If \(A = \begin{bmatrix}4&3\\2&-1\\1&0\end{bmatrix}\) and \(B^T = \begin{bmatrix}1&2&3\\2&1&-1\end{bmatrix}\), then \(A - B\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}3&1\\0&-2\\-2&1\end{bmatrix}\)
Since \((B^T)^T = B\), B is obtained by transposing \(B^T\) (its rows become columns):
\( \displaystyle B = \begin{bmatrix}1&2\\2&1\\3&-1\end{bmatrix} \)
Both A and B are 3 × 2, so they are subtracted entry by entry:
\( \displaystyle A - B = \begin{bmatrix}4 - 1&3 - 2\\2 - 2&-1 - 1\\1 - 3&0 - (-1)\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}3&1\\0&-2\\-2&1\end{bmatrix} \)