Exercise 1.2-Additional Problems - Chapter 1 Applications of Matrices and Determinants 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
$
\left[\begin{array}{rrr}
1 & 1 & -1 \\
3 & -2 & 3 \\
2 & -3 & 4
\end{array}\right]
$
Find the rank of the following matrices.
Solution:
Let
$
\begin{aligned}
A & =\left[\begin{array}{rrr}
1 & 1 & -1 \\
3 & -2 & 3 \\
2 & -3 & 4
\end{array}\right] \\
|A| & =\left[\begin{array}{rrr}
1 & 1 & -1 \\
3 & -2 & 3 \\
2 & -3 & 4
\end{array} \mid=1(-8+9)-1(12-6)-1(-9+4)=1-6+5=0\right.
\end{aligned}
$
Since $|\mathrm{A}|=0, \rho(\mathrm{A}) \neq 3$. i.e., $\rho(\mathrm{A})<3$
Consider a second order minor $\left|\begin{array}{cc}1 & 1 \\ 3 & -2\end{array}\right|=-2-3=-5 \neq 0$
A has at least one non-zero minor of order 2. $\rho(\mathrm{A})=2$
Question 2.
$
\left[\begin{array}{rrrr}
0 & 1 & 2 & 1 \\
2 & -3 & 0 & -1 \\
1 & 1 & -1 & 0
\end{array}\right]
$
Find the rank of the following matrices.

Solution:

Let
$
\begin{aligned}
A & =\left[\begin{array}{rrrr}
0 & 1 & 2 & 1 \\
2 & -3 & 0 & -1 \\
1 & 1 & -1 & 0
\end{array}\right] \sim\left[\begin{array}{rrrr}
1 & 1 & -1 & 0 \\
2 & -3 & 0 & -1 \\
0 & 1 & 2 & 1
\end{array}\right] \mathrm{R}_1 \leftrightarrow \mathrm{R}_3 \\
& \sim\left[\begin{array}{rrrr}
1 & 1 & -1 & 0 \\
0 & -5 & 2 & -1 \\
0 & 1 & 2 & 1
\end{array}\right] \mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\
& \sim\left[\begin{array}{rrrr}
1 & 1 & -1 & 0 \\
0 & -5 & 2 & -1 \\
0 & 0 & 12 & 4
\end{array}\right] \mathrm{R}_3 \rightarrow 5 \mathrm{R}_3+\mathrm{R}_2
\end{aligned}
$
The last equivalent matrix is in the echelon form. It has three non-zero rows.
$\therefore \rho(\mathrm{A})=3 ;$ Here $\mathrm{A}$ is of order $3 \times 4$
Question 3.
$
\left[\begin{array}{rrrr}
1 & -2 & 3 & 4 \\
-2 & 4 & -1 & -3 \\
-1 & 2 & 7 & 6
\end{array}\right]
$
Find the rank of the follwing matrix:

Solution:
Let
$
A=\left[\begin{array}{rrrr}
1 & -2 & 3 & 4 \\
-2 & 4 & -1 & -3 \\
-1 & 2 & 7 & 6
\end{array}\right] \sim\left[\begin{array}{rrrr}
1 & -2 & 3 & 4 \\
0 & 0 & 5 & 5 \\
0 & 0 & 10 & 10
\end{array}\right] \begin{aligned}
& \mathrm{R}_2 \rightarrow \mathrm{R}_2+2 \mathrm{R}_1 \\
& \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_1
\end{aligned}
$
$
=\left[\begin{array}{rrrr}
1 & -2 & 3 & 4 \\
0 & 0 & 5 & 5 \\
0 & 0 & 0 & 0
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_2
$
The last equivalent matrix is in the echelon form. The number of non-zero rows in this matrix is two. A is a matrix of order $3 \times 4 . \therefore \rho(\mathrm{A})=2$
Question 4.
Using elementary transformations find the inverse of the following matrix
$
\left[\begin{array}{ll}
4 & 7 \\
3 & 6
\end{array}\right] .
$

Solution:
Let
Now
$
A=\left[\begin{array}{ll}
4 & 7 \\
3 & 6
\end{array}\right]
$
Operating $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \quad\left[\begin{array}{ll}1 & 1 \\ 3 & 6\end{array}\right]=\left[\begin{array}{rr}1 & -1 \\ 0 & 1\end{array}\right] A$
Operating $R_2 \rightarrow R_2-3 R_1 \quad\left[\begin{array}{ll}1 & 1 \\ 0 & 3\end{array}\right]=\left[\begin{array}{rr}1 & -1 \\ -3 & 4\end{array}\right] A$
Operating $\mathrm{R}_2 \rightarrow \frac{1}{3} \mathrm{R}_2 \quad\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -1 & 4 / 3\end{array}\right] \mathrm{A}$
Operating $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \quad\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}2 & -7 / 3 \\ -1 & 4 / 3\end{array}\right] \mathrm{A}$
Thus,
$
A^{-1}=\left[\begin{array}{cc}
2 & -7 / 3 \\
-1 & 4 / 3
\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}
6 & -7 \\
-3 & 4
\end{array}\right]
$
Question 5.
Using elementary transformations find the inverse of the following matrices
$
\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]
$

Solution:
Let
Now
$
\begin{aligned}
A=\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right] & \\
A=\mathrm{IA} & \Rightarrow\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A}
\end{aligned}
$
Applying $R_2 \rightarrow R_2-2 R_1$
$
\left[\begin{array}{rr}
1 & -1 \\
0 & 5
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-2 & 1
\end{array}\right] \mathrm{A}
$
Applying $R_2 \rightarrow \frac{1}{5} R_2 \quad \Rightarrow\left[\begin{array}{rr}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ -2 / 5 & 1 / 5\end{array}\right] \mathrm{A}$ Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2 \quad \Rightarrow\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5\end{array}\right] \mathrm{A}$
Thus
$
\mathrm{A}^{-1}=\left[\begin{array}{cc}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}
3 & 1 \\
-2 & 1
\end{array}\right]
$
Question 6.

Using elementary transformations find the inverse of the following matrices 
$
\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]
$

Solution:
Let
Now
$
A=\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]
$
Applying $R_1 \leftrightarrow-R_2$
$A=I A$
$
\Rightarrow\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A}
$
$
\Rightarrow\left[\begin{array}{ll}
4 & -2 \\
3 & -1
\end{array}\right]=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right] A
$
Applying $R_1 \rightarrow R_1-R_2$ $\Rightarrow\left[\begin{array}{ll}1 & -1 \\ 3 & -1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 1 & 0\end{array}\right] A$
Applying $R_2 \rightarrow R_2-3 R_1$ $\Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right] A$
Applying $\mathrm{R}_2 \rightarrow \frac{1}{2} \mathrm{R}_2$ $\Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 2 & 3 / 2\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2$ $\Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 / 2 \\ 2 & 3 / 2\end{array}\right] \mathrm{A}$
Thus
$
A^{-1}=\left[\begin{array}{ll}
1 & 1 / 2 \\
2 & 3 / 2
\end{array}\right]
$
Question 7.
Using elementary transformations, find the inverse of the following matrices

$
\left[\begin{array}{ccc}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]
$
Solution:

Let $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Now
$
A=I A \quad \Rightarrow\left[\begin{array}{ccc}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] A
$
Applying $R_1 \leftrightarrow R_3$
$
\Rightarrow\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] \mathrm{A}
$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2$
$
\Rightarrow\left[\begin{array}{ccc}
1 & -4 & -1 \\
2 & 2 & 3 \\
2 & -3 & 3
\end{array}\right]=\left[\begin{array}{ccc}
0 & -1 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right] \mathrm{A}
$
Applying $R_2 \rightarrow R_2-2 R_1$ and $R_3 \rightarrow R_3-2 R_1 \Rightarrow\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 10 & 5 \\ 0 & 5 & 5\end{array}\right]=\left[\begin{array}{ccc}0 & -1 & 1 \\ 0 & 3 & -2 \\ 1 & 2 & -2\end{array}\right] A$
Applying $\mathrm{R}_2 \rightarrow \frac{1}{10} \mathrm{R}_2 \quad \Rightarrow\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 1 & 1 / 2 \\ 0 & 5 & 5\end{array}\right]=\left[\begin{array}{ccc}0 & -1 & 1 \\ 0 & 3 / 10 & -1 / 5 \\ 1 & 2 & -2\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_3 \rightarrow \mathrm{R}_3-5 \mathrm{R}_2$ $\Rightarrow\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 1 & 1 / 2 \\ 0 & 0 & 5 / 2\end{array}\right]=\left[\begin{array}{ccc}0 & -1 & 1 \\ 0 & 3 / 10 & -1 / 5 \\ 1 & 1 / 2 & -1\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_3 \rightarrow \frac{2}{5} \mathrm{R}_3$
$
\begin{aligned}
& \Rightarrow\left[\begin{array}{ccc}
1 & -4 & -1 \\
0 & 1 & 1 / 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
0 & -1 & 1 \\
0 & 3 / 10 & -1 / 5 \\
2 / 5 & 1 / 5 & -2 / 5
\end{array}\right] \mathrm{A} \\
& \Rightarrow\left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & 1 / 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 / 5 & 1 / 5 \\
0 & 3 / 10 & -1 / 5 \\
2 / 5 & 1 / 5 & -2 / 5
\end{array}\right] \mathrm{A}
\end{aligned}
$

$\begin{aligned}
& \text { Applying } R_1 \rightarrow R_1+4 R_2 \quad \Rightarrow\left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & 1 / 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 / 5 & 1 / 5 \\
0 & 3 / 10 & -1 / 5 \\
2 / 5 & 1 / 5 & -2 / 5
\end{array}\right] A \\
& \text { Applying } \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_3 \text { and } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\frac{1}{2} \mathrm{R}_3 \Rightarrow\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-2 / 5 & 0 & 3 / 5 \\
-1 / 5 & 1 / 5 & 0 \\
2 / 5 & 1 / 5 & -2 / 5
\end{array}\right] \mathrm{A} \\
& \mathrm{A}^{-1}=\left[\begin{array}{ccc}
-2 / 5 & 0 & 3 / 5 \\
-1 / 5 & 1 / 5 & 0 \\
2 / 5 & 1 / 5 & -2 / 5
\end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc}
-2 & 0 & 3 \\
-1 & 1 & 0 \\
2 & 1 & -2
\end{array}\right] \\
&
\end{aligned}$

Question 8.

Using elementary transformations, find the inverse of the following matrices

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

Solution:
Let
Now
$
A=\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]
$
Applying $R_2 \rightarrow R_2+3 R_1$ and $R_3 \rightarrow R_3-2 R_1 \Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] A$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2+8 \mathrm{R}_3 \quad \Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & -1 & 4\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ -13 & 1 & 8 \\ -2 & 0 & 1\end{array}\right] \mathrm{A}$

$\begin{aligned}
& \text { Applying } \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_2 \quad \Rightarrow\left[\begin{array}{ccc}
1 & 3 & -2 \\
0 & 1 & 21 \\
0 & 0 & 25
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-13 & 1 & 8 \\
-15 & 1 & 9
\end{array}\right] \mathrm{A} \\
& \text { Applying } \mathrm{R}_3 \rightarrow \frac{1}{25} \mathrm{R}_3 \quad \Rightarrow\left[\begin{array}{ccc}
1 & 3 & -2 \\
0 & 1 & 21 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-13 & 1 & 8 \\
-3 / 5 & 1 / 25 & 9 / 25
\end{array}\right] \mathrm{A} \\
& \text { Applying } R_1 \rightarrow R_1-3 R_2 \quad \Rightarrow\left[\begin{array}{ccc}
1 & 0 & -65 \\
0 & 1 & 21 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
40 & -3 & -24 \\
-13 & 1 & 8 \\
-3 / 5 & 1 / 25 & 9 / 25
\end{array}\right] \mathrm{A} \\
& \text { Applying } R_2 \rightarrow R_2-21 R_3 \quad \Rightarrow\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 / 5 & -3 / 5 \\
-2 / 5 & 4 / 25 & 11 / 25 \\
-3 / 5 & 1 / 25 & 9 / 25
\end{array}\right] \mathrm{A} \\
& \mathrm{A}^{-1}=\left[\begin{array}{ccc}
1 & -2 / 5 & -3 / 5 \\
-2 / 5 & 4 / 25 & 11 / 25 \\
-3 / 5 & 1 / 25 & 9 / 25
\end{array}\right] . \\
&
\end{aligned}$

Question 9.
Using elementary transformations, find the inverse of the following matrices

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

Let $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
Now $A=I A \quad \Rightarrow\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}$ $\begin{array}{ll}\text { Applying } R_1 \rightarrow 3 R_1-R_2 & \Rightarrow\left[\begin{array}{ccc}1 & -1 & -3 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \\ \text { Applying } R_2 \rightarrow R_2-5 R_1 & \Rightarrow\left[\begin{array}{ccc}1 & -1 & -3 \\ 0 & 6 & 15 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 0 \\ -15 & 6 & 0 \\ 0 & 0 & 1\end{array}\right] \text { A }\end{array}$
Applying $R_2 \rightarrow R_2-5 R_3$
Applying $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2$
Applying $\mathrm{R}_3 \rightarrow \frac{1}{3} \mathrm{R}_3$ $\Rightarrow\left[\begin{array}{ccc}1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 0 \\ -15 & 6 & -5 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}$ $\Rightarrow\left[\begin{array}{ccc}1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 0 \\ -15 & 6 & -5 \\ 15 & -6 & 6\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2$ $\Rightarrow\left[\begin{array}{ccc}1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 0 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] \mathrm{A}$ $\Rightarrow\left[\begin{array}{ccc}1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-12 & 5 & -5 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+3 \mathrm{R}_3 \quad \Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] \mathrm{A}$
Thus
$
A^{-1}=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]
$

Question 10.

Using elementary transformations, find the inverse of the following matrices
$
\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]
$
Solution:
Let
Now
$
A=\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]
$
$
\begin{array}{ll}
\text { Applying } \mathrm{R}_1 \rightarrow \frac{1}{2} \mathrm{R}_1 & \Rightarrow\left[\begin{array}{cc}
1 & 1 / 2 \\
4 & 2
\end{array}\right]=\left[\begin{array}{cc}
1 / 2 & 0 \\
0 & 1
\end{array}\right] \mathrm{A} \\
\text { Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-4 \mathrm{R}_1 & \Rightarrow\left[\begin{array}{cc}
1 & 1 / 2 \\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
1 / 2 & 0 \\
-2 & 1
\end{array}\right] \mathrm{A}
\end{array}
$
Since $R_2$ has all numbers zero, Thus inverse of matrix A does not exist.