Exercise 1.3 - 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

$\mathbf{E x} 1.3$
Question 1.
Solve the following system of linear equations by matrix inversion method:
(i) $2 x+5 y=-2, x+2 y=-3$
(ii) $2 x-y=8,3 x+2 y=-2$
(iii) $2 x+3 y-z=9, x+y+z=9,3 x-y-z=-1$
(iv) $x+y+z-2=0,6 x-4 y+5 z-31=0,5 x+2 y+2 z=13$
Solution:
(i) $2 x+5 y=-2, x+2 y=-3$
The matrix form of the above equations is $\left(\begin{array}{ll}2 & 5 \\ 1 & 2\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}-2 \\ -3\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B}$
Where
Now
$
\begin{aligned}
\mathrm{A} & =\left(\begin{array}{ll}
2 & 5 \\
1 & 2
\end{array}\right) ; \mathrm{X}=\left(\begin{array}{l}
x \\
y
\end{array}\right) \text { and } \mathrm{B}=\left(\begin{array}{l}
-2 \\
-3
\end{array}\right) \\
\mathrm{AX} & =\mathrm{B}
\end{aligned}
$

(ii) $2 x-y=8 ; 3 x+2 y=-2$
The above equations in matrix form is $\left(\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}8 \\ -2\end{array}\right)$ (i.e) $\quad \mathrm{AX}=\mathrm{B}$
Where
Now
$\mathrm{A}=\left(\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right) ; \mathrm{X}=\left(\begin{array}{l}x \\ y\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{c}8 \\ -2\end{array}\right)$
To find $\mathbf{A}^{-1}$ :
$
A X=B \quad \Rightarrow X=A^{-1} B
$
$|A|=\left|\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right|=4+3=7$
$
\begin{array}{rlrl} 
& & \operatorname{adj} \mathrm{A} & =\left(\begin{array}{cc}
2 & 1 \\
-3 & 2
\end{array}\right) \\
\therefore & \mathrm{A}^{-1} & =\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{7}\left(\begin{array}{cc}
2 & 1 \\
-3 & 2
\end{array}\right) \\
\text { Now } & \left(\begin{array}{l}
x \\
y
\end{array}\right) & =\mathrm{A}^{-1} \mathrm{~B}=\frac{1}{7}\left(\begin{array}{cc}
2 & 1 \\
-3 & 2
\end{array}\right) \mid\left(\begin{array}{c}
8 \\
-2
\end{array}\right)=\frac{1}{7}\left(\begin{array}{c}
16-2 \\
-24-4
\end{array}\right)=\frac{1}{7}\left(\begin{array}{c}
14 \\
-28
\end{array}\right)=\left(\begin{array}{c}
2 \\
-4
\end{array}\right) \\
\Rightarrow & x=2 ; y=-4
\end{array}
$
(iii) $2 x+3 y-z=9 ; x+y+z=9 ; 3 x-y-z=-1$
The above equations in matrix form is $\left(\begin{array}{ccc}2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)\left(\begin{array}{c}9 \\ 9 \\ -1\end{array}\right)$
(i.e)
$
A X=B \quad \Rightarrow X=\mathrm{A}^{-1} \mathrm{~B}
$
To find $A^{-1}$ :
$
\begin{aligned}
A & =\left(\begin{array}{ccc}
2 & 3 & -1 \\
1 & 1 & 1 \\
3 & -1 & -1
\end{array}\right) \\
|A| & =\left|\begin{array}{ccc}
2 & 3 & -1 \\
1 & 1 & 1 \\
3 & -1 & -1
\end{array}\right|=2(-1+1)-3(-1-3)-1(-1-3) \\
& =2(0)-3(-4)-1(-4)=12+4=16
\end{aligned}
$

(iv) $x+y+z-2=0, \quad 6 x-4 y+5 z-31=0, \quad 5 x+2 y+2 z=13$
The given equations are $x+y+z=2 ; 6 x-4 y+5 z=31 ; 5 x+2 y+2 z=13$.
The matrix form of the above equations is $\left(\begin{array}{ccc}1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}2 \\ 31 \\ 13\end{array}\right)$
(i.e)
$
\begin{aligned}
A X & =B \\
A & =\left(\begin{array}{ccc}
1 & 1 & 1 \\
6 & -4 & 5 \\
5 & 2 & 2
\end{array}\right) \quad \Rightarrow X=\mathrm{A}^{-1} \mathrm{~B} \\
|\mathrm{~A}| & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
6 & -4 & 5 \\
5 & 2 & 2
\end{array}\right|=1(-8-10)-1(12-25)+1(12+20) \\
& =1(-18)-1(-13)+32=-18+13+32=27
\end{aligned}
$

Question 2.
If $\mathrm{A}=\left[\begin{array}{ccc}-5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{lll}1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3\end{array}\right]$ find the products $\mathrm{AB}$ and $\mathrm{BA}$ and hence solve the system of equations $\mathrm{x}+\mathrm{y}+2 \mathrm{z}=1,3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}=7,2 \mathrm{x}+\mathrm{y}+3 \mathrm{z}=2$.
Solution:

Writing the given system of equations is matrix form, we get $\left(\begin{array}{lll}1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 7 \\ 2\end{array}\right)$ $\qquad x)(1)$
That is
$
\mathrm{B}\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
1 \\
7 \\
2
\end{array}\right)
$
$
\begin{aligned}
\Rightarrow \quad\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\mathrm{B}^{-1}\left(\begin{array}{l}
1 \\
7 \\
2
\end{array}\right)=\frac{1}{4}\left(\begin{array}{ccc}
-5 & 1 & 3 \\
7 & 1 & -5 \\
1 & -1 & 1
\end{array}\right)\left(\begin{array}{l}
1 \\
7 \\
2
\end{array}\right)\left(\therefore \mathrm{B}^{-1}=\frac{1}{4} \mathrm{~A}\right) \\
& =\frac{1}{4}\left(\begin{array}{c}
-5+7+6 \\
7+7-10 \\
1-7+2
\end{array}\right)=\frac{1}{4}\left(\begin{array}{r}
8 \\
4 \\
-4
\end{array}\right)=\left(\begin{array}{r}
2 \\
1 \\
-1
\end{array}\right) \\
\Rightarrow \quad x & =2 ; y=1 ; z=-1
\end{aligned}
$
Question 3.
A man is appointed in a job with a monthly salary of certain amount and a fixed amount of annual increment. If his salary was ₹ 19,800 per month at the end of the first month after 3 years of service and ₹ 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use matrix inversion method to solve the problem.)
Solution:
Let his monthly salary be ₹ $\mathrm{x}$ and his annual increment be ₹ $\mathrm{y}$
Given $x+3 y=19800$ and $x+9 y=23400$
Writing the above equations in matrix form, we get

$
\begin{aligned}
& \left(\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=\left(\begin{array}{l}
19800 \\
23400
\end{array}\right) \\
& \mathrm{AX}=\mathrm{B} \quad \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\
& \mathrm{~A}=\left(\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right) ; \mathrm{X}=\left(\begin{array}{l}
x \\
y
\end{array}\right) \text { and } \mathrm{B}=\left(\begin{array}{l}
19800 \\
23400
\end{array}\right) \\
& A=\left(\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right) \quad \Rightarrow|A|=\left|\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right|=9-3=6 \\
& \operatorname{adj} A=\left(\begin{array}{cc}
9 & -3 \\
-1 & 1
\end{array}\right) \\
&
\end{aligned}
$

$\begin{aligned}
& \operatorname{adj} A=\left(\begin{array}{cc}
9 & -3 \\
-1 & 1
\end{array}\right) \\
& \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{6}\left(\begin{array}{cc}
9 & -3 \\
-1 & 1
\end{array}\right) \\
& X=A^{-1} B=\frac{1}{6}\left(\begin{array}{cc}
9 & -3 \\
-1 & 1
\end{array}\right)\left(\begin{array}{c}
19800 \\
23400
\end{array}\right)=\left(\begin{array}{cc}
9 & -3 \\
-1 & 1
\end{array}\right)\left(\begin{array}{c}
\frac{19800}{6} \\
\frac{23400}{6}
\end{array}\right) \\
& =\overrightarrow{\left(\begin{array}{cc}
9 & -3 \\
-1 & 1
\end{array}\right)} \mid\left(\begin{array}{l}
3300 \\
3900
\end{array}\right)=\left(\begin{array}{c}
29700-11700 \\
-3300+3900
\end{array}\right)=\left(\begin{array}{c}
18000 \\
600
\end{array}\right) \\
& \mathrm{x}=₹ 18000 \text { and } \mathrm{y}=₹ 600 \\
& \text { (i.e) monthly salary }=₹ 18000 \text { and annual increment }=₹ 600 \\
&
\end{aligned}$

Question 4.
4 men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the same work by using matrix inversion method.
Solution:
Let the work done by man in 1 day be $\mathrm{x}$ and the work done by a woman in 1 day be $\mathrm{y}$
Now we are given
$4 \mathrm{x}+4 \mathrm{y}=\frac{1}{3} \Rightarrow 12 \mathrm{x}+12 \mathrm{y}=1$
and $2 x+5 y=\frac{1}{4} \Rightarrow 8 x+20 y=1$
The matrix form of the above equations is $\left(\begin{array}{cc}12 & 12 \\ 8 & 20\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}1 \\ 1\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B} \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$
Here

$
\begin{aligned}
\mathrm{A} & =\left(\begin{array}{cc}
12 & 12 \\
8 & 20
\end{array}\right) ;|\mathrm{A}|=\left|\begin{array}{cc}
12 & 12 \\
8 & 20
\end{array}\right|=240-96=144 \\
\operatorname{adj} \mathrm{A} & =\left(\begin{array}{rr}
20 & -12 \\
-8 & 12
\end{array}\right) \\
\mathrm{A}^{-1} & =\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{144}\left(\begin{array}{rr}
20 & -12 \\
-8 & 12
\end{array}\right) \\
\mathrm{X} & =\left(\begin{array}{l}
x \\
y
\end{array}\right)=\mathrm{A}^{-1} \mathrm{~B}=\frac{1}{144}\left(\begin{array}{rr}
20 & 12 \\
8 & 12
\end{array}\right)\left(\begin{array}{l}
1 \\
1
\end{array}\right) \\
& =\frac{1}{144}\left(\begin{array}{c}
20-12 \\
-8+12
\end{array}\right)=\frac{1}{144}\left(\begin{array}{l}
8 \\
4
\end{array}\right)=\left(\begin{array}{c}
18 \\
\frac{1}{36}
\end{array}\right) \quad \Rightarrow x=\frac{1}{18} ; y=\frac{1}{36}
\end{aligned}
$
(i.e) The work done by a man in 1 day $=x=\frac{1}{18}$
The time taken by a man to finish the work $=18$ days.
The work done by a woman in 1 day $=y=\frac{1}{36}$
The time is taken by a woman to finish the work $=36$ days.

Question 5 .
The prices of three commodities $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $₹ \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ per units respectively. A person $\mathrm{P}$ purchases 4 units of $\mathrm{B}$ and sells two units of $\mathrm{A}$ and 5 units of $\mathrm{C}$. Person $Q$ purchases 2 units of $\mathrm{C}$ and sells 3 units of $\mathrm{A}$ and one unit of $\mathrm{B}$. Person $\mathrm{R}$ purchases one unit of $\mathrm{A}$ and sells 3 unit of $\mathrm{B}$ and one unit of $\mathrm{C}$. In the process, $P, Q$ and $R$ earn $₹ 15,000$, ₹ 1,000 and ₹ 4,000 respectively. Find the prices per unit of $A, B$ and $C$. (Use matrix inversion method to solve the problem.)
Solution:
Price of $\mathrm{A}=₹ \mathrm{x} /$ /unit
Price of $\mathrm{B}=₹ \mathrm{y} / \mathrm{unit}$
Price of $\mathrm{C}=₹ \mathrm{z} /$ unit
We are given
$
\begin{aligned}
& 2 x-4 y+5 z=15000 \\
& 3 x+y-2 z=1000 \\
& -x+3 y+z=4000
\end{aligned}
$
The matrix form of the above equations is $\left(\begin{array}{rrr}2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{r}15000 \\ 1000 \\ 4000\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B} \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$
Here

$\begin{aligned}
& A=\left(\begin{array}{rrr}
2 & -4 & 5 \\
3 & 1 & -2 \\
-1 & 3 & 1
\end{array}\right) \\
& |A|=\left|\begin{array}{rrr}
2 & -4 & 5 \\
3 & 1 & -2 \\
-1 & 3 & 1
\end{array}\right|=2(1+6)+4(3-2)+5(9+1)=14+4+50=68 \\
& \left(A_{i j}\right)=\left[\begin{array}{rrr}
+\left|\begin{array}{rr}
1 & -2 \\
3 & 1
\end{array}\right| & -\left|\begin{array}{rr}
3 & -2 \\
-1 & 1
\end{array}\right| & +\left|\begin{array}{rr}
3 & 1 \\
-1 & 3
\end{array}\right| \\
-\left|\begin{array}{rr}
-4 & 5 \\
3 & 1
\end{array}\right| & +\left|\begin{array}{rr}
2 & 5 \\
-1 & 1
\end{array}\right| & -\left|\begin{array}{rr}
2 & -4 \\
-1 & 3
\end{array}\right| \\
+\left|\begin{array}{rr}
-4 & 5 \\
1 & -2
\end{array}\right| & -\left|\begin{array}{rr}
2 & 5 \\
3 & -2
\end{array}\right| & +\left|\begin{array}{rr}
2 & -4 \\
3 & 1
\end{array}\right|
\end{array}\right] \\
& =\left[\begin{array}{l}
+(1+6)-(3-2)+(9+1) \\
-(-4-15)+(2+5)-(6-4) \\
+\quad(8-5)-(-4-15)+(2+12)
\end{array}\right]=\left(\begin{array}{rrr}
7 & -1 & 10 \\
19 & 7 & -2 \\
3 & 19 & 14
\end{array}\right) \\
& \operatorname{Adj} A=\left(A_{i j}\right)^{\mathrm{T}}=\left(\begin{array}{rrr}
7 & 19 & 3 \\
-1 & 7 & 19 \\
10 & -2 & 14
\end{array}\right) \\
& \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{68}\left(\begin{array}{rrr}
7 & 19 & 3 \\
-1 & 7 & 19 \\
10 & -2 & 14
\end{array}\right) \\
& \mathrm{X}=\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\mathrm{A}^{-1} \mathrm{~B}=\frac{1}{68}\left(\begin{array}{rrr}
7 & 19 & 3 \\
-1 & 7 & 19 \\
10 & -2 & 14
\end{array}\right)\left(\begin{array}{r}
15000 \\
1000 \\
4000
\end{array}\right) \\
& =(1000) \frac{1}{68}\left[\begin{array}{rrr}
7 & 19 & 3 \\
-1 & 7 & 19 \\
10 & -2 & 14
\end{array}\right]\left(\begin{array}{r}
15 \\
1 \\
4
\end{array}\right)=\frac{1000}{68}\left[\begin{array}{c}
105+19+12 \\
-15+7+76 \\
150-2+56
\end{array}\right] \\
& =\frac{1000}{68}\left[\begin{array}{r}
136 \\
68 \\
204
\end{array}\right]=1000\left(\begin{array}{l}
2 \\
1 \\
3
\end{array}\right)=\left(\begin{array}{l}
2000 \\
1000 \\
3000
\end{array}\right) \\
& x=₹ 2000 ; y=₹ 1000 \text { and } z=₹ 3000 \\
&
\end{aligned}$