Question

If \( A = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 1 & 10 \\ -2 & -1 \end{bmatrix} \), then the matrix \( AB \) is:

• A. \( \begin{bmatrix} 1 & 10 \\ -1 & 0 \end{bmatrix} \)
• B. \( \begin{bmatrix} 15 & -11 \\ -2 & 4 \end{bmatrix} \)
• C. \( \begin{bmatrix} 3 & 49 \\ -2 & -20 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 9 \\ -2 & -20 \end{bmatrix} \)

Answer 1

The Correct Option is B

Given matrices:
\( A = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \)
\( B^T = \begin{bmatrix} 1 & 10 \\ -2 & -1 \end{bmatrix} \)
The transpose of \( B^T \) yields matrix \( B \):
\( B = \begin{bmatrix} 1 & -2 \\ 10 & -1 \end{bmatrix} \)
To compute matrix \( AB \), matrix multiplication is performed:
\[ AB = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & -2 \\ 10 & -1 \end{bmatrix} \]
The element at position (1,1) is computed as:
\((5 \times 1) + (1 \times 10) = 5 + 10 = 15\)
The element at position (1,2) is computed as:
\((5 \times -2) + (1 \times -1) = -10 - 1 = -11\)
The element at position (2,1) is computed as:
\((-2 \times 1) + (0 \times 10) = -2 + 0 = -2\)
The element at position (2,2) is computed as:
\((-2 \times -2) + (0 \times -1) = 4 + 0 = 4\)
Consequently, matrix \( AB \) is:
\( \begin{bmatrix} 15 & -11 \\ -2 & 4 \end{bmatrix} \)
Therefore, the result is:
\( \begin{bmatrix} 15 & -11 \\ -2 & 4 \end{bmatrix} \)