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 Cayley-Hamilton (characteristic equation of A) or verify directly by computing A² and A³.

Answer 1

Step 1: Key Formula or Approach:
Use the Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation \(\lambda^3 - (\text{tr }A)\lambda^2 + (\text{sum of principal }2\times2\text{ minors})\lambda - \det(A) = 0\), with \(A\) substituted for \(\lambda\) and \(I\) for the constant term.

Step 2: Finding the trace:
\[ \text{tr}(A) = 1+(-2)+1 = 0 \]

Step 3: Finding the sum of principal \(2\times2\) minors:
\[ M_{11}=\begin{vmatrix}-2&1\\2&1\end{vmatrix}=-2-2=-4,\quad M_{22}=\begin{vmatrix}1&3\\4&1\end{vmatrix}=1-12=-11,\quad M_{33}=\begin{vmatrix}1&2\\3&-2\end{vmatrix}=-2-6=-8 \]
\[ \text{Sum} = -4-11-8 = -23 \]

Step 4: Finding \(\det(A)\):
\[ \det(A) = 1[(-2)(1)-1(2)] - 2[3(1)-1(4)] + 3[3(2)-(-2)(4)] \]
\[ = 1(-2-2) - 2(3-4) + 3(6+8) = -4+2+42 = 40 \]

Step 5: Writing the characteristic equation:
\[ \lambda^3 - (0)\lambda^2 + (-23)\lambda - 40 = 0 \implies \lambda^3 - 23\lambda - 40 = 0 \]

Step 6: Applying Cayley-Hamilton:
By the Cayley-Hamilton theorem, \(A\) itself satisfies this equation:
\[ A^3 - 23A - 40I = 0 \]

Final Answer:
This proves the required identity. \[ \boxed{A^3-23A-40I=0} \]