Using elementary transformations- find the inverse of matrix - 2 1- 1 1 - if it exists-

Name: NCERT - Grade 12 - Mathematics - Matrices - Q71
Uploaded: 2022-07-04T10:36:42+05:30
Description: Let A= nbsp; [ 2 amp;1; nbsp; 1 amp;1 ] We know that A=IA ⇒ nbsp; nbsp; nbsp; [ 2 amp;1; nbsp; 1 amp;1 ]= nbsp;[ 1 amp;0; ...

Let A= nbsp; [ 2 amp;1; nbsp; 1 amp;1 ] We know that A=IA ⇒ nbsp; nbsp; nbsp; [ 2 amp;1; nbsp; 1 amp;1 ]= nbsp;[ 1 amp;0; ...

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Standard XII

Mathematics

Elementary Transformations

Using element...

Question

Using elementary transformations, find the inverse of matrix $[\begin{matrix} 2 & 1 1 & 1 \end{matrix}]$ , if it exists.

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Solution

Let $A = [\begin{matrix} 2 & 1 1 & 1 \end{matrix}]$
We know that $A = I A$
$\Rightarrow [\begin{matrix} 2 & 1 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A$
Applying $R_{1} \to R_{1} - R_{2}$
$\Rightarrow [\begin{matrix} 2 - 1 & 1 - 1 1 & 1 \end{matrix}] = [\begin{matrix} 1 - 0 & 0 - 1 0 & 1 \end{matrix}] A$
$\Rightarrow [\begin{matrix} 1 & 0 1 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 0 & 1 \end{matrix}] A$
Applying $R_{2} \to R_{2} - R_{1}$
$\Rightarrow [\begin{matrix} 1 & 0 1 - 1 & 1 - 0 \end{matrix}] = [\begin{matrix} 1 & - 1 0 - 1 & 1 - (- 1) \end{matrix}] A$
$\Rightarrow [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 - 1 & 2 \end{matrix}] A$
$\Rightarrow I = [\begin{matrix} 1 & - 1 - 1 & 2 \end{matrix}] A$
This is similar to $I = A^{- 1} A$
Thus $A^{- 1} = [\begin{matrix} 1 & - 1 - 1 & 2 \end{matrix}]$

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Elementary Transformations

Standard XII Mathematics

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Using elementary transformations, find the inverse of matrix [2111], if it exists.

Let A=[2111] We know that A=IA ⇒[2111]=[1001]A Applying R1→R1−R2 ⇒[2−11−111]=[1−00−101]A ⇒[1011]=[1−101]A Applying R2→R2−R1 ⇒[101−11−0]=[1−10−11−(−1)]A ⇒[1001]=[1−1−12]A ⇒I=[1−1−12]A This is similar to I=A−1A Thus A−1=[1−1−12]

Using elementary transformations, find the inverse of matrix $[\begin{matrix} 2 & 1 1 & 1 \end{matrix}]$ , if it exists.