Question

If \( A = \begin{pmatrix} x & y & y \\ y & x & y \\ y & y & x \end{pmatrix} \) and \( 5A^{-1} = \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix} \), then \( A^2 - 4A \) is:

• A. \( 5A^{-1} \)
• B. \( 5I \)
• C. \( 0 \)
• D. \( I \)

Hint:

Matrix operations such as inversion and multiplication can help simplify problems in linear algebra.

Answer 1

The Correct Option is B

Step 1: We are given that

\[ 5A^{-1} = \begin{pmatrix} -3 & 2 \\ 2 & -3 \end{pmatrix} \]

To find \( A^{-1} \), divide the matrix by 5:

\[ A^{-1} = \begin{pmatrix} -\frac{3}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{pmatrix} \]

Step 2: Now, multiply \( A \) by \( A^{-1} \) to obtain the identity matrix \( I \):

\[ A \cdot A^{-1} = I \]

Step 3: Next, compute \( A^2 - 4A \). The result of this calculation is:

\[ A^2 - 4A = 5I \]

Therefore, the correct answer is \( 5I \).