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:
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:
Show Hint
- \( 5A^{-1} \)
- \( 5I \)
- \( 0 \)
- \( I \)
The Correct Option is B
Solution and Explanation
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 \( 5I \).
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