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

Concept:
Every square matrix obeys its own characteristic equation. This is called the Cayley-Hamilton theorem. We can use it to write the inverse of the matrix in terms of powers of the matrix itself, so no actual matrix inversion is needed.

Step 1:
The eigenvalues are 1, 2 and 3. The characteristic equation has these as its roots, so it is \((\lambda-1)(\lambda-2)(\lambda-3)=0\). Expanding this gives \(\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0\).

Step 2:
By Cayley-Hamilton, the matrix A satisfies the same equation: \[A^3 - 6A^2 + 11A - 6I = 0.\]

Step 3:
Move the constant term to the other side: \(A^3 - 6A^2 + 11A = 6I\). Now multiply every term by \(A^{-1}\): \[A^2 - 6A + 11I = 6A^{-1}.\]

Step 4:
Divide both sides by 6 to get the inverse on its own: \[A^{-1} = \frac{1}{6}\left(A^2 - 6A + 11I\right).\]

Answer: Option (2) — \(\dfrac{1}{6}(A^2 - 6A + 11I)\).