If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that \[A^2 - 5A + 7I = O\]. Hence, find A−1.

\[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\]\[ \Rightarrow A^2 = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} = \begin{bmatrix}9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4\end{bmatrix} = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix}\]and\[ A^2 - 5A + 7I = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix} - \begin{bmatrix}15 & 5 \\ - 5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}\]\[ \Rightarrow A^2 - 5A + 7I = \begin{bmatrix}8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = O\]Now, \[ A^2 - 5A + 7I = 0\]\[ \Rightarrow A^2 - 5A = - 7I\]\[ \Rightarrow A^{- 1} A^2 - 5 A^{- 1} A = - 7 A^{- 1} I \left[\text{ Pre - multiplying both sides by }A^{- 1} \right]\]\[ \Rightarrow A - 5I = - 7 A^{- 1} \]\[ \Rightarrow A^{- 1} = - \frac{1}{7}\left( A - 5I \right)\]\[ \Rightarrow A^{- 1} = - \frac{1}{7}\left\{ \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} - \begin{bmatrix}5 & 0 \\ 0 & 5\end{bmatrix} \right\} = \frac{1}{7}\begin{bmatrix}2 & - 1 \\ 1 & 3\end{bmatrix}\]

If a = [ 3 1 − 1 2 ] , Show that a 2 − 5 a + 7 I = O . Hence, Find A−1. - Mathematics

प्रश्न

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$ , show that

A^{2} - 5 A + 7 I = O

. Hence, find A⁻¹.

उत्तर

$A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$
$\Rightarrow A^{2} = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] = [\begin{matrix} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{matrix}] = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}]$
and
$A^{2} - 5 A + 7 I = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - [\begin{matrix} 15 & 5 \\ - 5 & 10 \end{matrix}] + [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}]$
$\Rightarrow A^{2} - 5 A + 7 I = [\begin{matrix} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = O$
Now,
$A^{2} - 5 A + 7 I = 0$
$\Rightarrow A^{2} - 5 A = - 7 I$
$\Rightarrow A^{- 1} A^{2} - 5 A^{- 1} A = - 7 A^{- 1} I [Pre - multiplying both sides by A^{- 1}]$
$\Rightarrow A - 5 I = - 7 A^{- 1}$
$\Rightarrow A^{- 1} = - \frac{1}{7} (A - 5 I)$
$\Rightarrow A^{- 1} = - \frac{1}{7} {[\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] - [\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}]} = \frac{1}{7} [\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}]$

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Inverse of Matrix - Inverse of a Square Matrix by the Adjoint Method

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Q 19 Q 18 Q 20

अध्याय 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [पृष्ठ २४]

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अध्याय 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 19 | पृष्ठ २४

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If a = [ 3 1 − 1 2 ] , Show that a 2 − 5 a + 7 I = O . Hence, Find A−1. - Mathematics

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