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:
There are two ways to solve this:
1. First, calculate the product matrix \(C = AB\). Then, find the transpose of C, which is \(C^T\).
2. Use the property of transpose: \((AB)^T = B^T A^T\). First, find the transposes of A and B, then multiply them in reverse order.
We will use the first method as it is more direct.

Step 3: Detailed Explanation:

Step 3a: Calculate the product AB
\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \]
\[ AB = \begin{bmatrix} (1)(1) + (2)(1) & (1)(0) + (2)(0) \\ (3)(1) + (4)(1) & (3)(0) + (4)(0) \end{bmatrix} \]
\[ AB = \begin{bmatrix} 1 + 2 & 0 + 0 \\ 3 + 4 & 0 + 0 \end{bmatrix} \]
\[ AB = \begin{bmatrix} 3 & 0 \\ 7 & 0 \end{bmatrix} \]

Step 3b: Find the transpose of AB
The transpose of a matrix is found by interchanging its rows and columns.
Let \(C = AB = \begin{bmatrix} 3 & 0 \\ 7 & 0 \end{bmatrix}\).
Then the transpose of C is:
\[ C^T = (AB)^T = \begin{bmatrix} 3 & 7 \\ 0 & 0 \end{bmatrix} \]

Step 4: Final Answer:
The resulting matrix \((AB)^T\) is \(\begin{bmatrix} 3 & 7 \\ 0 & 0 \end{bmatrix}\).