Exercise 1.7 - 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 $1.7$
Question 1.
Solve the following system of homogeneous equations.
(i) $3 x+2 y+7 z=0,4 x-3 y-2 z=0,5 x+9 y+23 z=0$
(ii) $2 x+3 y-z=0, x-y-2 z=0,3 x+y+3 z=0$
Solution:
(i) The matrix form of the above equations is
$
\left(\begin{array}{ccc}
3 & 2 & 7 \\
4 & -3 & -2 \\
5 & 9 & 23
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
0 \\
0 \\
0
\end{array}\right)
$
The augmented matrix $[\mathrm{A}, \mathrm{B}]$ is
$
\begin{aligned}
(A, B) & =\left[\begin{array}{cccc}
3 & 2 & 7 & 0 \\
4 & -3 & -2 & 0 \\
5 & 9 & 23 & 0
\end{array}\right] \sim\left[\begin{array}{cccc}
3 & 2 & 7 & 0 \\
0 & -17 & -34 & 0 \\
0 & 17 & 34 & 0
\end{array}\right] \begin{array}{l}
R_2 \rightarrow 3 R_2-4 R_1 \\
R_3 \rightarrow 3 R_3-5 R_1
\end{array} \\
& \sim\left[\begin{array}{cccc}
3 & 2 & 7 & 0 \\
0 & -17 & -34 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] R_3 \rightarrow R_3+R_2
\end{aligned}
$
The above matrix is in echelon form.
Here $\rho(A, B)=\rho(A)<$ number of unknowns.
The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get $3 x+2 y+7 z=0$
$
-17 y-34 z=0 \ldots \ldots \text { (2) }
$
Taking $z=t$ in (2) we get $-17 y-34 t=0$
$
\begin{aligned}
& \Rightarrow-17 \mathrm{y}=34 \mathrm{t} \\
& \Rightarrow \mathrm{y}=-2 \mathrm{t}
\end{aligned}
$
Taking $\mathrm{z}=\mathrm{t}, \mathrm{y}=-2 \mathrm{t}$ in (1) we get
$
\begin{aligned}
& 3 \mathrm{x}+2(-2 \mathrm{t})+7 \mathrm{t}=0 \\
& \Rightarrow 3 \mathrm{x}-4 \mathrm{t}+7 \mathrm{t}=0 \\
& \Rightarrow 3 \mathrm{x}=-3 \mathrm{t} \\
& \Rightarrow \mathrm{x}=-\mathrm{t}
\end{aligned}
$
So the solution is $x=-t ; y=-2 t$; and $z=t, t \in R$
(ii) The matrix form of the equations is
$
\left(\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
0 \\
0 \\
0
\end{array}\right)
$

(i.e) $\mathrm{AX}=\mathrm{B}$
The augmented matrix $[\mathrm{A}, \mathrm{B}]$ is
$
\begin{aligned}
&(A, B)=\left(\begin{array}{cccc}
2 & 3 & -1 & 0 \\
1 & -1 & -2 & 0 \\
3 & 1 & 3 & 0
\end{array}\right) \sim\left(\begin{array}{cccc}
1 & -1 & -2 & 0 \\
2 & 3 & -1 & 0 \\
3 & 1 & 3 & 0
\end{array}\right) R_1 \leftrightarrow R_2 \\
& \sim\left(\begin{array}{cccc}
1 & -1 & -2 & 0 \\
0 & 5 & 3 & 0 \\
0 & 4 & 9 & 0
\end{array}\right) \mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\
& \mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_1 \\
& \sim\left(\begin{array}{cccc}
1 & -1 & -2 & 0 \\
0 & 5 & 3 & 0 \\
0 & 0 & 33 & 0
\end{array}\right) \mathrm{R}_3 \rightarrow 5 \mathrm{R}_3-4 \mathrm{R}_2
\end{aligned}
$
The above matrix is in echelon form also $\rho(A, B)=\rho(A)=3=$ number of unknowns The system is consistent with unique solution, $x=y=z=0$ (i.e) The system has trivial solution only.
Question 2.
Determine the values of $\lambda$ for which the following system of equations. $x+y+3 z=0,4 x+3 y+\lambda z=0,2 x+y+2 z=0$ has
(i) a unique solution
(ii) a non-trivial solution
Solution:
The matrix form of the equation is
$
\left(\begin{array}{ccc}
1 & 1 & 3 \\
4 & 3 & \lambda \\
2 & 1 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
0 \\
0 \\
0
\end{array}\right)
$
The augmented matrix $[\mathrm{A}, \mathrm{B}]$ is

$
\begin{aligned}
(A, B) & =\left[\begin{array}{cccc}
1 & 1 & 3 & 0 \\
4 & 3 & \lambda & 0 \\
2 & 1 & 2 & 0
\end{array}\right] \sim\left[\begin{array}{llll}
1 & 1 & 3 & 0 \\
2 & 1 & 2 & 0 \\
4 & 3 & \lambda & 0
\end{array}\right] \mathrm{R}_2 \leftrightarrow \mathrm{R}_3 \\
& \sim\left[\begin{array}{cccc}
1 & 1 & 3 & 0 \\
0 & -1 & -4 & 0 \\
0 & -1 & \lambda-12 & 0
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-4 \mathrm{R}_1-2 \mathrm{R}_1 \\
& \sim\left[\begin{array}{cccc}
1 & 1 & 3 & 0 \\
0 & -1 & -4 & 0 \\
0 & 0 & \lambda-8 & 0
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2
\end{aligned}
$
The above matrix is in echelon form
Case 1: When $\lambda \neq 8, \rho(A, B)=\rho(A)=3=$ number of unknowns The system is consistent with a unique solution.
Case 2: When $\lambda=8, \rho(A, B)=\rho(A)=2<$ number of unknowns.
The system is consistent with non-trivial solutions.

Question 3.
By using the Gaussian elimination method, balance the chemical reaction equation:
$
\mathrm{C}_2 \mathrm{H}_6+\mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2
$
Solution:
We are searching for positive integers $x_1, x_2, x_3$ and $x_4$ such that
$
\mathrm{x}_1 \mathrm{C}_2 \mathrm{H}_6+\mathrm{x}_2 \mathrm{O}_2=\mathrm{x}_3 \mathrm{H}_2 \mathrm{O}+\mathrm{x}_4 \mathrm{CO}_2 \ldots \ldots \text { (1) }
$
The number of carbon atoms on LHS of (1) should be equal to the number of carbon atoms on the RHS of (1).
So we get a linear homogeneous equation.
$
\begin{aligned}
& 2 \mathrm{x}_1=\mathrm{x}_4 \\
& \Rightarrow 2 \mathrm{x}_1-\mathrm{x}_4=0
\end{aligned}
$
Similarly considering hydrogen and oxygen atoms we get respectively.
$
\begin{aligned}
& 2 \mathrm{x}_2=\mathrm{x}_3+2 \mathrm{x}_4 \\
& \Rightarrow 2 \mathrm{x}_2-\mathrm{x}_3-2 \mathrm{x}_4=0
\end{aligned}
$
and $-2 \mathrm{x}_3+3 \mathrm{x}_4=0$
Equations (2), (3) and (4) constitute a homogeneous system of linear equations in four unknowns. The augmented matrix $(A, B)$ is
$
(A, B)=\left[\begin{array}{rrrr}
2 & 0 & 0 & -1 \\
0 & 2 & -1 & -2 \\
3 & 0 & -1 & 0
\end{array}\right] \sim\left[\begin{array}{rrrr}
2 & 0 & 0 & -1 \\
0 & 2 & -1 & -2 \\
0 & 0 & -2 & 3
\end{array}\right] R_3 \rightarrow 2 R_3-3 R_1
$
Now $\rho(A, B)=\rho(A)=3<$ number of unknowns.
So the system is consistent and has an infinite number of solutions.
Writing the equations using the echelon form we get

$
\begin{aligned}
& 2 \mathrm{x}_1-\mathrm{x}_4=0 \ldots \ldots \\
& 2 \mathrm{x}_2-\mathrm{x}_3-2 \mathrm{x}_4=0 \\
& -2 \mathrm{x}_3+3 \mathrm{x}_4=0
\end{aligned}
$
Taking $x_4=t,(t \neq 0)$ in (7) we get
$
\begin{aligned}
-2 x_3+3 t & =0 \\
2 x_3 & =3 t \quad \Rightarrow x_3=\frac{3}{2} t
\end{aligned}
$

Taking $x_4=t$, in (5) we get
$
\begin{aligned}
2 x_1-t & =0 \\
2 x_1 & =t \quad \Rightarrow x_1=\frac{t}{2}
\end{aligned}
$
Taking $x_3=\frac{3}{2} t, x_4=t$ in (6) we get
$
\begin{gathered}
2 x_2-\frac{3}{2} t-2 t=0 \quad \Rightarrow 2 x_2=2 t+\frac{3}{2} t=\frac{7}{2} t \\
x_2=\frac{7}{4} t
\end{gathered}
$
Now
$
x_1=\frac{t}{2}, x_2=\frac{7}{4} t, x_3=\frac{3}{2} t \text { and } x_4=t
$
Since $x_1, x_2, x_3$ and $x_4$ are positive integers. Let us choose $t=4 t$.
Then we get $x_1=2, x_2=7, x_3=6$, and $x_4=4$
So the balanced equation is $2 \mathrm{C}_2 \mathrm{H}_6+7 \mathrm{O}_2 \rightarrow 6 \mathrm{H}_2 \mathrm{O}+4 \mathrm{CO}_2$.