Question:
If \(A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\) and \(A^{-1}=xA+yI\), then the values of \(x\) and \(y\) are...
If \(A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\) and \(A^{-1}=xA+yI\), then the values of \(x\) and \(y\) are...
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Find a quadratic relation satisfied by A and multiply it by the inverse of A.
Updated On: Nov 12, 2026
- \(x=\frac{1}{5},y=\frac{-4}{5}\)
- \(x=\frac{-1}{5},y=\frac{4}{5}\)
- \(x=\frac{1}{5},y=\frac{4}{5}\)
- \(x=1,y=-4\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
We need a relation among \(A^2\), \(A\) and \(I\), then multiply by \(A^{-1}\).
Step 2: Compute A squared
\(A^2=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}\).
Notice that \(A^2=4A+5I\), since \(4A+5I\) has diagonal \(4+5=9\) and off-diagonal entries \(8\).
Step 3: Multiply by the inverse
Multiply \(A^2=4A+5I\) by \(A^{-1}\):
\(A=4I+5A^{-1}\), so \(A^{-1}=\dfrac15A-\dfrac45I\).
Step 4: Compare with xA + yI
So \(x=\dfrac15\) and \(y=-\dfrac45\), which matches the first option.
Final Answer:
The values are \(x=\dfrac15\) and \(y=-\dfrac45\). \[ \boxed{\text{(A) }x=\dfrac{1}{5},\,y=\dfrac{-4}{5}} \]
We need a relation among \(A^2\), \(A\) and \(I\), then multiply by \(A^{-1}\).
Step 2: Compute A squared
\(A^2=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}\).
Notice that \(A^2=4A+5I\), since \(4A+5I\) has diagonal \(4+5=9\) and off-diagonal entries \(8\).
Step 3: Multiply by the inverse
Multiply \(A^2=4A+5I\) by \(A^{-1}\):
\(A=4I+5A^{-1}\), so \(A^{-1}=\dfrac15A-\dfrac45I\).
Step 4: Compare with xA + yI
So \(x=\dfrac15\) and \(y=-\dfrac45\), which matches the first option.
Final Answer:
The values are \(x=\dfrac15\) and \(y=-\dfrac45\). \[ \boxed{\text{(A) }x=\dfrac{1}{5},\,y=\dfrac{-4}{5}} \]
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