Q.Find $A^2 - 5A + 6I$, if $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$.

Concept: Matrix Polynomial Evaluation — substitute the matrix into the polynomial, compute powers and scalar multiples, then combine.

First, compute $A^2$:

Now compute $5A$ and $6I$:

Finally, combine: …

Here $A$ acts like a variable, but with matrix multiplication and addition. The constant term $6I$ is $6$ times the $3\times 3$ identity, which keeps every term a $3\times 3$ matrix.

Step 1: Compute $A^2 = A\cdot A$

Each entry $(i,j)$ is row $i$ of $A$ dotted with column $j$ of $A$:

• $(1,1)=2\cdot2+0\cdot2+1\cdot1=5$
• $(1,2)=2\cdot0+0\cdot1+1\cdot(-1)=-1$
• $(1,3)=2\cdot1+0\cdot3+1\cdot0=2$
• $(2,1)=2\cdot2+1\cdot2+3\cdot1=9$
• $(2,2)=2\cdot0+1\cdot1+3\cdot(-1)=-2$
• $(2,3)=2\cdot1+1\cdot3+3\cdot0=5$
• $(3,1)=1\cdot2+(-1)\cdot2+0\cdot1=0$
• $(3,2)=1\cdot0+(-1)\cdot1+0\cdot(-1)=-1$
• $(3,3)=1\cdot1+(-1)\cdot3+0\cdot0=-2$

Step 2: Compute $-5A$ and $6I$

$-5A = \begin{bmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{bmatrix},\qquad 6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$ …

Method: Evaluating a matrix polynomial $p(A)$

To compute an expression like $A^2-5A+6I$, treat $A$ as the variable but replace the constant term with that constant times the identity matrix, then add the terms entry-wise.

Steps

Step 1: Compute the required powers of $A$

Find $A^2 = A\cdot A$ by matrix multiplication (not by squaring individual entries), and any higher powers needed.

Step 2: Form each term to the same order …

Common Mistakes

Mistake 1: Squaring entries instead of multiplying the matrix by itself

Why it's wrong: $A^2$ means $A\cdot A$ via the row-by-column rule, not squaring each entry; the two give completely different matrices. Correct approach: perform the full matrix multiplication.

Mistake 2: Adding the bare constant 6 instead of $6I$ …