Question

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

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

Hint:

Remember the "reversal law" for the transpose of a product: \((AB)^T = B^T A^T\). This property is very useful and extends to more matrices, e.g., \((ABC)^T = C^T B^T A^T\). While direct multiplication worked well here, knowing this rule is essential for more complex problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two 2x2 matrices, A and B. We need to find the transpose of their product, (AB)^T.

Step 2: Key Formula or Approach (Alternate Method):
Instead of multiplying A and B first then transposing, use the reverse-order property of transpose: (AB)^T = B^T × A^T.

Step 3: Detailed Explanation:
Given: A = [[1, 2], [3, 4]], B = [[1, 0], [1, 0]]. First, find B^T by swapping rows and columns of B: B^T = [[1, 1], [0, 0]]. Next, find A^T by swapping rows and columns of A: A^T = [[1, 3], [2, 4]]. Now compute B^T × A^T: Element (1,1): 1(1) + 1(2) = 3. Element (1,2): 1(3) + 1(4) = 7. Element (2,1): 0(1) + 0(2) = 0. Element (2,2): 0(3) + 0(4) = 0. So, (AB)^T = [[3, 7], [0, 0]].

Step 4: Final Answer:
The resulting matrix (AB)^T is [[3, 7], [0, 0]].