Question

If \[ A = \begin{bmatrix} 2 & \sqrt{2} & 0 \\ 3 & -2 & \frac{2}{5} \end{bmatrix} \] then \[ A^t = \] The correct answer is:

• A. \( \begin{bmatrix} 2 & 0\\ \sqrt{2} & 2 \\ 3 & \frac{2}{5} \end{bmatrix} \)
• B. \( \begin{bmatrix} 2 & 0 \\ \sqrt{2} & 2 \\ 3 & \frac{2}{5} \end{bmatrix} \)
• C. \( \begin{bmatrix} 2 & \sqrt{2} & 0 \\ 3 & -2 & \frac{2}{5} \end{bmatrix} \)
• D. \( \begin{bmatrix} 3 & 2 \\ -2 & \frac{2}{5} \\ 0 & 2 \end{bmatrix} \)

Hint:

To find the transpose of a matrix, simply swap its rows with columns. This operation does not change the order of multiplication for square matrices, but be careful when performing operations with non-square matrices.

Answer 1

The Correct Option is C

To find the transpose of a matrix \( A \), we interchange its rows and columns. The transpose is denoted by \( A^T \).

The given matrix is: \[ A = \begin{bmatrix} 2 & \sqrt{2} & 0 \\ 3 & -2 & \frac{2}{5} \end{bmatrix}. \]

The first row of \( A \), \( [2, \sqrt{2}, 0] \), becomes the first column of \( A^T \). The second row, \( [3, -2, \frac{2}{5}] \), becomes the second column of \( A^T \).

Therefore, the transpose of \( A \) is: \[ A^T = \begin{bmatrix} 2 & 3 \\ \sqrt{2} & -2 \\ 0 & \frac{2}{5} \end{bmatrix}. \]

Hence, the correct answer is: \[ \boxed{ A^T = \begin{bmatrix} 2 & 3 \\ \sqrt{2} & -2 \\ 0 & \frac{2}{5} \end{bmatrix}. } \]

Correct Answer:

(C) \( \begin{bmatrix} 2 & 3 \\ \sqrt{2} & -2 \\ 0 & \frac{2}{5} \end{bmatrix} \)