KC Sinha Mathematics Solution Class 12 Chapter 5 आव्यूह ( Matrices ) Exercise 5.4

Ex-5.4

---

Exercise 5.4

[using elementary transformations (operations), find the inverse of the following matrices, if it exist]
Question 1
$\begin{bmatrix}1&3\\2&7\end{bmatrix}$
Sol :
A=$\begin{bmatrix}1&3\\2&7\end{bmatrix}$

A=IA

$\begin{bmatrix}1&3\\2&7\end{bmatrix}$=$\begin{bmatrix}1&0\\0&1\end{bmatrix}A$

R₂→R₂-2R₁

$\begin{bmatrix}1&3\\0&1\end{bmatrix}$=$\begin{bmatrix}1&0\\-2&1\end{bmatrix}A$

R₁→R₁-3R₂

$\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix}A$

$A^{-1}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix}$

Question 2

$\begin{bmatrix}2&1\\7&4\end{bmatrix}$

Sol :

Let A=$\begin{bmatrix}2&1\\7&4\end{bmatrix}$

A=IA

$\begin{bmatrix}2&1\\7&4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$

R_1↔R₂

$\begin{bmatrix}7&4\\2&1\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}A$

R₁→R₁-3R₂

$\begin{bmatrix}1&1\\2&1\end{bmatrix}=\begin{bmatrix}-3&1\\1&0\end{bmatrix}A$

R₂→R₂-2R₁

$\begin{bmatrix}1&1\\0&-1\end{bmatrix}=\begin{bmatrix}-3&1\\7&-2\end{bmatrix}A$

R₁→-1R₂

$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}-3&1\\-7&2\end{bmatrix}A$

R₁→R₁-R₂

$\begin{bmatrix}1&1\\0&-1\end{bmatrix}=\begin{bmatrix}-3&1\\7&-2\end{bmatrix}A$

R₂→-1R₂

$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}-3&1\\-7&2\end{bmatrix}A$

R₁→R₁-R₂

$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}A$

$A^{-1}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}$

Question 3

$\begin{bmatrix}2&1\\1&1\end{bmatrix}$

Sol :

Let A=$\begin{bmatrix}2&1\\1&1\end{bmatrix}$

A=IA

$\begin{bmatrix}2&1\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$

R₁→R₁-R₂

$\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}A$

R₂→R₂-R₁

$\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}A$

$A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}$

Question 4

$\begin{bmatrix}\dfrac{1}{5}&\dfrac{2}{5}\\\dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}$

Sol :

A=$\begin{bmatrix}\dfrac{1}{5}&\dfrac{2}{5}\\\dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}$

A=IA

$\begin{bmatrix}\dfrac{1}{5}&\dfrac{2}{5}\\\dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$

R₁→5R₁

$\begin{bmatrix}1&2\\ \dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}=\begin{bmatrix}5&0\\0&1\end{bmatrix}A$

$\begin{bmatrix}1&2\\ 0&-1\end{bmatrix}=\begin{bmatrix}5&0\\-2&1\end{bmatrix}A$

R₂→-R₂

$\begin{bmatrix}1&2\\ 0&1\end{bmatrix}=\begin{bmatrix}5&0\\2&-1\end{bmatrix}A$

R₁→R₁-2R₂

$\begin{bmatrix}1&0\\ 0&1\end{bmatrix}=\begin{bmatrix}1&2\\2&-1\end{bmatrix}A$

$A^{-1}=\begin{bmatrix}1&2\\2&-1\end{bmatrix}$

Question 5

$\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]$
Sol :
$A=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$

A=IA

$\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A$

 R₂→R₂-2R₁

$\left[\begin{array}{cc}1 & 2 \\ 0 & -5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]A$

$\left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ \frac{2}{5} & \frac{-1}{5}\end{array}\right]A$

 R₁→R₁-2R₂

$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right]A$

$A^{-1}=\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right]$