Question

If \[ A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \] then \((AB)^T=\)

• A. \(\begin{bmatrix}0 & 0 \\ 3 & 4\end{bmatrix}\)
• B. \(\begin{bmatrix}0 & 0 \\ 3 & 7\end{bmatrix}\)
• C. \(\begin{bmatrix}3 & 7 \\ 0 & 0\end{bmatrix}\)
• D. \(\begin{bmatrix}3 & 6 \\ 0 & 0\end{bmatrix}\)

Hint:

For \((AB)^T\), first calculate \(AB\), then interchange rows and columns.

Answer 1

The Correct Option is C

To find \((AB)^T\), we first need to calculate the product \(AB\) and then take the transpose of the resulting matrix.

Given:

\(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}\)

Step 1: Compute the matrix multiplication \(AB\).

\(AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} (1 \cdot 1 + 2 \cdot 1) & (1 \cdot 0 + 2 \cdot 0) \\ (3 \cdot 1 + 4 \cdot 1) & (3 \cdot 0 + 4 \cdot 0) \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 7 & 0 \end{bmatrix}\)

Step 2: Take the transpose of matrix \(AB\).

\((AB)^T = \begin{bmatrix} 3 & 0 \\ 7 & 0 \end{bmatrix}^T = \begin{bmatrix} 3 & 7 \\ 0 & 0 \end{bmatrix}\)

Therefore, the matrix \((AB)^T\) is:

\(\begin{bmatrix} 3 & 7 \\ 0 & 0 \end{bmatrix}\)

Conclusion: The correct answer is \(\begin{bmatrix} 3 & 7 \\ 0 & 0 \end{bmatrix}\), which matches with the given option.