If $A=\begin{vmatrix}4&3\\2&-1\\1&0\end{vmatrix}$ and $B^T=\begin{vmatrix}1&2&3\\2&1&-1\end{vmatrix}$, then $A – B$ is equal to

$\begin{vmatrix}3&1\\0&-2\\-2&1\end{vmatrix}$

The correct answer is Option (3) → $\begin{vmatrix}3&1\\0&-2\\-2&1\end{vmatrix}$

Given:

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

$B^T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & -1 \end{bmatrix}$

So, $B = (B^T)^T = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & -1 \end{bmatrix}$

Now compute $A - B$:

$A - B = \begin{bmatrix} 4 & 3 \\ 2 & -1 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 4 - 1 & 3 - 2 \\ 2 - 2 & -1 - 1 \\ 1 - 3 & 0 - (-1) \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 0 & -2 \\ -2 & 1 \end{bmatrix}$