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

Concept:
First multiply the matrices \(A\) and \(B\), then take transpose of the product.

Step 1: Given: \[ A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \]

Step 2: Find \(AB\): \[ AB= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \]

Step 3: Multiply row by column: \[ AB= \begin{bmatrix} 1(1)+2(1) & 1(0)+2(0) \\ 3(1)+4(1) & 3(0)+4(0) \end{bmatrix} \] \[ AB= \begin{bmatrix} 3 & 0 \\ 7 & 0 \end{bmatrix} \]

Step 4: Now take transpose: \[ (AB)^T= \begin{bmatrix} 3 & 7 \\ 0 & 0 \end{bmatrix} \] \[ \boxed{\begin{bmatrix}3 & 7 \\ 0 & 0\end{bmatrix}} \]