Let A = \left[\right. 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \left]\right. , then the correct statement is

Engineering

Mathematics

Inverse of a Matrix

Question

Let $A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}]$ , then the correct statement is

A^–1 = $\frac{1}{5}$ (A – 4I₃)

A² – 4A – 5I₃ = 0

A³ is not invertible

A² is invertible

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$A^{2} = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] = [\begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix}];$

Also det A = 1(1 – 4) – 2(2 – 4) + 2(4 – 2)

= – 3 + 4 + 4 = 5

$∴$ det A $\neq$ 0 ⇒ A is invertible

We have A² – 4A – 5I₃

$= [\begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix}] - 4 [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] - 5 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = 0$

⇒ 5I₃ = A² – 4A = A(A – 4I₃)

⇒ I₃ = A $[\frac{1}{5} (A - 4 I_{3})]$

⇒ A^–1 = $\frac{1}{5} (A - 4 I_{3})$

Note that |A| = 5. Since |A³| = |A|³ = 5³ $\neq$ 0, A³ is invertible

Similarly, A² is invertible.