Question

The Eigenvalues of \(3\times 3\) real matrix A are 1, 2, 3 then \(A^{-1} =\)

• A. \(A^2 - 6A + 11I\)
• B. \(\dfrac{1}{6}(A^2 - 6A + 11I)\)
• C. \(\dfrac{1}{6}(A^3 - 6I)\)
• D. \(A^3 - 6I\)

Hint:

Every square matrix obeys its own characteristic equation. This is called the Cayley-Hamilton theorem.

Answer 1

The Correct Option is B

Step 1: Note what the eigenvalues give us.
The eigenvalues of A are 1, 2 and 3. The characteristic polynomial of a matrix is just \((\lambda-1)(\lambda-2)(\lambda-3)\), because the roots of that polynomial are exactly the eigenvalues.

Step 2: Expand the polynomial.
Multiply it out: \((\lambda-1)(\lambda-2)(\lambda-3)=\lambda^3-6\lambda^2+11\lambda-6\).

Step 3: Use Cayley-Hamilton.
The theorem says the matrix satisfies its own characteristic equation. So replace \(\lambda\) by A: \[A^3-6A^2+11A-6I=0.\]
Step 4: Pull out a factor of A.
Move the constant across and group the rest: \[A(A^2-6A+11I)=6I.\]
Step 5: Divide by A using the inverse.
Multiply both sides by \(A^{-1}\) and then by \(\tfrac16\): \[A^{-1}=\frac{1}{6}\left(A^2-6A+11I\right).\]
Step 6: Pick the matching option.
This is exactly option 2, and the inverse exists since no eigenvalue is zero.
\[ \boxed{A^{-1}=\tfrac{1}{6}\left(A^2-6A+11I\right)} \]