Question

The eigen values of a \(3\times3\) real matrix \(A\) are \(1,-2,3\). Then

• A. \(A^{-1}=\dfrac{1}{6}(5I+2A-A^2)\)
• B. \(A^{-1}=\dfrac{1}{6}(5I-2A+A^2)\)
• C. \(A^{-1}=\dfrac{1}{6}(5I-2A-A^2)\)
• D. \(A^{-1}=\dfrac{1}{6}(5I+2A+A^2)\)

Hint:

For eigenvalue questions, \[ \boxed{ \text{Characteristic Polynomial} \rightarrow \text{Cayley--Hamilton} \rightarrow A^{-1} } \] Always multiply the Cayley--Hamilton equation by \(A^{-1}\) when inverse is required.

Answer 1

The Correct Option is A

Concept:

Use the Cayley–Hamilton Theorem, which states that every square matrix satisfies its own characteristic equation.

Since the eigenvalues are \(1,\,-2,\;3\), the characteristic polynomial is:

\[ (\lambda-1)(\lambda+2)(\lambda-3) \]

Step 1: Find the characteristic polynomial.

\[ \begin{aligned} (\lambda-1)(\lambda+2) &=\lambda^2+\lambda-2,\\ (\lambda^2+\lambda-2)(\lambda-3) &=\lambda^3-2\lambda^2-5\lambda+6. \end{aligned} \]

Hence, the characteristic equation is:

\[ \lambda^3-2\lambda^2-5\lambda+6=0. \]

Step 2: Apply the Cayley–Hamilton theorem.

Replacing \(\lambda\) by \(A\):

\[ A^3-2A^2-5A+6I=0. \]

Therefore,

\[ A^3-2A^2-5A=-6I. \]

Step 3: Obtain \(A^{-1}\).

Since none of the eigenvalues is zero, \(A^{-1}\) exists.

Multiplying both sides by \(A^{-1}\):

\[ A^2-2A-5I+6A^{-1}=0. \]

Hence,

\[ 6A^{-1}=5I+2A-A^2. \]

Therefore,

\[ A^{-1}=\frac{1}{6}(5I+2A-A^2). \]

Hence, the correct option is (A).