KC Sinha Mathematics Solution Class 12 Chapter 5 आव्यूह ( Matrices ) Exercise 5.4 (Q11-Q15)

Ex-5.4

---

Exercise 5.4

Question 11

$\left[\begin{array}{cr}2 & -3 \\ -1 & 2\end{array}\right]$

Sol :

Question 12

$\left[\begin{array}{cc}10 & -2 \\ -5 & 1\end{array}\right]$

Sol :

Question 13

$\left[\begin{array}{ll}3 & 9 \\ 1 & 3\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}3 & 9 \\ 1 & 3\end{array}\right]$

A=IA

$\left[\begin{array}{ll}3 & 9 \\ 1 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$

R₁→R₁-3R₂

$\left[\begin{array}{ll}0 & 0 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right] A$

∴ A^-1 का अस्तित्व नही हैं

Question 14

$\left[\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
Sol :
$A=\left[\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

A=IA

$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]A$

R₁→R₁-R₂ 

$\left[\begin{array}{ccc}5 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$

R₁→R₁-2R₂ 

$\left[\begin{array}{ccc}1 & 1 & 2 \\ 2 & 0 & -1 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]A$

R₂→R₂-2R₁

$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]a$

$R_{3} \rightarrow R_{3}+\frac{1}{2} R_{2}$

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

R₃→2R₃

$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 5 & -2 & 0 \\ 5 & -2 & 20\end{array}\right]A$

R₂→R₂+5R₃

$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 30 & -12 & 10 \\ 5 & -2 & 2\end{array}\right]A$

R₁→R₁-2R₃ 

$\left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-12 & 5 & -4 \\ 30 & -12 & 10 \\ 5 & -2 & 2\end{array}\right]A$

$R_{1} \rightarrow R_{1}+\frac{1}{2} R_{2}$

$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 1 \\ 30 & -12 & 10 \\ 5 & -2 & 2\end{array}\right]A$

$R_{2} \rightarrow-\frac{1}{2} R_{2}$

$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{cccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$

$\therefore A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

Question 15

Using elementary transformations (operations) , find the inverse of the following matrices , if it exist
$\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Sol :
Let $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$

A=IA

$\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$


$\left[\begin{array}{ccc}3 & -2 & 2 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$


$\left[\begin{array}{rrr}1 & -4 & -1 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$

$R_{1}\leftrightarrow R_{3}$

$\left[\begin{array}{rrr}3 & -2 & 2 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$

$R_{1} \rightarrow R_{1}-R_{2}$

$\left[\begin{array}{rrr}1 & -4 & -1 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$

$R_{2} \rightarrow R_{2}-R_{3}$

$\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 5 & 0 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{ccc}0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$

$R_{3} \rightarrow R_{3}-2 R_{1}$

$\left[\begin{array}{rrr}1 & -4 & -1 \\ 0 & 5 & 0 \\ 0 & 5 & 5\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 2 & -2\end{array}\right]A$

$R_{3} \rightarrow R_{3}-R_{2}$

$\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ -1 & 1 & 0 \\ 2 & 1 & -2\end{array}\right]A$

$R_{3} \rightarrow \frac{1}{5} R_{3}$

$\left[\begin{array}{rrr}1 & -4 & -1 \\ 0 & 5 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ -1 & 1 & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]A$

$R_{1} \rightarrow R_{1}+R_{3}$

$\left[\begin{array}{rrrr}1 & -4 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}\frac{2}{5} & -\frac{4}{5} & \frac{3}{5} \\ -1 & 1 & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]A$

$R_{2} \rightarrow \frac{1}{5} R_{2}$

$\left[\begin{array}{ccc}1 & -4 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{2}{5} & -\frac{4}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{8}\end{array}\right]A$

$R_1 \longrightarrow R_{1}+4 R_{2}$

$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{ccc}-\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]$

$\theta^{-1}=\left[\begin{array}{rrr}-\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]$