If $A = \begin{bmatrix}2&-2&1\\0&4&2\end{bmatrix}$ and $B=\begin{bmatrix}1&-2&7\\2&0&6\end{bmatrix}$ are two matrices such that $3A - 2B + 4C = 0$, then matrix $C$ is equal to:

$\begin{bmatrix}-1&1/2&11/4\\1&-3&3/2\end{bmatrix}$

The correct answer is Option (2) → $\begin{bmatrix}-1&1/2&11/4\\1&-3&3/2\end{bmatrix}$

Given matrices:

A = $\begin{bmatrix} 2 & -2 & 1 \\ 0 & 4 & 2 \end{bmatrix}$, B = $\begin{bmatrix} 1 & -2 & 7 \\ 2 & 0 & 6 \end{bmatrix}$

Equation: 3A - 2B + 4C = 0

Solve for C:

4C = 2B - 3A ⇒ C = $\frac{1}{4}(2B - 3A)$

Compute 2B - 3A:

2B = $\begin{bmatrix} 2 & -4 & 14 \\ 4 & 0 & 12 \end{bmatrix}$

3A = $\begin{bmatrix} 6 & -6 & 3 \\ 0 & 12 & 6 \end{bmatrix}$

2B - 3A = $\begin{bmatrix} 2-6 & -4 -(-6) & 14 -3 \\ 4-0 & 0-12 & 12-6 \end{bmatrix} = \begin{bmatrix} -4 & 2 & 11 \\ 4 & -12 & 6 \end{bmatrix}$

Divide by 4:

C = $\frac{1}{4} \begin{bmatrix} -4 & 2 & 11 \\ 4 & -12 & 6 \end{bmatrix} = \begin{bmatrix} -1 & 0.5 & 11/4 \\ 1 & -3 & 3/2 \end{bmatrix}$