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 solve for \((BA)^T\), we first need to find the product \(BA\) and then take its transpose. Given: 

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

The product \(BA\) is given by:

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

Calculating each entry:

• First row, first column: \((-1 \times 3) + (0 \times -1) = -3\).
• First row, second column: \((-1 \times 2) + (0 \times 1) = -2\).
• Second row, first column: \((2 \times 3) + (5 \times -1) = 6 - 5 = 1\).
• Second row, second column: \((2 \times 2) + (5 \times 1) = 4 + 5 = 9\).
• Third row, first column: \((3 \times 3) + (4 \times -1) = 9 - 4 = 5\).
• Third row, second column: \((3 \times 2) + (4 \times 1) = 6 + 4 = 10\).

Hence,

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

The transpose of a matrix is obtained by swapping its rows with columns. Thus, the transpose \((BA)^T\) is:

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

This corresponds to the option:

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

Answer 2

To find $(BA)^T$, we compute $BA$ first, then take its transpose.
Multiplying $B$ and $A$ The matrices $B$ and $A$ are:
$B = \begin{bmatrix} -1 & 0 \\ 2 & 5 \\ 3 & 4 \end{bmatrix}, \quad A = \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix}$
The product $BA$ is calculated as:
$BA = \begin{bmatrix} -1 & 0 \\ 2 & 5 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix}$
Perform the multiplication row by row:
1.First row of $B$ with both columns of $A$:
  $[-1 \cdot 3 + 0 \cdot (-1), \ -1 \cdot 2 + 0 \cdot 1] = [-3, -2]$
2.Second row of $B$ with both columns of $A$:
  $[2 \cdot 3 + 5 \cdot (-1), \ 2 \cdot 2 + 5 \cdot 1] = [6 - 5, 4 + 5] = [1, 9]$
3.Third row of $B$ with both columns of $A$:
 $[3 \cdot 3 + 4 \cdot (-1), \ 3 \cdot 2 + 4 \cdot 1] = [9 - 4, 6 + 4] = [5, 10]$
Thus,
$BA = \begin{bmatrix} -3 & -2 \\ 1 & 9 \\ 5 & 10 \end{bmatrix}$
Transposing $BA$ The transpose of $BA$ is obtained by interchanging rows and columns:
$(BA)^T = \begin{bmatrix} -3 & 1 & 5 \\ -2 & 9 & 10 \end{bmatrix}$
Final Answer: The matrix $(BA)^T$ is:
$\begin{bmatrix} -3 & 1 & 5 \\ -2 & 9 & 10 \end{bmatrix}$