If \(A =\begin{bmatrix}1&2&3\\-4&-5&-2\end{bmatrix}, B=\begin{bmatrix}2&-3\\4&-5\\2&-1\end{bmatrix}\) and \(BA = [b_{ij}]\), then \((b_{23}-b_{31})\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 16
B is of order 3 × 2 and A is of order 2 × 3, so BA is 3 × 3. Its entry \(b_{ij}\) is row i of B times column j of A, so only the two entries asked for are needed.
Row 2 of B is (4, −5) and column 3 of A is (3, −2):
\(b_{23} = 4 \times 3 + (-5)(-2) = 12 + 10 = 22\)
Row 3 of B is (2, −1) and column 1 of A is (1, −4):
\(b_{31} = 2 \times 1 + (-1)(-4) = 2 + 4 = 6\)
So
\(b_{23} - b_{31} = 22 - 6 = 16\)