Exercise 1.4-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 non-homogeneous system of linear equations by determinant method: $3 x+2 y=5$, $x$ $+3 y=4$.
Solution:
Now
$
\Delta=\left|\begin{array}{ll}
3 & 2 \\
1 & 3
\end{array}\right|=9-2=7 \neq 0
$
$\because \quad \Delta \neq 0$, the system has unique solution.
$
\begin{aligned}
\Delta_x & =\left|\begin{array}{ll}
5 & 2 \\
4 & 3
\end{array}\right|=15-8=7 \\
\Delta_y & =\left|\begin{array}{ll}
3 & 5 \\
1 & 4
\end{array}\right|=12-5=7 \\
x & =\frac{\Delta_x}{\Delta}=\frac{7}{7}=1 \\
y & =\frac{\Delta_y}{\Delta}=\frac{7}{7}=1
\end{aligned}
$
Solution is $(\mathrm{x}, \mathrm{y})=(1,1)$.
Question 2 .
Solve the following non-homogeneous system of linear equations by determinant method: $x+y+z=4$; $\mathrm{x}-\mathrm{y}+\mathrm{z}=2 ; 2 \mathrm{x}+\mathrm{y}-\mathrm{z}=1$
Solution:
$
\Delta=\left|\begin{array}{rrr}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|=1(1-1)-1(-1-2)+1(1+2)=0+3+3=6 \neq 0
$
Since $\Delta \neq 0$, the system has unique solution

Now
$
\text { Now } \begin{aligned}
\Delta_x & =\left|\begin{array}{rrr}
4 & 1 & 1 \\
2 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|=4(1-1)-1(-2-1)+1(2+1) \\
& =0+3+3=6 \\
\Delta_y & =\left|\begin{array}{rrr}
1 & 4 & 1 \\
1 & 2 & 1 \\
2 & 1 & -1
\end{array}\right|=1(-2-1)-4(-1-2)+1(1-4) \\
& =-3+12-3=6 \\
\Delta_z & =\left|\begin{array}{lrr}
1 & 1 & 4 \\
1 & -1 & 2 \\
2 & 1 & 1
\end{array}\right|=1(-1-2)-1(1-4)+4(1+2) \\
& =-3+3+12=12 \\
x & =\frac{\Delta_x}{\Delta}=\frac{6}{6}=1 ; \\
y & =\frac{\Delta_y}{\Delta} \frac{6}{6}=1 ; \\
z & =\frac{\Delta_z}{\Delta}=\frac{12}{6}=2
\end{aligned}
$
Solution is $\mathrm{x}, \mathrm{y}, \mathrm{z}=1,1,2$
Question 3.
Solve the following non-homogeneous system of linear equations by determinant method: $3 x+y-z=2$; $2 x-y+2 z=6 ; 2 x+y-2 z=-2$
Solution:

$
\Delta=\left|\begin{array}{rrr}
3 & 1 & -1 \\
2 & -1 & 2 \\
2 & 1 & -2
\end{array}\right|=3(0)-1(-8)-1(4)=8-4=4
$
Since $\Delta \neq 0$, the system has unique solution.
$
\begin{aligned}
\Delta_x & =\left|\begin{array}{rrr}
2 & 1 & -1 \\
6 & -1 & 2 \\
-2 & 1 & -2
\end{array}\right|=2(0)-1(-8)-1(4)=8-4=4 \\
\Delta_y & =\left|\begin{array}{rrr}
3 & 2 & -1 \\
2 & 6 & 2 \\
2 & -2 & -2
\end{array}\right|=3(-8)-2(-8)-1(-16)=-24+16+16=8 \\
\Delta_z & =\left|\begin{array}{rrr}
3 & 1 & 2 \\
2 & -1 & 6 \\
2 & 1 & -2
\end{array}\right|=3(-4)-1(-16)+2(4)=-12+16+8=12 \\
\therefore \quad x & =\frac{\Delta_x}{\Delta}=\frac{4}{4}=1 ; y=\frac{\Delta_y}{\Delta} \frac{8}{4}=2 ; z=\frac{\Delta_z}{\Delta}=\frac{12}{4}=3
\end{aligned}
$
Solution is $\mathrm{x}, \mathrm{y}, \mathrm{z}=1,2,3$
Question 4.
Solve the following non-homogeneous system of linear equations by determinant method:
$
\frac{1}{x}+\frac{2}{y}-\frac{1}{z}=1 ; \frac{2}{x}+\frac{4}{y}+\frac{1}{z}=5 ; \frac{3}{x}-\frac{2}{y}-\frac{2}{z}=0 .
$
Solution:
Taking $\frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c$, we get
$
\begin{aligned}
& \mathrm{a}+2 \mathrm{~b}-\mathrm{c}=1 \text {... (1) } \\
& 2 \mathrm{a}+4 \mathrm{~b}+\mathrm{c}=5 \\
& 3 \mathrm{a}-2 \mathrm{~b}-2 \mathrm{c}=0 \\
& \Delta=\left|\begin{array}{rrr}
1 & 2 & -1 \\
2 & 4 & 1 \\
3 & -2 & -2
\end{array}\right|=1(-6)-2(-7)-1(-16)=-6+14+16=24 \neq 0 \\
&
\end{aligned}
$

$\therefore$ The system has unique solution.
$
\begin{aligned}
& \Delta_a=\left|\begin{array}{rrr}
1 & 2 & -1 \\
5 & 4 & 1 \\
0 & -2 & -2
\end{array}\right|=1(-6)-5(-6)=-6+30=24 \\
& \Delta_b=\left|\begin{array}{rrr}
1 & 1 & -1 \\
2 & 5 & 1 \\
3 & 0 & -2
\end{array}\right|=3(6)-0(3)-2(3)=18-6=12 \\
& \Delta_c=\left|\begin{array}{rrr}
1 & 2 & 1 \\
2 & 4 & 5 \\
3 & -2 & 0
\end{array}\right|=1(-16)-5(-8)=-16+40=24 \\
& \Delta=24, \Delta_a=24, \Delta_b=12, \Delta_c=24 \\
& \therefore \quad a=\frac{\Delta_a}{\Delta}=\frac{24}{24}=1 ; b=\frac{\Delta_b}{\Delta}=\frac{12}{24}=\frac{1}{2} ; c=\frac{\Delta_c}{\Delta}=\frac{24}{24}=1 \text {. } \\
& \text { but } \\
& \frac{1}{x}=a \quad \Rightarrow x=\frac{1}{a}=\frac{1}{1}=1 \\
& \frac{1}{y}=b \quad \Rightarrow y=\frac{1}{b}=\frac{1}{\frac{1}{2}}=2 \\
& \frac{1}{z}=c \quad \Rightarrow z=\frac{1}{c}=\frac{1}{1}=1 \\
&
\end{aligned}
$
$\therefore$ Solution is $\mathrm{x}, \mathrm{y}, \mathrm{z}=1,2,1$.