Question

Let \(A=\begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}\) and \(A^{-1}=xA+yI\), then the pair \((x,y)=\cdots\)

• A. \(\left(\frac{1}{11},\frac{2}{11}\right)\)
• B. \(\left(\frac{1}{11},-\frac{2}{11}\right)\)
• C. \(\left(-\frac{1}{11},-\frac{2}{11}\right)\)
• D. \(\left(-\frac{1}{11},\frac{2}{11}\right)\)

Hint:

Use the Cayley-Hamilton theorem for A to write the inverse in terms of A and the identity matrix.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a 2 by 2 matrix, the Cayley-Hamilton theorem says \(A^{2}-(\text{tr}A)A+|A|I=0\).

Step 2: Find trace and determinant
\(\text{tr}A=1+1=2\) and \(|A|=1\cdot 1-2\cdot(-5)=11\).
So \(A^{2}-2A+11I=0\).

Step 3: Multiply by the inverse
Multiplying by \(A^{-1}\): \(A-2I+11A^{-1}=0\), so \(A^{-1}=\frac{1}{11}(2I-A)=-\frac{1}{11}A+\frac{2}{11}I\).

Final Answer:
x = -1/11 and y = 2/11. \[ \boxed{\text{(D) }\left(-\dfrac{1}{11},\dfrac{2}{11}\right)} \]