Q.If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 - 5A + 7I = O$. Hence find $A^{-1}$.

The key idea is the Cayley-Hamilton Theorem: every square matrix satisfies its own characteristic equation.

Step 1 – Characteristic equation

For $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,

$\det(A - \lambda I) = \det\begin{bmatrix} 3-\lambda & 1 \\ -1 & 2-\lambda \end{bmatrix} = (3-\lambda)(2-\lambda) + 1 = \lambda^2 - 5\lambda + 7 = 0$.

Step 2 – Apply Cayley-Hamilton

The matrix satisfies its own equation:

$A^2 - 5A + 7I = O$.

Step 3 – Find $A^{-1}$

From $A^2 - 5A + 7I = O$, multiply by $A^{-1}$:

$A - 5I + 7A^{-1} = O \quad\Rightarrow\quad 7A^{-1} = 5I - A$. …

The problem asks two things: first, verify that $A^2 - 5A + 7I$ is the zero matrix; second, use that result to find $A^{-1}$. The connection between these two parts is the Cayley-Hamilton theorem — one of the most elegant results in matrix algebra.

The Core Idea

Every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$, the characteristic equation is $\lambda^2 - (\text{trace})\lambda + \det(A) = 0$. The theorem says that if you replace $\lambda$ by $A$ in that polynomial, you get the zero matrix. That polynomial is exactly $A^2 - 5A + 7I$ here. Once we confirm it, we can rearrange to solve for $A^{-1}$.

Let’s work through it step by step.

1. Compute $A^2$ directly.

We multiply $A$ by itself:

First row, first column: $(3)(3) + (1)(-1) = 9 - 1 = 8$

First row, second column: $(3)(1) + (1)(2) = 3 + 2 = 5$

Second row, first column: $(-1)(3) + (2)(-1) = -3 - 2 = -5$

Second row, second column: $(-1)(1) + (2)(2) = -1 + 4 = 3$

So:

2. Compute $5A$ and $7I$.

3. Form $A^2 - 5A + 7I$ and simplify.

Subtract $5A$ from $A^2$, then add $7I$:

Now add $7I$:

So indeed $A^2 - 5A + 7I = O$. This is the Cayley-Hamilton relation for this matrix.

4. Use the relation to find $A^{-1}$.

We have $A^2 - 5A + 7I = O$. Rearrange:

$A^2 - 5A = -7I$ …

Method: Finding a Matrix Inverse via the Cayley-Hamilton Theorem

This method finds $A^{-1}$ without computing cofactors at all, by using the fact that every square matrix satisfies its own characteristic equation. (This route is beyond the core CBSE syllabus but is a standard JEE Advanced technique for small matrices.)

Steps

Step 1: Find the characteristic equation of $A$

For a $2\times2$ matrix, the characteristic equation is

where $\operatorname{tr}A$ is the sum of the diagonal entries and $\det A$ is the determinant. Compute both first.

Step 2: Apply the Cayley-Hamilton theorem

The theorem states that substituting the matrix $A$ itself for $\lambda$ in its characteristic polynomial gives the zero matrix:

This is exactly the identity the question asks you to show — it follows immediately once the characteristic equation is known, though the direct entry-by-entry check ($A^2$, then combine) is what a board answer should actually show. …

Common Mistakes

Mistake 1: Dropping the minus sign when rearranging $A(A-5I)=-7I$ to solve for $A^{-1}$

Why it's wrong: multiplying both sides by $-\tfrac{1}{7}$ gives $A^{-1}=-\tfrac{1}{7}(A-5I)=\tfrac{1}{7}(5I-A)$ — writing $A^{-1}=\tfrac{1}{7}(A-5I)$ instead (forgetting the sign) produces a matrix that is the negative of the correct inverse, and would fail the check $AA^{-1}=I$. Correct approach: keep the equation in the form $A(\dots)=-7I$ explicit, and carry the negative sign through every step of the rearrangement rather than dropping it partway.

Mistake 2: Quoting the Cayley-Hamilton result without actually computing $A^2$ and substituting numerically …