Exercise 1.6 - 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

Ex $1.6$
Question 1.
Test for consistency and if possible, solve the following systems of equations by rank method.
(i) $x-y+2 z=2,2 x+y+4 z=7,4 x-y+z=4$
(ii) $3 x+y+z=2, x-3 y+2 z=1,7 x-y+4 z=5$
(iii) $2 x+2 y+z=5, x-y+z=1,3 x+y+2 z=4$
(iv) $2 x-y+z=2,6 x-3 y+3 z=6,4 x-2 y+2 z=4$
Solution:
(i) Here the number of unknowns $=3$.
The matrix form of the system is $\mathrm{AX}=\mathrm{B}$ where
$
\begin{aligned}
& \mathrm{A}=\left(\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 4 \\
4 & -1 & 1
\end{array}\right) ; \mathrm{X}=\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) \text { and } \mathrm{B}=\left(\begin{array}{l}
2 \\
7 \\
4
\end{array}\right) \\
& \text { (i.e) } \mathrm{AX}=\mathrm{B} \\
&
\end{aligned}
$
The augmented matrix $(\mathrm{A}, \mathrm{B})$ is
$
(\mathrm{A}, \mathrm{B})=\left(\begin{array}{cccc}
1 & -1 & 2 & 2 \\
2 & 1 & 4 & 7 \\
4 & -1 & 1 & 4
\end{array}\right)
$
Applying Gaussian elimination method on $[\mathrm{A}, \mathrm{B}]$ we get
$
\begin{aligned}
{[A, B] } & \sim\left[\begin{array}{rrrr}
1 & -1 & 2 & 2 \\
0 & 3 & 0 & 3 \\
0 & 3 & -7 & -4
\end{array}\right] \begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-4 R_1
\end{array} \\
& \sim\left[\begin{array}{rrrr}
1 & -1 & 2 & 2 \\
0 & 3 & 0 & 3 \\
0 & 0 & -7 & -7
\end{array}\right] \quad R_3 \rightarrow R_3-R_2
\end{aligned}
$
The above matrix is in echelon form also $\rho(\mathrm{A}, \mathrm{B})=\rho(\mathrm{A})=3=$ number of unknowns The system of equations is consistent with a unique solution. To find the solution.
Now writing the equivalent equations we get

$
\begin{aligned}
& \left(\begin{array}{ccc}
1 & -1 & 2 \\
0 & 3 & 0 \\
0 & 0 & -7
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{r}
2 \\
3 \\
-7
\end{array}\right) \\
& \mathrm{x}-\mathrm{y}+2 \mathrm{z}=2 \\
& 3 \mathrm{y}=3 \Rightarrow \mathrm{y}=1 \\
& 7 \mathrm{z}=-7 \Rightarrow \mathrm{z}=1
\end{aligned}
$
Substituting $z=y=1$ in (1) we get
$
\begin{aligned}
& \mathrm{x}-1+2=2 \Rightarrow \mathrm{x}=1 \\
& \Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=1
\end{aligned}
$
(ii) Here the number of unknowns is 3 .
The matrix form of the given system of equations is:
$
\left(\begin{array}{ccc}
3 & 1 & 1 \\
1 & -3 & 2 \\
7 & -1 & 4
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
2 \\
1 \\
5
\end{array}\right)
$
(i.e) Now the augmented matrix $[A, B]$ is

$
\begin{aligned}
& {[A, B] }=\left(\begin{array}{cccc}
3 & 1 & 1 & 2 \\
1 & -3 & 2 & 1 \\
7 & -1 & 4 & 5
\end{array}\right) \sim\left(\begin{array}{rrrr}
1 & -3 & 2 & 1 \\
3 & 1 & 1 & 2 \\
7 & -1 & 4 & 5
\end{array}\right)\left(R_1 \leftrightarrow R_2\right) \\
& \sim\left(\begin{array}{rrrr}
1 & -3 & 2 & 1 \\
0 & 10 & -5 & -1 \\
0 & 20 & -10 & -2
\end{array}\right) \mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1 \\
& \mathrm{R}_3 \rightarrow \mathrm{R}_3-7 \mathrm{R}_1 \\
& \sim\left(\begin{array}{rrrr}
1 & -3 & 2 & 1 \\
0 & 10 & -5 & -1 \\
0 & 0 & 0 & 0
\end{array}\right) \mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_2
\end{aligned}
$
The above matrix is in echelon form also $\rho(A, B)=\rho(A)=2<$ number of unknowns
The system of equations is consistent with the infinite number of solutions. To find the solution:
Now writing the equivalent equations we get
$
\begin{aligned}
&\left(\begin{array}{rrr}
1 & -3 & 2 \\
0 & 10 & -5 \\
0 & 0 & 0
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{r}
1 \\
-1 \\
0
\end{array}\right) \\
& \text { (i.e) } x-3 y+2 z=1
\end{aligned}
$

$
10 y-5 z=-2
$
Taking $z=k,(k \in \mathrm{R})$ in (2)
$
\begin{array}{rlrl} 
& & 10 y-5 k & =-1 \\
\Rightarrow & y & =\frac{5 k-1}{10}
\end{array} \quad \Rightarrow 10 y=5 k-1
$
Taking $z=k, y=\frac{5 k-1}{10}$ in (1)
(i.e)
$
x=1+3 y-2 z \quad \text { (i.e) } x=1+3\left(\frac{5 k-1}{10}\right)-2 k
$
(i.e) $x=\frac{10+15 k-3-20 k}{10}$ $x=\frac{7-5 k}{10}$
So the solution is $\quad x=\frac{7-5 k}{10}, y=\frac{5 k-1}{10}, z=k$ where $k \in \mathrm{R}$

Here the number of unknowns is 3 .
The matrix form of the given equation is
$
\begin{aligned}
& \left(\begin{array}{ccc}
2 & 2 & 1 \\
1 & -1 & 1 \\
3 & 1 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
5 \\
1 \\
4
\end{array}\right) \\
& \mathrm{AX}=\mathrm{B} \\
& {[A, B]=\left(\begin{array}{rrrr}
2 & 2 & 1 & 5 \\
1 & -1 & 1 & 1 \\
3 & 1 & 2 & 4
\end{array}\right) \sim\left(\begin{array}{rrrr}
1 & -1 & 1 & 1 \\
2 & 2 & 1 & 5 \\
3 & 1 & 2 & 4
\end{array}\right) R_1 \leftrightarrow R_2} \\
& \sim\left(\begin{array}{rrrr}
1 & -1 & 1 & 1 \\
0 & 4 & -1 & 3 \\
0 & 4 & -1 & 1
\end{array}\right) \begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-3 R_1
\end{array} \\
& \sim\left(\begin{array}{rrrr}
1 & -1 & 1 & 1 \\
0 & 4 & -1 & 3 \\
0 & 0 & 0 & -2
\end{array}\right) R_3 \rightarrow R_3-R_2 \\
&
\end{aligned}
$
The above matrix is in echelon form.
Here $\rho(\mathrm{A}, \mathrm{B})=3 ; \rho(\mathrm{A})=2$
So $\rho(A, B) \neq \rho(A)$
The system of equations is inconsistent with no solution.
(iv) Here the number of unknowns is 3 .
The matrix form of the given equation is

$
\begin{aligned}
& \left(\begin{array}{lll}
2 & -1 & 1 \\
6 & -3 & 3 \\
4 & -2 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
2 \\
6 \\
4
\end{array}\right) \\
& \mathrm{AX}=\mathrm{B}
\end{aligned}
$
The augmented matrix $[\mathrm{A}, \mathrm{B}]$ is
$
[A, B]=\left[\begin{array}{rrrr}
2 & -1 & 1 & 2 \\
6 & -3 & 3 & 6 \\
4 & -2 & 2 & 4
\end{array}\right] \sim\left[\begin{array}{rrrr}
2 & -1 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] \begin{aligned}
& R_2 \rightarrow R_2-3 R_1 \\
& R_3 \rightarrow R_3-2 R_1
\end{aligned}
$
The above matrix is in echelon form also $\rho(\mathrm{A}, \mathrm{B})=\rho(\mathrm{A})=1<$ number of unknowns The system of equations is consistent with the infinite number of solutions. To find the Solution Now writing the equivalent equations we get

$
\left(\begin{array}{rrr}
2 & -1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
2 \\
0 \\
0
\end{array}\right)
$
(i.e) $2 x-y+z=2$
Taking $z=t$ and $y=s(t, s \in \mathrm{R})$ in (1) we get
$
\begin{aligned}
2 x-s+t & =2 \\
2 x & =2+s-t \\
x & =\frac{1}{2}(2+s-t)
\end{aligned}
$
So the solution is $x=\frac{1}{2}(2+s-t), y=s, z=t, s, t \in \mathrm{R}$

Question 2.
Find the value of $k$ for which the equations $k x-2 y+z=1, x-2 k y+z=-2, x-2 y+k z=1$ have
(i) no solution
(ii) unique solution
(iii) infinitely many solution
Solution:
The matrix form of the given system of equation is
$
\left(\begin{array}{ccc}
k & -2 & 1 \\
1 & -2 k & 1 \\
1 & -2 & k
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right)
$
(i.e) $\mathrm{AX}=\mathrm{B}$
The augmented matrix $(A, B)$ is
$
\begin{aligned}
{[\mathrm{A}, \mathrm{B}] } & =\left[\begin{array}{cccc}
k & -2 & 1 & 1 \\
1 & -2 k & 1 & -2 \\
1 & -2 & k & 1
\end{array}\right] \sim\left[\begin{array}{cccc}
1 & -2 k & 1 & -2 \\
k & -2 & 1 & 1 \\
1 & -2 & k & 1
\end{array}\right] \mathrm{R}_1 \leftrightarrow \mathrm{R}_2 \\
& \sim\left[\begin{array}{cccc}
1 & -2 k & 1 & -2 \\
0 & 2 k^2-2 & 1-k & 1+2 k \\
0 & -2+2 k & k-1 & 3
\end{array}\right] \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-k \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & -2 k & 1 & -2 \\
0 & 2\left(k^2-1\right) & 1-k & 1+2 k \\
0 & 2(k-1) & k-1 & 3
\end{array}\right) \sim\left(\begin{array}{cccc}
1 & -2 k & 1 & -2 \\
0 & 2(k+1) & 1-k & 1+2 k \\
0 & (k-1) & & \\
0 & 2(k-1) & k-1 & 3
\end{array}\right)
\end{aligned}
$

$
\begin{aligned}
& \sim\left(\begin{array}{cccc}
1 & -2 k & 1 & -2 \\
0 & 2(k+1) & 1-k & 1+2 k \\
& (k-1) & & \\
0 & 0 & k^2+k-2 & k+2
\end{array}\right) \mathrm{R}_3 \rightarrow(k+1) \mathrm{R}_3-\mathrm{R}_2 \\
& \mathrm{k}^2+\mathrm{k}-2=(\mathrm{k}+2)(\mathrm{k}-1)
\end{aligned}
$
The above matrix is in echelon form
Case 1: when $\mathrm{k}=1, \rho(\mathrm{A}, \mathrm{B})=3, \rho(\mathrm{A})=2$ (i.e) $\rho(\mathrm{A}, \mathrm{B}) \neq \rho(\mathrm{A})$ The system is inconsistent and the system has no solution.
Case 2 : when $k \neq 1, k \neq-2$, then $\rho(A, B)=\rho(A)=3=$ number of unknowns The system is consistent with unique solution.
Case 3: When $\mathrm{k}=-2$ then $\rho(\mathrm{A})=\rho(\mathrm{A}, \mathrm{B})=2<$ number of unknowns.
The system is consistent with the infinite number of solutions.

Question 3.
Investigate the values of $\lambda$ and $\mu$ the system of linear equations. $2 x+3 y+5 z=9,7 x+3 y-5 z=8,2 x+3 y+\lambda z=\mu$, have
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.
Solution:
The matrix form of the above equation is
$
\begin{aligned}
& \left(\begin{array}{ccc}
2 & 3 & 5 \\
7 & 3 & -5 \\
2 & 3 & \lambda
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
9 \\
8 \\
\mu
\end{array}\right) \\
& \text { (i.e) } A X=B \\
&
\end{aligned}
$
The augmented matrix $[\mathrm{A}, \mathrm{B}]$ is
$
[A, B]=\left[\begin{array}{cccc}
2 & 3 & 5 & 9 \\
7 & 3 & -5 & 8 \\
2 & 3 & \lambda & \mu
\end{array}\right] \sim\left[\begin{array}{cccc}
2 & 3 & 5 & 9 \\
0 & -15 & -45 & -47 \\
0 & 0 & \lambda-5 & \mu-9
\end{array}\right] \begin{aligned}
& R_2 \rightarrow 2 R_2-7 R_1 \\
& R_3 \rightarrow R_3-R_1
\end{aligned}
$
The above matrix is in echelon form
Case 1: When $\lambda=5, \mu \neq 9$
$\rho(A)=2, \rho(A, B)=3$ (i.e) $\rho(A, B) \neq p(A)$
The system is inconsistent and it has no solution.
Case 2: When $\lambda \neq 5, \mu \in R$,
$\rho(A, B)=\rho(A)=3=$ number of unknowns
The system is consistent with a unique solution.
Case 3: When $\lambda=5, \mu=9$,
then $\rho(A, B)=\rho(A)=2<$ number of unknowns
The system is consistent with infinite number of solutions.