If $A = \begin{bmatrix}1&5\\7&12\end{bmatrix},B=\begin{bmatrix}9&1\\7&8\end{bmatrix}$ and $C$ are three matrices such that $3A +5B+2C = 0$, then the matrix $C$ is equal to

$\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$

The correct answer is Option (2) → $\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$

Given: $A=\begin{bmatrix}1 & 5\\7 & 12\end{bmatrix},\ B=\begin{bmatrix}9 & 1\\7 & 8\end{bmatrix}$

and $3A + 5B + 2C = 0$

⇒ $2C = -(3A + 5B)$

⇒ $C = -\frac{1}{2}(3A + 5B)$

$3A = \begin{bmatrix}3 & 15\\21 & 36\end{bmatrix}$

$5B = \begin{bmatrix}45 & 5\\35 & 40\end{bmatrix}$

$3A + 5B = \begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}$

$C = -\frac{1}{2}\begin{bmatrix}48 & 20\\56 & 76\end{bmatrix} = \begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix}$

Final Answer:

$C = \begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix}$