Q.Show that the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ satisfies the equation $A^2 - 4A + I = O$, where $I$ is $2 \times 2$ identity matrix and $O$ is $2 \times 2$ zero matrix. Using this equation, find $A^{-1}$.

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

Step 1 – Compute $A^2$:

Step 2 – Compute $A^2 - 4A + I$:

Thus $A$ satisfies $A^2 - 4A + I = O$.

Step 3 – Rearrange to find $A^{-1}$:

Multiply both sides by $-1$:

$A(4I - A) = I$ …

The core idea here is the Cayley-Hamilton Theorem: every square matrix satisfies its own characteristic polynomial. For a $2 \times 2$ matrix, the characteristic equation is $\lambda^2 - (\text{trace})\lambda + \det = 0$. If we can show $A$ obeys $A^2 - 4A + I = O$, then we can rearrange that matrix equation to isolate $A^{-1}$ without doing any row reduction.

Let’s walk through it.

1. Verify the given equation directly. Compute $A^2$:

Now compute $4A$:

Then $A^2 - 4A + I$ is:

So indeed $A^2 - 4A + I = O$ holds. This is the matrix version of the characteristic equation $\lambda^2 - 4\lambda + 1 = 0$.

2. Rearrange to find $A^{-1}$. From $A^2 - 4A + I = O$, bring the $I$ term to the other side:

Factor $A$ on the left:

Multiply both sides by $-1$:

$-A(A - 4I) = I \quad \Rightarrow \quad A(4I - A) = I.$ …

Method: Using a Given Matrix-Polynomial Identity to Find the Inverse Algebraically

This method verifies a given quadratic matrix equation by direct computation, then rearranges that same equation algebraically to isolate the inverse — without computing a determinant or adjoint at all.

Steps

Step 1: Compute $A^2$ directly

Multiply $A$ by itself using ordinary matrix multiplication, entry by entry.

Step 2: Substitute into the given equation and simplify

Compute $A^2 - cA + dI$ (using the specific numbers given in the equation) entry by entry and confirm the result is the zero matrix — this verifies the identity holds for this particular $A$.

Step 3: Isolate the identity-matrix term algebraically

Rearrange the verified equation so the constant (identity-matrix) term stands alone on one side:

Step 4: Factor out $A$ from the left side

Step 5: Scale to make the right side the identity matrix

Multiply both sides by $-\tfrac1d$ (valid since $d\neq 0$ is exactly what makes $A$ invertible here):

$A\left(\frac{cI - A}{d}\right) = I$ …

Common Mistakes

Mistake 1: Sign error when moving the identity-matrix term across the equation

Why it's wrong: from $A^2-4A+I=O$, moving $I$ to the other side gives $A^2-4A=-I$ (not $+I$) — dropping the negative sign here breaks every subsequent step, including the final formula for $A^{-1}$. Correct approach: treat $I$ exactly like a number when rearranging — moving it across the equals sign flips its sign, same as any other term.

Mistake 2: Forgetting to multiply both sides by $-1$ before factoring out $A$ …