If \(A =\begin{bmatrix}2&3&1\\2&-1&0\end{bmatrix}\) and \(B^T =\begin{bmatrix}4&4\\6&-2\\2&0\end{bmatrix}\), then \(4A + B\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3B
B is the transpose of \(B^T\):
\( \displaystyle B = \begin{bmatrix}4&6&2\\4&-2&0\end{bmatrix} \)
\( \displaystyle B = 2\begin{bmatrix}2&3&1\\2&-1&0\end{bmatrix} = 2A \)
Both A and B are \(2 \times 3\), so the sum exists:
\(4A + B = 4A + 2A = 6A\)
\(6A = 3(2A) = 3B\)
Options (2) and (3) are the same matrix, \(4B = 8A\), which is not 6A.