Using elementary transformations- find the inverse of matrix - 1 3- 2 7 - if it exists-

Name: NCERT - Grade 12 - Mathematics - Matrices - Q72
Uploaded: 2022-07-04T10:36:42+05:30
Description: Let A = nbsp; [ 1 amp; 3 nbsp;; nbsp; 2 amp; 7 ] nbsp; We know that A=IA nbsp;⇒ nbsp; nbsp; nbsp;[ 1 amp; 3 nbsp;; nbsp; 2 amp; 7 ...

Let A = nbsp; [ 1 amp; 3 nbsp;; nbsp; 2 amp; 7 ] nbsp; We know that A=IA nbsp;⇒ nbsp; nbsp; nbsp;[ 1 amp; 3 nbsp;; nbsp; 2 amp; 7 ...

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Standard XII

Mathematics

Elementary Transformations

Using element...

Question

Using elementary transformations, find the inverse of matrix $[\begin{matrix} 1 & 3 2 & 7 \end{matrix}]$ , if it exists.

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Solution

Let $A = [\begin{matrix} 1 & 3 2 & 7 \end{matrix}]$
We know that $A = I A$
$\Rightarrow [\begin{matrix} 1 & 3 2 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A$
Applying $R_{2} \to R_{2} - 2 R_{1}$
$\Rightarrow [\begin{matrix} 1 & 3 2 - 2 & 7 - 6 \end{matrix}] = [\begin{matrix} 1 & 0 0 - 2 & 1 - 0 \end{matrix}] A$
$\Rightarrow [\begin{matrix} 1 & 3 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 - 2 & 1 \end{matrix}] A$
Applying $R_{1} \to R_{1} - 3 R_{2}$
$\Rightarrow [\begin{matrix} 1 - 3 (0) & 3 - 3 (1) 0 & 1 \end{matrix}] = [\begin{matrix} 1 - (- 2 \times 3) & 0 - 1 (3) - 2 & 1 \end{matrix}] A$
$\Rightarrow [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 7 & - 3 - 2 & 1 \end{matrix}] A$
This is similar to $I = A^{- 1} A$
Thus, $A^{- 1} = [\begin{matrix} 7 & - 3 - 2 & 1 \end{matrix}]$

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Elementary Transformations

Standard XII Mathematics

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Using elementary transformations, find the inverse of matrix [1327], if it exists.

Let A=[1327] We know that A=IA ⇒[1327]=[1001]A Applying R2→R2−2R1 ⇒[132−27−6]=[100−21−0]A ⇒[1301]=[10−21]A Applying R1→R1−3R2 ⇒[1−3(0)3−3(1)01]=[1−(−2×3)0−1(3)−21]A ⇒[1001]=[7−3−21]A This is similar to I=A−1A Thus, A−1=[7−3−21]

Using elementary transformations, find the inverse of matrix $[\begin{matrix} 1 & 3 2 & 7 \end{matrix}]$ , if it exists.