Question

If \(A=\begin{bmatrix}1 & 2 & 3\\3 & -2 & 1\\4 & 2 & 1\end{bmatrix}\), then show that \(A^3-23A-40I=0\).

Hint:

Use the Cayley-Hamilton theorem: find the characteristic equation of A, then substitute A for x in that equation.

Answer 1

Step 1: Understanding What Is to Be Shown:
We must show the matrix expression \(A^3-23A-40I\) equals the zero matrix, where I is the 3 by 3 identity matrix.
The direct way is to compute \(A^2\), then \(A^3\), and finally combine the terms and check every entry becomes 0.

Step 2: Compute A Squared:
Multiply A by itself row by column.
\[ A^2=AA=\begin{bmatrix}19 & 4 & 8\\1 & 12 & 8\\14 & 6 & 15\end{bmatrix} \]

Step 3: Compute A Cubed:
Multiply the result by A once more.
\[ A^3=A^2A=\begin{bmatrix}63 & 46 & 69\\69 & -6 & 23\\92 & 46 & 63\end{bmatrix} \]

Step 4: Form the Full Expression:
Compute \(23A\) and \(40I\) separately, then subtract both from \(A^3\), entry by entry.
\[ 23A=\begin{bmatrix}23 & 46 & 69\\69 & -46 & 23\\92 & 46 & 23\end{bmatrix}, \qquad 40I=\begin{bmatrix}40 & 0 & 0\\0 & 40 & 0\\0 & 0 & 40\end{bmatrix} \]
\[ A^3-23A-40I=\begin{bmatrix}63-23-40 & 46-46-0 & 69-69-0\\69-69-0 & -6+46-40 & 23-23-0\\92-92-0 & 46-46-0 & 63-23-40\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} \]

Final Answer:
Every entry of the computed expression is 0, so the matrix identity holds and this proves the required result.
\[ \boxed{A^3-23A-40I=O} \]