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

First, compute \( B \) by transposing \( B^T \):

\[ B = \begin{bmatrix} 1 & -2 \\ 10 & -1 \end{bmatrix}. \]

Now compute \( AB \) using matrix multiplication:

\[ AB = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & -2 \\ 10 & -1 \end{bmatrix}. \]

Perform the row-column multiplication:

First row:

\[ \begin{bmatrix} 5 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 10 \end{bmatrix} = (5)(1) + (1)(10) = 5 + 10 = 15, \]

First row:

\[ \begin{bmatrix} 5 & 1 \end{bmatrix} \cdot \begin{bmatrix} -2 \\ -1 \end{bmatrix} = (5)(-2) + (1)(-1) = -10 - 1 = -11. \]

Second row:

\[ \begin{bmatrix} -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 10 \end{bmatrix} = (-2)(1) + (0)(10) = -2, \]

Second row:

\[ \begin{bmatrix} -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} -2 \\ -1 \end{bmatrix} = (-2)(-2) + (0)(-1) = 4. \]

Thus, the resulting matrix is:

\[ AB = \begin{bmatrix} 15 & -11 \\ -2 & 4 \end{bmatrix}. \]

Hence, the correct answer is:

\[ \text{(2)} \begin{bmatrix} 15 & -11 \\ -2 & 4 \end{bmatrix}. \]