Let $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix}, C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}$. Find a matrix $D$ such that $CD - AB = O$.

$\begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$

The correct answer is Option (2) → $\begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$ ##

Since $A, B, C$ are all square matrices of order $2$, and $CD - AB$ is well defined, $D$ must be a square matrix of order $2$.

Let $D = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then $CD - AB = O$ gives

$\begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = \text{O}$

or $\begin{bmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

or $\begin{bmatrix} 2a + 5c - 3 & 2b + 5d \\ 3a + 8c - 43 & 3b + 8d - 22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

By equality of matrices, we get

$2a + 5c - 3 = 0 \quad \dots (1)$

$3a + 8c - 43 = 0 \quad \dots (2)$

$2b + 5d = 0 \quad \dots (3)$

and $3b + 8d - 22 = 0 \quad \dots (4)$

Solving (1) and (2), we get $a = -191, c = 77$. Solving (3) and (4), we get $b = -110, d = 44$.

Therefore $D = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$