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Equality of Matrices

For the matri...

Question

For the matrix $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦$ , show that $A^{3} - {6 A}^{2} + 5 A + 11 I = 0$ . Hence find $A^{- 1}$

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Solution

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

To Prove : $A^{3} - 6 A^{2} + 5 A + 11 I = O$

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

$= ⎡ ⎢ ⎣ \begin{matrix} 1 + 1 + 2 & 1 + 2 - 1 & 1 - 3 + 3 1 + 2 - 6 & 1 + 4 + 3 & 1 - 6 - 9 2 - 1 + 6 & 2 - 2 - 3 & 2 + 3 + 9 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 1 - 3 & 8 & - 14 7 & - 3 & 14 \end{matrix} ⎤ ⎥ ⎦$

Now, $A^{3} = A^{2} A$
$= ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 1 - 3 & 8 & - 14 7 & - 3 & 14 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 4 + 2 + 2 & 4 + 4 - 1 & 4 - 6 + 3 - 3 + 8 - 28 & - 3 + 16 + 14 & - 3 - 24 - 42 7 - 3 + 28 & 7 - 6 - 14 & 7 + 9 + 42 \end{matrix} ⎤ ⎥ ⎦$

$A^{3} = ⎡ ⎢ ⎣ \begin{matrix} 8 & 7 & 1 - 23 & 27 & - 69 32 & - 13 & 58 \end{matrix} ⎤ ⎥ ⎦$

Consider, $A^{3} - 6 A^{2} + 5 A + 11 I$

$= ⎡ ⎢ ⎣ \begin{matrix} 8 & 7 & 1 - 23 & 27 & - 69 32 & - 13 & 58 \end{matrix} ⎤ ⎥ ⎦ - 6 ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 1 - 3 & 8 & - 14 7 & - 3 & 14 \end{matrix} ⎤ ⎥ ⎦ + 5 ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦ + 11 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 8 & 7 & 1 - 23 & 27 & - 69 32 & - 13 & 58 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 24 & 12 & 6 - 18 & 48 & - 84 42 & - 18 & 84 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 & 5 5 & 10 & - 15 10 & - 5 & 15 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 11 & 0 & 0 0 & 11 & 0 0 & 0 & 11 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 & 1 - 6 + 5 - 23 + 18 + 5 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 32 - 42 + 10 & - 13 + 18 - 5 & 58 - 84 + 15 + 11 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{3} - 6 A^{2} + 5 A + 11 I = O$ .....(1)

$A^{- 1}$ :
Multiplying eqn (1) by $A^{- 1}$ , we get
$A^{3} A^{- 1} - 6 A^{2} A^{- 1} + 5 A A^{- 1} + 11 I A^{- 1} = O A^{- 1}$
$\Rightarrow A^{2} - 6 A + 5 I + 11 A^{- 1} = O$
$\Rightarrow 11 A^{- 1} = - A^{2} + 6 A - 5 I$
Substituting the values in above eqn
$11 A^{- 1} = - ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 1 - 3 & 8 & - 14 7 & - 3 & 14 \end{matrix} ⎤ ⎥ ⎦ + 6 ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦ - 5 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 2 & - 1 3 & - 8 & 14 - 7 & 3 & - 14 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 6 & 6 & 6 6 & 12 & - 18 12 & - 6 & 18 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 5 & 0 & 0 0 & 5 & 0 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦$

$11 A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} - 4 + 6 - 5 & - 2 + 6 & - 1 + 6 3 + 6 & - 8 + 12 - 5 & 14 - 18 - 7 + 12 & 3 - 6 & - 14 + 18 - 5 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A^{- 1} = \frac{1}{11} ⎡ ⎢ ⎣ \begin{matrix} - 3 & 4 & 5 9 & - 1 & - 4 5 & - 3 & - 1 \end{matrix} ⎤ ⎥ ⎦$

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A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦

A^{3} - 6 A^{2} + 5 A + 11 I = 0

A^{- 1}

For the matrixshow that A³ − 6A² + 5A + 11 I = O. Hence, find A^−1.

Q. For the matrix show that A 3 − 6 A 2 + 5 A + 11 I = O. Hence, find A −1.

For the matrix $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦$ . Show that $A^{3} - 6 A^{2} + 5 A + 11 I = 0$ . Hence, find $A^{- 1}$

Q. For the matrix

A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]

A^{- 3} - 6 A^{2} + 5 A + 11 I_{3} = O

. Hence, find A⁻¹.

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