If A = […] and B^T = […] , then 4A + B is

The correct answer is Option (4) → 3B

Given matrices:

$A = \begin{bmatrix} 2 & 3 & 1 \\ 2 & -1 & 0 \end{bmatrix}$

$B^T = \begin{bmatrix} 4 & 4 \\ 6 & -2 \\ 2 & 0 \end{bmatrix}$

First, find $B$ by taking transpose of $B^T$:

$B = \begin{bmatrix} 4 & 6 & 2 \\ 4 & -2 & 0 \end{bmatrix}$

Compute $4A$:

$4A = 4 \cdot \begin{bmatrix} 2 & 3 & 1 \\ 2 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 8 & 12 & 4 \\ 8 & -4 & 0 \end{bmatrix}$

Now, $4A + B = \begin{bmatrix} 8+4 & 12+6 & 4+2 \\ 8+4 & -4+(-2) & 0+0 \end{bmatrix} = \begin{bmatrix} 12 & 18 & 6 \\ 12 & -6 & 0 \end{bmatrix} = 3B$