Exercise 1.3-Additional Problems - 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

Additional Problems
Question 1.
Using matrix method, solve the following system of equations:
$
x+2 y+z=7, x+3 z=11,2 x-3 y=1
$
Solution:
The system of equations can be written in the form $\mathrm{AX}=\mathrm{B}$, where
$
\mathrm{A}=\left[\begin{array}{ccc}
1 & 2 & 1 \\
1 & 0 & 3 \\
2 & -3 & 0
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}
7 \\
11 \\
1
\end{array}\right]
$
Now
$
|A|=\left|\begin{array}{ccc}
1 & 2 & 1 \\
1 & 0 & 3 \\
2 & -3 & 0
\end{array}\right|=1(0+9)-2(0-6)+1(-3-0)
$
$=9+12-3=18 \neq 0$ Non-singular matrix.
$A_{11}=(0+9)=9, A_{12}=-(0-6)=6$,
$A_{13}=(-3-0)=-3$
$A_{21}=-(0+3)=-3, A_{22}=(0-2)=-2$
$A_{23}=-(-3-4)=7$
$A_{31}=(6-0)=6, A_{32}=-(-3-1)=-2$,
$A_{33}=(0-2)=-2$
$\operatorname{adj} A=\left[\begin{array}{rrr}9 & 6 & -3 \\ -3 & -2 & 7 \\ 6 & -2 & -2\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{rrr}9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{18}\left[\begin{array}{ccc}9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2\end{array}\right]$
$\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$
$
\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{18}\left[\begin{array}{ccc}
9 & -3 & 6 \\
6 & -2 & -2 \\
-3 & 7 & -2
\end{array}\right]\left[\begin{array}{c}
7 \\
11 \\
1
\end{array}\right]=\frac{1}{18}\left[\begin{array}{c}
63-33+6 \\
42-22-2 \\
-21+77-2
\end{array}\right]=\frac{1}{18}\left[\begin{array}{l}
36 \\
18 \\
54
\end{array}\right]=\left[\begin{array}{l}
2 \\
1 \\
3
\end{array}\right]
$

Thus $x=2, y=1$ and $z=3$.
Question 2.
Using matrices, solve the following system of linear equations: $x+2 y-3 z=-4,2 x+3 y+2 z=2,3 x-3 y$ $-4 z=11$.
Solution:
The system of equations can be written in the form $\mathrm{AX}=\mathrm{B}$, where
$
\begin{aligned}
& \mathrm{A}=\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right], \mathrm{X}=\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right] \\
& |A|=\left|\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right|=1(-12+6)-2(-8-6)-3(-6-9) \\
& =-6+28+45=67 \neq 0 \text { Non-singular matrix } \\
& A_{11}=(-12+6)=-6, A_{12}=-(-8-6)=14, A_{13}=(-6-9)=-15 \\
& A_{21}=-(-8-9)=17, A_{22}=(-4+9)=5, A_{23}=-(-3-6)=9 \\
& A_{31}=(4+9)=13, A_{32}=-(2+6)=-8, A_{33}=(3-4)=-1 \\
& \operatorname{adj} A=\left[\begin{array}{ccc}
-6 & 14 & -15 \\
17 & 5 & 9 \\
13 & -8 & -1
\end{array}\right]^T=\left[\begin{array}{ccc}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{67}\left[\begin{array}{ccc}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right] \\
& \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\
& {\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{67}\left[\begin{array}{ccc}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right]} \\
& =\frac{1}{67}\left[\begin{array}{c}
24+34+143 \\
-56+10-88 \\
60+18-11
\end{array}\right] \\
& =\frac{1}{67}\left[\begin{array}{c}
201 \\
-134 \\
67
\end{array}\right]=\left[\begin{array}{c}
3 \\
-2 \\
1
\end{array}\right] \\
&
\end{aligned}
$

Thus $\mathrm{x}=3$,
$
y=-2 \text { and } z=1
$
Question 3 .
$
\mathbf{A}=\left(\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right),
$
Find $A^{-1}$. Hence using $A^{-1}$ solve the system of equations.
$
\begin{aligned}
& 2 x-3 y+5 z=11 \\
& 3 x+2 y-4 z=-5 \\
& x+y-2 z=-3
\end{aligned}
$
Solution:
$
\begin{aligned}
& A=\left(\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right) \\
& \Rightarrow \quad|A|=2(0)+3(-2)+5(1)=-1 \neq 0 \\
& A_{11}=0, A_{12}=2, A_{13}=1 \\
& A_{21}=-1, A_{22}=-9, A_{23}=-5 \\
& A_{31}=2, A_{32}=+23, A_{33}=1.3 \\
& \Rightarrow \quad A^{-1}=-1\left(\begin{array}{ccc}
0 & 2 & 1 \\
-1 & -9 & -5 \\
2 & 23 & 13
\end{array}\right)^{\mathrm{T}}=-1\left(\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right)=\left(\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right) \\
& \text { Given equations can be written as }\left(\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right) \\
& \text { or } \quad \mathrm{AX}=\mathrm{B} \quad \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\
& \Rightarrow \quad\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right)\left(\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right)=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right] \\
& \Rightarrow \quad x=1, y=2 \text { and } z=3 \\
&
\end{aligned}
$

Question 4.
Use product $\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{rrr}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations
$
\begin{aligned}
& x+3 z=9 \\
& -x+2 y-2 z=4 \\
& 2 x-3 y+4 z=-3
\end{aligned}
$
Solution:
$
\begin{aligned}
& \left(\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right)\left(\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right)=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) \\
& \mathrm{AB}=\mathrm{I} \quad \Rightarrow \mathrm{B}=\mathrm{A}^{-1} \\
& \left(\begin{array}{rrr}
1 & 0 & 3 \\
-1 & 2 & -2 \\
2 & -3 & 4
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
9 \\
4 \\
-3
\end{array}\right) \\
&
\end{aligned}
$
Writing the given system of equations in matrix form, we get
$
\begin{aligned}
\mathrm{A}^{\prime} \mathrm{X} & =\mathrm{C} \\
\mathrm{X} & =\left(\mathrm{A}^{\prime}\right)^{-1} \mathrm{C}=\left(\mathrm{A}^{-1}\right)^{\prime} \mathrm{C}=\mathrm{B}^{\prime} \mathrm{C} \\
\mathrm{X} & =\left(\begin{array}{ccc}
-2 & 9 & 6 \\
0 & 2 & 1 \\
1 & -3 & -2
\end{array}\right)\left(\begin{array}{c}
9 \\
4 \\
-3
\end{array}\right)=\left(\begin{array}{c}
-18+36-18 \\
0+8-3 \\
9-12+6
\end{array}\right) \\
x & =0, y=5 \text { and } z=3
\end{aligned}
$

Question 5.
Use product $\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]_{\text {to solve the system of equation }}$
$
\begin{aligned}
& -y+2 z=1 \\
& 2 y-3 z=1 \\
& x-2 y+4 z=2
\end{aligned}
$
Solution:

Let
$
\begin{aligned}
& A=\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \text { and } B=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right] \\
& A B=\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right] \\
& =\left[\begin{array}{ccc}
-2-9+12 & 0-2+2 & 1+3-4 \\
0+18-18 & 0+4-3 & 0-6+6 \\
-6-18+24 & 0-4+4 & 3+6-8
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \therefore \quad \mathrm{AB}=\mathrm{I}_3 \quad \therefore \mathrm{A}^{-1}=\mathrm{B} \\
& A^{-1}=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right] \\
&
\end{aligned}
$
The system of equations can be written in the form $\mathrm{AX}=\mathrm{C}$, where
$
\begin{aligned}
\mathrm{A} & =\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right) \text { and } \mathrm{C}=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
\mathrm{X} & =\mathrm{A}^{-1} \mathrm{C} \\
\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
-2+0+2 \\
9+2-6 \\
6+1-4
\end{array}\right]=\left[\begin{array}{l}
0 \\
5 \\
3
\end{array}\right]
\end{aligned}
$
Question 6.
An amount of $₹ 7000$ is invested in three types of investments $x, y$ and $z$ at the rate of $3 \%, 4 \%$ and $5 \%$ interest respectively. The total annual income is ₹ 280 . If the combined income from $\mathrm{x}$ and $\mathrm{y}$ is ₹ 80 more than that from $z$, then
(i) Represent the above situation in form of linear equations .
(ii) Is it possible to frame the given linear equations in the form of matrix to obtain the three values $\mathrm{x}, \mathrm{y}$ and $z$ using matrix multiplication? If yes, find.
(iii) Which value is more beneficial to invest?
Solution:

$
\begin{aligned}
& \text { (i) } \quad x+y+z=7000 \\
& \Rightarrow \quad \frac{3 x}{100}+\frac{4 y}{100}+\frac{5 z}{100}=280 \\
& \Rightarrow \quad 3 x+4 y+5 z=28000 \\
& \frac{3 x}{100}+\frac{4 y}{100}=\frac{5 z}{100}+80 \\
& \Rightarrow \quad 3 x+4 y-5 z=8000 \\
&
\end{aligned}
$
(ii) Yes, it is possible to frame the given equation in matrix form.
Let
$
\mathrm{A}=\left[\begin{array}{rrr}
1 & 1 & 1 \\
3 & 4 & 5 \\
3 & 4 & -5
\end{array}\right], \mathrm{X}=\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) \text { and } \mathrm{C}=\left[\begin{array}{r}
7000 \\
28000 \\
8000
\end{array}\right]
$
$
\begin{aligned}
& |\mathrm{A}|=1\left|\begin{array}{cc}
4 & 5 \\
4 & -5
\end{array}\right|-1\left|\begin{array}{cc}
3 & 5 \\
3 & -5
\end{array}\right|+1\left|\begin{array}{cc}
3 & 4 \\
3 & 4
\end{array}\right| \\
& \Rightarrow-40+30+0=-10 \neq 0 \\
& \text { Cofactors } C_{11}=+(-20-20)=-40 \\
& \mathrm{C}_{21}=-(5-4)=9 \\
& \mathrm{C}_{31}=+(5-4)=1 \\
& \mathrm{C}_{12}=+(-15-15)=-30 \\
& \mathrm{C}_{22}=+(-5-3)=-8 \\
& \mathrm{C}_{32}=-(5-3)=-2 \\
& \mathrm{C}_{13}=+(12-12)=0 \\
& C_{23}=-(4-3)=-1 \\
& C_{33}=+(4-3)=1 \\
& \therefore \quad\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{-1}{10}\left[\begin{array}{rrr}
-40 & 9 & 1 \\
30 & -8 & -2 \\
0 & -1 & 1
\end{array}\right]\left[\begin{array}{r}
7000 \\
28000 \\
8000
\end{array}\right] \\
& {\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{-1}{10}\left[\begin{array}{r}
-280000+252000+8000 \\
210000-224000-16000 \\
0-28000+8000
\end{array}\right]=\frac{-1}{10}\left[\begin{array}{l}
-20000 \\
-30000 \\
-20000
\end{array}\right]} \\
&
\end{aligned}
$

Thus, the three values $x=₹ 2000, y=₹ 3000$ and $z=₹ 2000$

(iii) The value of $y$ is more beneficial to invest

Question 7.
If $\mathbf{A}=\left(\begin{array}{rrr}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right)$, find $A^{-1}$. Using $A^{-1}$ solve the system of equations.
$
\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=2 ; \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=5 ; \frac{6}{x}+\frac{9}{y}+\frac{20}{z}=-4
$
Solution:
$
\begin{aligned}
& A=\left(\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right) \quad \Rightarrow|\mathrm{A}|=\left|\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right| \\
& =2(120-45)-3(-80-30)+10(36+36)=2(75)-3(-110)+10(72) \\
& =150+330+720=1200 \neq 0 \\
& \mathrm{C}_{i j}=(-1)^{i+j} \mathrm{M}_{i j} \\
& \mathrm{C}_{11}=75 \quad \mathrm{C}_{21}=150 \quad \mathrm{C}_{31}=75 \\
& \mathrm{C}_{12}=110 \quad \mathrm{C}_{22}=-100 \quad \mathrm{C}_{32}=30 \\
& \mathrm{C}_{13}=72 \quad \mathrm{C}_{23}=0 \quad \mathrm{C}_{33}=-24 \\
& \operatorname{adj} A=\left(\begin{array}{ccc}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right) \\
& A^{-1}=\frac{\operatorname{adjA}}{|A|}=\frac{1}{1200}\left(\begin{array}{rrr}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right) \\
& \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=2, \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=5, \frac{6}{x}+\frac{9}{y}+\frac{20}{z}=-4 \\
&
\end{aligned}
$
In matrix form
$
\left(\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right)\left(\begin{array}{l}
\frac{1}{x} \\
\frac{1}{y} \\
\frac{1}{z}
\end{array}\right)=\left(\begin{array}{r}
2 \\
5 \\
-4
\end{array}\right)
$