Question

If \(A = \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix}  \quad \text{and} \quad  B = \begin{bmatrix} -1 & 0 \\ 2 & 5 \\ 3 & 4 \end{bmatrix},\)then \((BA)^T\) is equal to:

• A. \(\begin{bmatrix} -3 & 1 & 5 \\ -2 & 9 & 10 \end{bmatrix}\)
• B. \(\begin{bmatrix} 3 & 1 & 5 \\ 2 & 9 & 10 \end{bmatrix}\)
• C. \(\begin{bmatrix} -3 & -2 \\ 1 & 9 \\ 5 & 10 \end{bmatrix}\)
• D. \(\begin{bmatrix} 3 & 2 \\ 1 & 9 \\ 5 & 10 \end{bmatrix}\)

Answer 1

The Correct Option is A

To determine \((BA)^T\), we first compute the product \(BA\) and subsequently its transpose. The given matrices are:

\(A = \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix}\) and \(B = \begin{bmatrix} -1 & 0 \\ 2 & 5 \\ 3 & 4 \end{bmatrix}\).

The matrix product \(BA\) is calculated as:

\(BA = \begin{bmatrix} -1 & 0 \\ 2 & 5 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix}\).

The entries of \(BA\) are computed as follows:

• Entry (1,1): \((-1)(3) + (0)(-1) = -3\).
• Entry (1,2): \((-1)(2) + (0)(1) = -2\).
• Entry (2,1): \((2)(3) + (5)(-1) = 6 - 5 = 1\).
• Entry (2,2): \((2)(2) + (5)(1) = 4 + 5 = 9\).
• Entry (3,1): \((3)(3) + (4)(-1) = 9 - 4 = 5\).
• Entry (3,2): \((3)(2) + (4)(1) = 6 + 4 = 10\).

Therefore, the product \(BA\) is:

\(BA = \begin{bmatrix} -3 & -2 \\ 1 & 9 \\ 5 & 10 \end{bmatrix}\).

The transpose \((BA)^T\) is obtained by interchanging the rows and columns of \(BA\):

\((BA)^T = \begin{bmatrix} -3 & 1 & 5 \\ -2 & 9 & 10 \end{bmatrix}\).

This result matches the option:

\(\begin{bmatrix} -3 & 1 & 5 \\ -2 & 9 & 10 \end{bmatrix}\)