Q.Find $A^{-1}$ if $A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$ and show that $A^{-1} = \dfrac{A^2 - 3I}{2}$.

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

Step 1: Find the characteristic polynomial of $A$.

$|A - \lambda I| = \begin{vmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \end{vmatrix} = -\lambda^3 + 3\lambda + 2$.

Step 2: By Cayley-Hamilton, $A$ satisfies

$-A^3 + 3A + 2I = 0 \quad\Rightarrow\quad A^3 = 3A + 2I$. …

The problem asks us to find $A^{-1}$ and then verify a given expression. The direct approach — computing the inverse via cofactors or row reduction — would work, but the problem is clearly designed to illustrate a deeper idea: the Cayley-Hamilton theorem. This theorem says that every square matrix satisfies its own characteristic equation. That means we can replace the matrix $A$ in its characteristic polynomial and get the zero matrix. Once we have that polynomial, we can rearrange it to solve for $A^{-1}$ as a polynomial in $A$ — no messy fractions or cofactor expansions needed.

Let’s walk through it.

1. Find the characteristic polynomial of $A$.

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

we have

Compute the determinant. A neat trick: add all rows to the first row, or notice the pattern. Let’s do it systematically:

Each $2\times2$ determinant:

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

So

Simplify:

Thus

It’s often nicer to multiply by $-1$:

2. Apply the Cayley-Hamilton theorem.

The theorem states that $A$ satisfies its own characteristic equation:

So

We want $A^{-1}$. Multiply both sides of $A^3 - 3A = 2I$ by $A^{-1}$ (which exists because $\det A \neq 0$ — we’ll check later). That gives:

Therefore

That’s exactly the expression we needed to show. So the second part is done — it follows directly from Cayley-Hamilton.

3. Compute $A^2$ explicitly.

Compute entry by entry:

• Row 1, Col 1: $(0)(0) + (1)(1) + (1)(1) = 0 + 1 + 1 = 2$.

• Row 1, Col 2: $(0)(1) + (1)(0) + (1)(1) = 0 + 0 + 1 = 1$.

• Row 1, Col 3: $(0)(1) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1$.

• Row 2, Col 1: $(1)(0) + (0)(1) + (1)(1) = 0 + 0 + 1 = 1$.

• Row 2, Col 2: $(1)(1) + (0)(0) + (1)(1) = 1 + 0 + 1 = 2$.

• Row 2, Col 3: $(1)(1) + (0)(1) + (1)(0) = 1 + 0 + 0 = 1$.

• Row 3, Col 1: $(1)(0) + (1)(1) + (0)(1) = 0 + 1 + 0 = 1$.

• Row 3, Col 2: $(1)(1) + (1)(0) + (0)(1) = 1 + 0 + 0 = 1$.

• Row 3, Col 3: $(1)(1) + (1)(1) + (0)(0) = 1 + 1 + 0 = 2$.

So …

Method: Finding a Matrix Inverse via the Adjoint (and Verifying a Given Identity Directly)

Use this method whenever you're asked to find $A^{-1}$ for a $3\times3$ matrix and then confirm it matches some other expression built from $A$ (such as $\tfrac12(A^2-3I)$). The CBSE Class 12 syllabus route is the adjoint method — not the Cayley–Hamilton theorem, which is a useful shortcut but sits outside the core syllabus (it's a JEE-Advanced-level extension topic).

Steps

Step 1: Confirm $A$ is invertible

Compute $|A|$ by cofactor expansion along any row or column. If $|A|\neq 0$, the inverse exists and can be found.

Step 2: Build the cofactor matrix, then the adjoint

For each entry $a_{ij}$, compute the cofactor $C_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the determinant left after deleting row $i$ and column $j$. Collect all nine cofactors into the cofactor matrix, then transpose it to get the adjoint:

Step 3: Assemble the inverse

Step 4 (Applying to this problem): Verify a claimed identity by direct computation, not by invoking a shortcut theorem …

Common Mistakes

Mistake 1: Sign errors expanding $\det(A-\lambda I)$

Why it's wrong: with a matrix full of $0$s, $1$s and $-\lambda$s, it's easy to drop a minus sign in one of the three $2\times2$ cofactors, producing a wrong characteristic polynomial (e.g. $-\lambda^3-3\lambda+2$ instead of $-\lambda^3+3\lambda+2$) — which then gives a completely wrong Cayley–Hamilton relation and a wrong $A^{-1}$.

Mistake 2: Multiplying the Cayley–Hamilton equation by $A^{-1}$ without checking invertibility first …