Q.If $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$. Verify that $A^3 - 6A^2 + 9A - 4I = O$ and hence find $A^{-1}$.

Concept: Cayley-Hamilton Theorem — every square matrix satisfies its own characteristic equation.

Step 1 – Find the characteristic equation.

For $A$, compute $\det(A - \lambda I)$:

Simplifying:

So the characteristic equation is:

Step 2 – Apply Cayley-Hamilton.

$A$ satisfies its own equation:

$A^3 - 6A^2 + 9A - 4I = O$ …

We are given the matrix

The problem asks us to verify that $A^3 - 6A^2 + 9A - 4I = O$ and then use this to find $A^{-1}$.

The key idea is the Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. So if we find the characteristic polynomial of $A$, the theorem guarantees that plugging $A$ into that polynomial gives the zero matrix. The given cubic expression looks suspiciously like a characteristic polynomial — we just need to check.

1. Find the characteristic polynomial of $A$.

The characteristic polynomial is $p(\lambda) = \det(\lambda I - A)$. Compute:

Take the determinant. A neat trick: add all rows to the first row, or notice the pattern. Let's do it directly:

Compute each $2\times2$ determinant:

• First: $(\lambda-2)^2 - 1 = \lambda^2 - 4\lambda + 4 - 1 = \lambda^2 - 4\lambda + 3$.
• Second: $1\cdot(\lambda-2) - 1\cdot(-1) = \lambda - 2 + 1 = \lambda - 1$.
• Third: $1\cdot 1 - (\lambda-2)(-1) = 1 + \lambda - 2 = \lambda - 1$.

So:

Simplify:

Expand $(\lambda-2)(\lambda^2 - 4\lambda + 3)$:

• $\lambda \cdot \lambda^2 = \lambda^3$
• $\lambda \cdot (-4\lambda) = -4\lambda^2$
• $\lambda \cdot 3 = 3\lambda$
• $(-2) \cdot \lambda^2 = -2\lambda^2$
• $(-2) \cdot (-4\lambda) = 8\lambda$
• $(-2) \cdot 3 = -6$

Sum: $\lambda^3 - 6\lambda^2 + 11\lambda - 6$.

Now subtract $2(\lambda-1) = 2\lambda - 2$:

Thus the characteristic polynomial is:

2. Apply Cayley-Hamilton theorem.

The theorem states that $p(A) = O$. That is:

This is exactly what we needed to verify. So the given equation holds — no need to compute powers of $A$ explicitly.

3. Use the polynomial to find $A^{-1}$.

We have:

We want $A^{-1}$. Multiply the entire equation by $A^{-1}$ (assuming $A$ is invertible — we'll check later):

Now solve for $A^{-1}$:

So:

We need $A^2$ to compute this explicitly.

4. Compute $A^2$.

Compute entry by entry:

• Row 1, Col 1: $2\cdot2 + (-1)(-1) + 1\cdot1 = 4 + 1 + 1 = 6$.

• Row 1, Col 2: $2\cdot(-1) + (-1)\cdot2 + 1\cdot(-1) = -2 -2 -1 = -5$.

• Row 1, Col 3: $2\cdot1 + (-1)(-1) + 1\cdot2 = 2 + 1 + 2 = 5$.

• Row 2, Col 1: $(-1)\cdot2 + 2\cdot(-1) + (-1)\cdot1 = -2 -2 -1 = -5$. …

Method: Cayley-Hamilton for a 3×3 Matrix, Then Solving for the Inverse

Same technique as any Cayley-Hamilton "verify and hence find the inverse" question: derive the characteristic cubic, confirm the matrix satisfies it, then rearrange for $A^{-1}$.

Steps

Step 1: Compute $\operatorname{tr}A$, the sum of principal $2\times2$ minors, and $\det A$

These three numbers are exactly the coefficients of the characteristic cubic

Step 2: Invoke Cayley-Hamilton to get the matrix identity

Check that the coefficients computed in Step 1 match the ones printed in the question — this is the verification the question is asking for.

Step 3: Isolate $A^{-1}$ by factoring …

Common Mistakes

Mistake 1: An arithmetic error computing $A^2$

Why it's wrong: this matrix has repeated entries and several negative values, so a dot-product entry is easy to mis-add — and the whole cubic identity, plus the final inverse, depends on $A^2$ being exactly right. Correct approach: compute and write out each of the nine entries of $A^2$ individually as a labelled dot product, rather than combining several in your head.

Mistake 2: Mismatching the coefficients of the characteristic cubic against the given identity …