Q.If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$, then show that $A^3 - 23A - 40I = O$.

To check a matrix polynomial relation we simply compute the required powers of $A$ by direct multiplication and substitute — no special theorem is needed. Here $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$.

First, $A^2 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}$, and then $A^3 = A^2 A = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}$. …

We are given $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$ and must show that the matrix expression $A^3 - 23A - 40I$ equals the zero matrix $O$. Here $I$ is the $3\times3$ identity matrix, and the constant term becomes $40I$ because a plain number cannot be added to a matrix. We do this by honest, direct computation — find $A^2$, then $A^3$, then substitute.

Step 1 — Compute $A^2 = A\cdot A$

Each entry is (row of $A$) $\cdot$ (column of $A$):

For instance the $(1,1)$ entry is $1(1)+2(3)+3(4)=19$, and the $(2,2)$ entry is $3(2)+(-2)(-2)+1(2)=12$.

Step 2 — Compute $A^3 = A^2\cdot A$

For instance the $(1,1)$ entry is $19(1)+4(3)+8(4)=63$.

Step 3 — Substitute into $A^3 - 23A - 40I$ …

Method: Cayley–Hamilton verification for a $3\times3$ matrix

Use this to show a $3\times3$ matrix satisfies $A^3 + pA^2 + qA + rI = O$ via its characteristic polynomial.

Steps

Step 1: Build the characteristic polynomial.

For a $3\times3$ matrix it is

where $M$ is the sum of the three principal $2\times2$ minors.

Step 2: Apply Cayley–Hamilton. …

Common Mistakes

Mistake 1: Wrong sign or value for the minor-sum $M$ or $\det A$.

Why it's wrong: the coefficients depend on $\operatorname{tr}A$, $M$ (sum of principal $2\times2$ minors) and $\det A$; a slip gives the wrong polynomial. Correct approach: compute each principal minor and the determinant carefully — here $M = -23$, $\det A = 40$.

Mistake 2: Treating the constant term as a scalar, not $rI$. …