Question:medium

If \( A = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix} \), then \( B^T A^T \) is:

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\((AB)^T = B^T A^T\) — always use this property to simplify.
Updated On: Apr 17, 2026
  • null matrix
  • an identity matrix
  • scalar but not an identity matrix
  • such that \( \text{Tr}(B^T A^T) = 4 \)
Show Solution

The Correct Option is B

Solution and Explanation

To solve the problem and compute \( B^T A^T \), we begin by finding the transpose of matrices \( A \) and \( B \). Given:

\( A = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \)\( B = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix} \)

Step 1: Compute the transpose of matrix \( A \):

\( A^T = \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} \)

Step 2: Compute the transpose of matrix \( B \):

\( B^T = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \)

Step 3: Multiply \( B^T \) and \( A^T \) to find \( B^T A^T \):

The product \( B^T A^T \) is calculated as:

\( B^T A^T = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} \)

The element in the first row and first column of the result is:

\((4 \times 2) + (7 \times -1) = 8 - 7 = 1\)

The element in the first row and second column of the result is:

\((4 \times -7) + (7 \times 4) = -28 + 28 = 0\)

The element in the second row and first column of the result is:

\((1 \times 2) + (2 \times -1) = 2 - 2 = 0\)

The element in the second row and second column of the result is:

\((1 \times -7) + (2 \times 4) = -7 + 8 = 1\)

Thus, \( B^T A^T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), which is an identity matrix.

Conclusion: The matrix \( B^T A^T \) is an identity matrix, confirming the correct answer is "an identity matrix".

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