{bmatrix}.

Hence, the value of $\text{A} = \begin{bmatrix$}[r] 1 & 0 \ 2 & -1 \end{bmatrix} \text{and B} = \begin{bmatrix}[r] 0 & 1 \ 2 & -1 \end{bmatrix}.