Q.For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$.

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

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

Step 2: By Cayley-Hamilton, $A$ satisfies $A^2 - 4A + I = O$. …

The problem asks for numbers $a$ and $b$ such that $A^2 + aA + bI = O$, where $O$ is the zero matrix. This is exactly the form of a matrix satisfying its own characteristic equation — the Cayley-Hamilton theorem tells us that such $a$ and $b$ exist and are uniquely determined by the eigenvalues of $A$.

Why Cayley-Hamilton works here: For any $2 \times 2$ matrix $A$, its characteristic polynomial is $\lambda^2 - (\text{tr }A)\lambda + \det A$. The theorem states that $A^2 - (\text{tr }A)A + (\det A)I = 0$. So if we match $A^2 + aA + bI = 0$, we get $a = -\text{tr }A$ and $b = \det A$.

Let’s compute.

1. Find the trace and determinant.

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

   $\text{tr }A = 3 + 1 = 4$,

   $\det A = (3)(1) - (2)(1) = 3 - 2 = 1$.

2. Write the characteristic equation.

   The characteristic polynomial is $\lambda^2 - 4\lambda + 1 = 0$.

   By Cayley-Hamilton, $A^2 - 4A + I = 0$.

3. Match to the given form.

   We have $A^2 + aA + bI = 0$. Compare with $A^2 - 4A + I = 0$:

   $a$ must be $-4$, and $b$ must be $1$. …

Method: Finding Unknown Coefficients via the Cayley-Hamilton Theorem

This method finds the coefficients $a,b$ in a matrix polynomial equation like $A^2+aA+bI=O$ by matching it against the matrix's own characteristic equation — no need to guess-and-check numbers.

Steps

Step 1: Compute the trace and determinant of $A$

Step 2: Write down the characteristic equation

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

Step 3: Apply Cayley-Hamilton to get the matrix identity $A$ satisfies …

Common Mistakes

Mistake 1: Writing $a=+\operatorname{tr}A$ instead of $a=-\operatorname{tr}A$

Why it's wrong: the characteristic equation is $\lambda^2-(\operatorname{tr}A)\lambda+\det A=0$, so matching it against the target form $A^2+aA+bI=O$ carries a sign flip: $a=-\operatorname{tr}A$, not $+\operatorname{tr}A$. Here $\operatorname{tr}A=4$, so the correct value is $a=-4$, and writing $a=4$ is a very common slip. Correct approach: write the characteristic equation out in full first ($\lambda^2-4\lambda+1=0$, giving $A^2-4A+I=O$), then read $a$ and $b$ off by direct comparison with the given form, rather than trying to recall the sign rule from memory.

Mistake 2: Skipping the verification step …