Q.Show that A = [[1, 2, 2], [2, 1, 2], [2, 2, 1]] matrix satisfies the equation A² - 4A - 5I₃ = 0. Hence find A⁻¹. [I₃ = [[1,0,0],[0,1,0],[0,0,1]]] OR If A = [[1, -1, 0], [-1, 2, 1], [0, 1, 1]] and B = [[1, 1, -1], [0, 1, -1], [0, 0, 1]] then show that BᵀAB is a diagonal matrix.

Compute $A^2$ directly, verify it equals $4A+5I_3$, then use that matrix equation (the Cayley–Hamilton-style trick) to solve for $A^{-1}$ without doing a full cofactor expansion.

Given $A=\begin{pmatrix}1&2&2\\2&1&2\\2&2&1\end{pmatrix}$.

Step 1 — compute $A^2 = A\cdot A$:

Row 1: $(1\cdot1+2\cdot2+2\cdot2,\ 1\cdot2+2\cdot1+2\cdot2,\ 1\cdot2+2\cdot2+2\cdot1) = (9,8,8)$

Row 2: $(2\cdot1+1\cdot2+2\cdot2,\ 2\cdot2+1\cdot1+2\cdot2,\ 2\cdot2+1\cdot2+2\cdot1) = (8,9,8)$

Row 3: $(2\cdot1+2\cdot2+1\cdot2,\ 2\cdot2+2\cdot1+1\cdot2,\ 2\cdot2+2\cdot2+1\cdot1) = (8,8,9)$

$A^2 = \begin{pmatrix}9&8&8\\8&9&8\\8&8&9\end{pmatrix}$

Step 2 — compute $4A+5I_3$:

$4A = \begin{pmatrix}4&8&8\\8&4&8\\8&8&4\end{pmatrix}$, $5I_3=\begin{pmatrix}5&0&0\\0&5&0\\0&0&5\end{pmatrix}$

$4A+5I_3 = \begin{pmatrix}9&8&8\\8&9&8\\8&8&9\end{pmatrix} = A^2$

So $A^2 = 4A+5I_3$, i.e. $A^2-4A-5I_3=0$, as required.

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