Find the matrix A, if B = $\begin{bmatrix$}[r] 2 & 1 \ 0 & 1 \end{bmatrix} \text{ and } B^2 = B + \dfrac{1}{2}A.

Given,

$\Rightarrow B^2 = B + \dfrac{1}{2}A$

Substituting value of B in above equation we get,

$\Rightarrow \begin{bmatrix$}[r] 2 & 1 \ 0 & 1 \end{bmatrix}\begin{bmatrix}[r] 2 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 2 & 1 \ 0 & 1 \end{bmatrix} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times 1\ 0 \times 2 + 1 \times 0 & 0 \times 1 + 1 \times 1 \end{bmatrix} = \begin{bmatrix}[r] 2 & 1 \ 0 & 1 \end{bmatrix} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix}[r] 4 + 0 & 2 + 1 \ 0 & 0 + 1 \end{bmatrix} = \begin{bmatrix}[r] 2 & 1 \ 0 & 1 \end{bmatrix} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 2 & 1 \ 0 & 1 \end{bmatrix} + \dfrac{1}{2}A \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} - \begin{bmatrix}[r] 2 & 1 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix}[r] 4 - 2 & 3 - 1 \ 0 - 0 & 1 - 1 \end{bmatrix} \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix}[r] 2 & 2 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow A = 2\begin{bmatrix}[r] 2 & 2 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix}[r] 4 & 4 \ 0 & 0 \end{bmatrix}.

Hence, A = $\begin{bmatrix$}[r] 4 & 4 \ 0 & 0 \end{bmatrix}.