Question

If \( S = \begin{bmatrix} 0 & 1 \\ 0 & 1 \\ 1 & 0 \end{bmatrix} \) and \( A = \frac{1}{2} \begin{bmatrix} b + c & c - a \\ a + b & b - c \end{bmatrix} \), then \( SAS^{-1} \) is:

• A. \( \begin{bmatrix} a & 0 & 0 \\0 & b & 0 \\0 & 0 & c \end{bmatrix} \)
• B. \( \begin{bmatrix} a & 0 & 0 \\0 & 0 & b \\0 & 0 & c \end{bmatrix} \)
• C. \( \begin{bmatrix} a & 0 & 0 \\0 & b & 0 \\0 & 0 & 0 \end{bmatrix} \)
• D. \( \begin{bmatrix} a & b & c \\b & c & a \\c & a & b \end{bmatrix} \)

Hint:

Remember to carefully compute the matrix multiplication and check the order of the matrices.

Answer 1

The Correct Option is A

We are given the matrix equation \( S \) and \( A \), and we are asked to find \( SAS^{-1} \).
Step 1: Write out the matrices for \( S \) and \( A \). \[ S = \begin{bmatrix} 0 & 1 \\ 0 & 1 \\ 1 & 0 \end{bmatrix} \] \[ A = \frac{1}{2} \begin{bmatrix} b + c & c - a \\ a + b & b - c \end{bmatrix} \]
Step 2: Calculate \( SAS^{-1} \). The inverse of \( S \), denoted \( S^{-1} \), can be calculated. In this case, \( S^{-1} \) is: \[ S^{-1} = \begin{bmatrix} 0 & 1 \\ 0 & 1 \\ 1 & 0 \end{bmatrix} \] Multiplying the matrices \( SAS^{-1} \), we get: \[ SAS^{-1} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \]
Thus, the correct answer is option (1).