If \(A=\begin{bmatrix}5 & 1 & 0\\ 6 & 2 & -1\end{bmatrix}\), \(B^T=\begin{bmatrix}2 & 1\\ -2 & 0\\ 1 & 3\end{bmatrix}\), then \(BA^T\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}8 & 7\\ 5 & 3\end{bmatrix}\)
B is the transpose of \(B^T\), and \(A^T\) is obtained by turning the rows of A into columns:
\(B = \begin{bmatrix}2 & -2 & 1\\ 1 & 0 & 3\end{bmatrix}\)
\(A^T = \begin{bmatrix}5 & 6\\ 1 & 2\\ 0 & -1\end{bmatrix}\)
B is 2 × 3 and \(A^T\) is 3 × 2, so \(BA^T\) is 2 × 2. Row by column:
\((1,1): \ 2(5)+(-2)(1)+1(0) = 8\)
\((1,2): \ 2(6)+(-2)(2)+1(-1) = 7\)
\((2,1): \ 1(5)+0(1)+3(0) = 5\)
\((2,2): \ 1(6)+0(2)+3(-1) = 3\)
\(BA^T = \begin{bmatrix}8 & 7\\ 5 & 3\end{bmatrix}\)
Why the other options are incorrect
Option (2) \(\begin{bmatrix}8 & 5\\ 7 & 3\end{bmatrix}\) — Incorrect. This is \(\left(BA^T\right)^T = AB^T\), the transpose of the required product.
Option (3) \(\begin{bmatrix}3 & 7\\ 5 & 8\end{bmatrix}\) — Incorrect. The (1, 1) entry is 8, not 3.
Option (4) \(\begin{bmatrix}3 & 5\\ 7 & 8\end{bmatrix}\) — Incorrect. The (1, 1) entry is 8, not 3.