Step 1: Skip the theorem and verify by direct matrix multiplication instead:
First compute $A^2=A\cdot A$ entry by entry. Row 1 of $A^2$: $[1(1)+2(3)+3(4),\ 1(2)+2(-2)+3(2),\ 1(3)+2(1)+3(1)]=[19,4,8]$. Row 2: $[3(1)+(-2)(3)+1(4),\ 3(2)+(-2)(-2)+1(2),\ 3(3)+(-2)(1)+1(1)]=[1,12,8]$. Row 3: $[4(1)+2(3)+1(4),\ 4(2)+2(-2)+1(2),\ 4(3)+2(1)+1(1)]=[14,6,15]$.
Step 2: So $A^2=\begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix}$. Now compute $A^3=A^2\cdot A$:
Row 1: $[19(1)+4(3)+8(4),\ 19(2)+4(-2)+8(2),\ 19(3)+4(1)+8(1)]=[63,46,69]$. Row 2: $[1(1)+12(3)+8(4),\ 1(2)+12(-2)+8(2),\ 1(3)+12(1)+8(1)]=[69,-6,23]$. Row 3: $[14(1)+6(3)+15(4),\ 14(2)+6(-2)+15(2),\ 14(3)+6(1)+15(1)]=[92,46,63]$.
Step 3: So $A^3=\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix}$. Now compute $23A+40I$:
$23A=\begin{bmatrix}23&46&69\\69&-46&23\\92&46&23\end{bmatrix}$, and $40I=\begin{bmatrix}40&0&0\\0&40&0\\0&0&40\end{bmatrix}$, so $23A+40I=\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix}$.
Step 4: Compare $A^3$ with $23A+40I$:
Both matrices are identical entry for entry, so $A^3-23A-40I$ is the zero matrix.
Final Answer:
Direct computation confirms $A^3-23A-40I=0$.
\[ \boxed{A^3-23A-40I=0} \]