If I is the unit matrix of order 2 × 2; find the matrix M such that :

5M + 3I = $4\begin{bmatrix$}[r] 2 & -5 \ 0 & -3 \end{bmatrix}

Given,

$\Rightarrow 5M + 3I = 4\begin{bmatrix$}[r] 2 & -5 \ 0 & -3 \end{bmatrix} \\[1em] \Rightarrow 5M + 3\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 8 & -20 \ 0 & -12 \end{bmatrix} \\[1em] \Rightarrow 5M + \begin{bmatrix}[r] 3 & 0 \ 0 & 3 \end{bmatrix} = \begin{bmatrix}[r] 8 & -20 \ 0 & -12 \end{bmatrix} \\[1em] \Rightarrow 5M = \begin{bmatrix}[r] 8 & -20 \ 0 & -12 \end{bmatrix} - \begin{bmatrix}[r] 3 & 0 \ 0 & 3 \end{bmatrix} \\[1em] \\[1em] \Rightarrow 5M = \begin{bmatrix}[r] 8 - 3 & -20 - 0 \ 0 - 0 & -12 - 3 \end{bmatrix} \\[1em] \Rightarrow 5M = \begin{bmatrix}[r] 5 & -20 \ 0 & -15 \end{bmatrix} \\[1em] \Rightarrow M = \dfrac{1}{5}\begin{bmatrix}[r] 5 & -20 \ 0 & -15 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix}[r] 1 & -4 \ 0 & -3 \end{bmatrix}.

Hence, M = $\begin{bmatrix$}[r] 1 & -4 \ 0 & -3 \end{bmatrix}.