Name: Finding Inverse by Elementary Transformations
Uploaded: 2019-07-19
Duration: 12 min 5 s
Description: In this video, we will learn to find the inverse of matrices using elementary transformations of matrices. We will learn how to find the inverse of a matrix,...

Question

Show that the matrix $A = [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}]$ , satisfies the equation $x^{2} + 4 x - 42 = 0$ , and hence find $A^{- 1}$

Accepted Answer

Given: $A = [\begin{array}{l} - 8 & 5 \\ 2 & 4 \end{array}]$

To show: Matrix A satisfies the equation $x^{2} + 4 x - 42 = 0$

If Matrix $A$ , satisfies the given equation then $A^{2} + 4 A - 42 = 0$

Firstly, we find the $A^{2}$

$A^{2} = A . A = [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}]$

$= [\begin{matrix} 64 + 10 & - 40 + 20 \\ - 16 + 8 & 10 + 16 \end{matrix}]$

$= [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}]$

Taking $L H S$ , of the given equation.i.e. $A^{2} + 4 A - 42$

$= [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] + 4 [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] - 42 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$= [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] + [\begin{matrix} - 32 & 20 \\ 8 & 16 \end{matrix}] - [\begin{matrix} 42 & 0 \\ 0 & 42 \end{matrix}]$

$= [\begin{matrix} 74 - 32 & - 20 + 20 \\ - 8 + 8 & 26 + 16 \end{matrix}] - [\begin{matrix} 42 & 0 \\ 0 & 42 \end{matrix}]$

$= [\begin{matrix} 42 & 0 \\ 0 & 42 \end{matrix}] - [\begin{matrix} 42 & 0 \\ 0 & 42 \end{matrix}]$

$= [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}]$

$= 0$

$= R H S$

$∴$ $L H S = R H S$

Hence, matrix A satisfies the given equation $x^{2} + 4 x - 42 = 0$

Now, we have to find $A^{- 1}$

Finding $A^{- 1}$ , using given equation $A^{2} + 4 A - 42 = O$

Post multiplying by $A^{- 1}$ , both sides, we get

$(A^{2} + 4 A - 42) A^{- 1} = {OA}^{- 1}$

$\Rightarrow A^{2} \cdot A^{- 1} + 4 A \cdot A^{- 1} - 42 \cdot A^{- 1} = O [{OA}^{- 1} = O]$

$\Rightarrow A . ({AA}^{- 1}) + 4 I - 42 A^{- 1} = O [{AA}^{- 1} = I]$

$\Rightarrow A (I) + 4 I - 42 A^{- 1} = O$

$\Rightarrow A + 4 I - 42 A^{- 1} = O$

$\Rightarrow A + 4 I - O = 42 A^{- 1}$

$\Rightarrow A^{- 1} = \frac{1}{42} (A + 4 I)$

$\Rightarrow A^{- 1} = \frac{1}{42} \{[\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] + 4 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]\}$

$\Rightarrow A^{- 1} = \frac{1}{42} \{[\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] + [\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}]\}$

$\Rightarrow A^{- 1} = \frac{1}{42} \{\begin{matrix} - 8 + 4 & 5 + 0 \\ 2 + 0 & 4 + 4 \end{matrix}\}$

$\Rightarrow A^{- 1} = \frac{1}{42} [\begin{matrix} - 4 & 5 \\ 2 & 8 \end{matrix}]$

R S Aggarwal Solutions for Exercise 1: EXERCISE

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE

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