Question

If \(A=\begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}\), then \(A^{-1}=\)

• A. \(\frac{1}{6}(A-4I)\)
• B. \(\frac{1}{6}(A+5I)\)
• C. \(\frac{1}{6}(A+4I)\)
• D. \(\frac{1}{6}(A-5I)\)

Hint:

Write the characteristic equation of the matrix and multiply by the inverse to express it in terms of the matrix itself.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
By the Cayley-Hamilton theorem, a matrix satisfies its own characteristic equation \(A^{2}-(\text{trace})A+(\det A)I=0\).

Step 2: Form the equation
For \(A=\begin{bmatrix}4&5\\2&1\end{bmatrix}\), the trace is \(5\) and \(|A|=4-10=-6\).
So \(A^{2}-5A-6I=0\).

Step 3: Find the inverse
Multiply by \(A^{-1}\): \(A-5I-6A^{-1}=0\), which gives \(A^{-1}=\frac{1}{6}(A-5I)\).

Final Answer:
The inverse is \(\frac{1}{6}(A-5I)\). \[ \boxed{\text{(D) }\dfrac{1}{6}(A-5I)} \]