Exercise 1.6-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.
Discuss the solutions of the system of equations for all values of $\lambda$.
$
\begin{aligned}
& x+y+z=2 \\
& 2 x+y-2 z=2 \\
& \lambda x+y+4 z=2
\end{aligned}
$
Solution:
The augmented matrix is
$
\begin{aligned}
& {[A, B]=\left[\begin{array}{rrrr}
1 & 1 & 1 & 2 \\
2 & 1 & -2 & 2 \\
\lambda & 1 & 4 & 2
\end{array}\right]} \\
& \mathrm{C}_1 \leftrightarrow \mathrm{C}_2 \sim\left[\begin{array}{rrrr}
1 & 1 & 1 & 2 \\
1 & 2 & -2 & 2 \\
1 & \lambda & 4 & 2
\end{array}\right] \sim\left[\begin{array}{rrrr}
0 & -1 & 3 & 0 \\
0 & 2-\lambda & -6 & 0 \\
1 & \lambda & 4 & 2
\end{array}\right] \begin{array}{l}
\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\end{array} \\
& \sim\left[\begin{array}{rrrr}
1 & \lambda & 4 & 2 \\
0 & 2-\lambda & -6 & 0 \\
0 & -1 & 3 & 0
\end{array}\right] \mathrm{R}_1 \leftrightarrow \mathrm{R}_3 \\
& \sim\left[\begin{array}{rrrr}
1 & \lambda & 4 & 2 \\
0 & 2-\lambda & -6 & 0 \\
0 & -\lambda & 0 & 0
\end{array}\right] \mathrm{R}_3 \rightarrow 2 \mathrm{R}_3+\mathrm{R}_2 \\
& \sim\left[\begin{array}{rrrr}
1 & 0 & 4 & 2 \\
0 & 2-\lambda & -6 & 0 \\
0 & -\lambda & 0 & 0
\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_3 \\
&
\end{aligned}
$

Case 1:
If $\lambda \neq 0$, then we get $\rho(A, B)=\rho(A)=3=$ number of unknowns.
$\therefore$ The system has unique solution.
Case 2: If $\lambda=0$, then $\rho(A, B)=\rho(A)=2<$ the number of unknowns $\Rightarrow$ the system has more number of solutions. Taking $\lambda=0$ we get the systems of equations as
$
\begin{aligned}
& x+4 z=2 \\
& 2 y-6 z=0
\end{aligned}
$
Taking $z=k$ in (2), we get
$
\begin{aligned}
& 2 \mathrm{y}-6 \mathrm{k}=0 \Rightarrow 2 \mathrm{y}=6 \mathrm{k} \\
& \mathrm{y}=3 \mathrm{k}
\end{aligned}
$
Taking $\mathrm{z}=\mathrm{A}$ in (1) we get,
$
\mathrm{x}+4 \mathrm{k}=2 \Rightarrow \mathrm{x}=2-4 \mathrm{k}
$
$\therefore$ The solution is $\mathrm{x}=2-4 \mathrm{k}, \mathrm{y}=3 \mathrm{k}, \mathrm{z}=\mathrm{k}$
Question 2.
For what values of $k$, the system of equations $\mathrm{kx}+\mathrm{y}+\mathrm{z}=1, \mathrm{x}+\mathrm{ky}+\mathrm{z}=1, \mathrm{x}+\mathrm{y}+\mathrm{kz}=1$ have
(i) unique solution,
(ii) more than one solution,
(iii) no solution.
Solution:
The augmented matrix is
$
\begin{aligned}
{[\mathrm{A}, \mathrm{B}] } & =\left[\begin{array}{rrrr}
k & 1 & 1 & 1 \\
1 & k & 1 & 1 \\
1 & 1 & k & 1
\end{array}\right] \sim\left[\begin{array}{rrrr}
k & 1 & 1 & 1 \\
0 & k-1 & 1-k & 0 \\
1-k & 0 & k-1 & 0
\end{array}\right] \mathrm{R}_2 \rightarrow \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_3 \\
& \sim\left[\begin{array}{rrrr}
k & 1 & 1 & 1 \\
0 & k-1 & 1-k & 0 \\
1-k & 0 & 0 & 0
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_2
\end{aligned}
$
Case (i): When $\mathrm{k} \neq 1$, the system has a unique solution
Case (ii): When $\mathrm{k}=1$, the system reduces to a single equation $\mathrm{x}+\mathrm{y}+\mathrm{z}=1$. The system can have more than one solution.
Case (iii) : When $k=-2$, the system is inconsistent
$\therefore \rho(A, B) \neq \rho(A) ; \therefore$ the system has no solution.