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Equality of Matrices

If A=[ 2 -1...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$ , verify that $A^{3} - {6 A}^{2} + 9 A - 4 I = 0$ . Hence find $A^{- 1}$

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$

To Prove : $A^{3} - 6 A^{2} + 9 A - 4 I = O$

Consider, $A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 4 + 1 + 1 & - 2 - 2 - 1 & 2 + 1 + 2 - 2 - 2 - 1 & 1 + 4 + 1 & - 1 - 2 - 2 2 + 1 + 2 & - 1 - 2 - 2 & 1 + 1 + 4 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 6 & - 5 & 5 - 5 & 6 & - 5 5 & - 5 & 6 \end{matrix} ⎤ ⎥ ⎦$

Now, $A^{3} = A^{2} A$
$= ⎡ ⎢ ⎣ \begin{matrix} 6 & - 5 & 5 - 5 & 6 & - 5 5 & - 5 & 6 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 12 + 5 + 5 & - 6 - 10 - 5 & 6 + 5 + 10 - 10 - 6 - 5 & 5 + 12 + 5 & - 5 - 6 - 10 10 + 5 + 6 & - 5 - 10 - 6 & 5 + 5 + 12 \end{matrix} ⎤ ⎥ ⎦$

$A^{3} = ⎡ ⎢ ⎣ \begin{matrix} 22 & - 21 & 21 - 21 & 22 & - 21 21 & - 21 & 22 \end{matrix} ⎤ ⎥ ⎦$

Consider, $A^{3} - 6 A^{2} + 9 A - 4 I$

$= ⎡ ⎢ ⎣ \begin{matrix} 22 & - 21 & 21 - 21 & 22 & - 21 21 & - 21 & 22 \end{matrix} ⎤ ⎥ ⎦ - 6 ⎡ ⎢ ⎣ \begin{matrix} 6 & - 5 & 5 - 5 & 6 & - 5 5 & - 5 & 6 \end{matrix} ⎤ ⎥ ⎦ + 9 ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦ - 4 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 22 & - 21 & 21 - 21 & 22 & - 21 21 & - 21 & 22 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 36 & - 30 & 30 - 30 & 36 & - 30 30 & - 30 & 36 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 18 & - 9 & 9 - 9 & 18 & - 9 9 & - 9 & 18 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 0 0 & 4 & 0 0 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 22 - 36 + 18 - 4 & - 21 + 30 - 9 & 21 - 30 + 9 - 21 + 30 - 9 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 21 - 30 + 9 & - 21 + 30 - 9 & 22 - 36 + 18 - 4 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{3} - 6 A^{2} + 9 A - 4 I = O$ .....(1)

$A^{- 1}$ :
Multiplying eqn (1) by $A^{- 1}$ , we get
$A^{3} A^{- 1} - 6 A^{2} A^{- 1} + 9 A A^{- 1} - 4 I A^{- 1} = O A^{- 1}$
$\Rightarrow A^{2} - 6 A + 9 I - 4 A^{- 1} = O$
$\Rightarrow 4 A^{- 1} = A^{2} - 6 A + 9 I$
Substituting the values in above eqn
$\Rightarrow 4 A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 6 & - 5 & 5 - 5 & 6 & - 5 5 & - 5 & 6 \end{matrix} ⎤ ⎥ ⎦ - 6 ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦ + 9 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 4 A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 6 & - 5 & 5 - 5 & 6 & - 5 5 & - 5 & 6 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 12 & - 6 & 6 - 6 & 12 & - 6 6 & - 6 & 12 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 9 & 0 & 0 0 & 9 & 0 0 & 0 & 9 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 4 A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 6 - 12 + 9 & - 5 + 6 & 5 - 6 - 5 + 6 & 6 - 12 + 9 & - 5 + 6 5 - 6 & - 5 + 6 & 6 - 12 + 9 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A^{- 1} = \frac{1}{4} ⎡ ⎢ ⎣ \begin{matrix} 3 & 1 & - 1 1 & 3 & 1 - 1 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦$

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If verify that A³ − 6A² + 9A − 4I = O and hence find A⁻¹

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