Exercise 1.7-Additional Problems - Chapter 1 - Applications of Matrices and Determinants - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Solve the following homogeneous linear equations.
$
\begin{aligned}
& x+2 y-5 z=0 \\
& 3 x+4 y+6 z=0 \\
& x+y+z=0
\end{aligned}
$
Solution:
The given system of equations can be written in the form of matrix equation
$
\left[\begin{array}{rrr}
1 & 2 & -5 \\
3 & 4 & 6 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
$

$\mathrm{AX}=\mathrm{B}$
The augmented matrix is
$
\begin{aligned}
& {[A, B]=\left[\begin{array}{cccc}
1 & 2 & -5 & 0 \\
3 & 4 & 6 & 0 \\
1 & 1 & 1 & 0
\end{array}\right]} \\
& \sim\left[\begin{array}{cccc}
1 & 2 & -5 & 0 \\
0 & -2 & 21 & 0 \\
0 & -1 & 6 & 0
\end{array}\right] \begin{array}{llll}
R_2 & \rightarrow & R_2 & -3 R_1 \\
R_3 & \rightarrow & R_3 & -R_1
\end{array} \\
& \sim\left[\begin{array}{cccc}
1 & 2 & -5 & 0 \\
0 & -1 & 6 & 0 \\
0 & -2 & 21 & 0
\end{array}\right] R_2 \leftrightarrow R_3 \\
& \sim\left[\begin{array}{cccc}
1 & 2 & -5 & 0 \\
0 & -1 & 6 & 0 \\
0 & 0 & 9 & 0
\end{array}\right] R_3 \rightarrow R_3-2 R_2 \\
&
\end{aligned}
$
This is in the echelon form.
Clearly $\rho[\mathrm{A}, \mathrm{B}]=3$ and $\rho(\mathrm{A})=3$
$\therefore \rho(\mathrm{A})=\rho[\mathrm{A}, \mathrm{B}]=3=$ number of unknowns
$\therefore$ The given system of equations is consistent and has a unique solution. i.e., trivial solution
$\therefore \mathrm{x}=0, \mathrm{y}=0$ and $\mathrm{z}=0$
Note: Since $\rho(\mathrm{A})=3,|\mathrm{~A}| \neq 0$ i.e. $\mathrm{A}$ is non-singular;
$\therefore$ The given system has only trivial solution $\mathrm{x}=0, \mathrm{y}=0, \mathrm{z}=0$
Question 2.
For what value of $\mathrm{n}$ the equations.
$
\begin{aligned}
& x+y+3 z=0 \\
& 4 x+3 y+\mu z=0 \\
& 2 x+y+2 z=0 \text { have a }
\end{aligned}
$
(i) trivial solution,
(ii) non-trivial solution.
Solution:
The system of equations can be written as $A X=B$
$
\left[\begin{array}{lll}
1 & 1 & 3 \\
4 & 3 & \mu \\
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]
$

$
\begin{aligned}
& \sim\left[\begin{array}{cccc}
1 & 1 & 3 & 0 \\
0 & -1 & \mu-12 & 0 \\
0 & 0 & 8-\mu & 0
\end{array}\right] R_3 \rightarrow R_3-R_2 \\
&
\end{aligned}
$
Case (i): If $\mu \neq 8$ then $8-\mu \neq 0$ and hence there are three non-zero rows.
$\therefore \rho[\mathrm{A}]=\rho[\mathrm{A}, \mathrm{B}]=3=$ the number of unknowns.
$\therefore$ The system has the trivial solution $\mathrm{x}=0, \mathrm{y}=0, \mathrm{z}=0$
Case (ii): If $\mu=8$ then.
$\rho[\mathrm{A}, \mathrm{B}]=2$ and $\rho(\mathrm{A})=2$
$\therefore \rho(\mathrm{A})=\rho[\mathrm{A}, \mathrm{B}]=2<$ number of unknowns.
The given system is equivalent to
$
\begin{aligned}
& \mathrm{x}+\mathrm{y}+3 \mathrm{z}=0 ; \mathrm{y}+4 \mathrm{z}=0 \\
& \therefore \mathrm{y}=-4 \mathrm{z} ; \mathrm{x}=\mathrm{z}
\end{aligned}
$
Taking $\mathrm{z}=\mathrm{k}$, we get $\mathrm{x}=\mathrm{k}, \mathrm{y}=-4 \mathrm{k}, \mathrm{z}=\mathrm{k}[\mathrm{k} \in \mathrm{R}-\{0\}]$ which are non-trivial solutions.
Thus the system is consistent and has infinitely many non-trivial solutions.
Note: In case (ii) the system also has trivial solution. For only non-trivial solutions we removed $k=0$.