Q.If $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, prove that $A^3 - 6A^2 + 7A + 2I = 0$.

The relation is proved by computing the powers of $A$ directly and substituting. For $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$,

$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix},\qquad A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}.$ …

We must prove $A^3 - 6A^2 + 7A + 2I = O$ for $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, with $I$ the $3\times3$ identity and $O$ the zero matrix. We compute the powers of $A$ directly and substitute.

Step 1 — Compute $A^2 = A\cdot A$

For instance the $(1,1)$ entry is $1(1)+0(0)+2(2)=5$ and the $(3,3)$ entry is $2(2)+0(1)+3(3)=13$.

Step 2 — Compute $A^3 = A^2\cdot A$

For instance the $(1,1)$ entry is $5(1)+0(0)+8(2)=21$.

Step 3 — Form each term …

Method: Proving a matrix polynomial identity equals the zero matrix

To prove a relation such as $A^3-6A^2+7A+2I=O$, substitute the matrix into the polynomial and show every entry of the result is zero — the reliable, syllabus-standard route is direct computation of the powers.

Steps

Step 1: Compute the successive powers

Find $A^2 = A\cdot A$, then $A^3 = A^2\cdot A$ by matrix multiplication.

Step 2: Form every term to the same order

Compute $-6A^2$ and $7A$, and write the constant as $2I$ (2 on the diagonal, 0 elsewhere).

Step 3: Add and confirm the zero matrix …

Common Mistakes

Mistake 1: Computing $A^3$ as the entry-wise cube

Why it's wrong: powers of a matrix are repeated matrix multiplication ($A^3=A^2\cdot A$), never cubing individual entries. Correct approach: compute $A^2$ first, then multiply by $A$ again.

Mistake 2: Writing the constant term as $2$ instead of $2I$ …