Exercise 1.4 - 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

Ex $1.4$
Question 1.
Solve the following systems of linear equations by Cramer's rule:
(i) $5 x-2 y+16=0, x+3 y-7=0$
(ii) $\frac{3}{x}+2 y=12, \frac{2}{x}+3 y=13$
(iii) $3 x+3 y-z=11,2 x-y+2 z=9,4 x+3 y+2 z=25$
(iv) $\frac{3}{x}-\frac{4}{y}-\frac{2}{z}-1=0, \frac{1}{x}+\frac{2}{y}+\frac{1}{z}-2=0, \frac{2}{x}-\frac{5}{y}-\frac{4}{z}+1=0$
Solution:
(i) $5 x-2 y+16=0, x+3 y-7=0$
The above equations are $5 x-2 y=-16$ and $x+3 y=-7$
The matrix form of two above equations is
(i.e)
$
\left(\begin{array}{rr}
5 & -2 \\
1 & 3
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=\left(\begin{array}{r}
-16 \\
7
\end{array}\right)
$
Now $A X=B$
$|A|=\Delta=\left|\begin{array}{rr}5 & -2 \\ 1 & 3\end{array}\right|=15+2=17 \neq 0$
$\Delta x=\left|\begin{array}{rr}-16 & -2 \\ 7 & 3\end{array}\right|=-48+14=-34$
$\Delta y=\left|\begin{array}{rr}5 & -16 \\ 1 & 7\end{array}\right|=35+16=51$
Now
$
\begin{aligned}
& x=\frac{\Delta_x}{\Delta}=\frac{-34}{17}=-2 \\
& y=\frac{\Delta_y}{\Delta}=\frac{51}{17}=3
\end{aligned}
$
So, $\quad x=-2 ; y=3$
(ii) $\frac{3}{x}+2 y=12, \frac{2}{x}+3 y=13$
Put $\quad \frac{1}{x}=a$
$\Rightarrow \quad 3 a+2 y=12$
$2 a+3 y=13$

Writing the above equations in matrix form we get
$
\left(\begin{array}{ll}
3 & 2 \\
2 & 3
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=\left(\begin{array}{l}
12 \\
13
\end{array}\right) \quad \text { (i.e) } \mathrm{AX}=\mathrm{B}
$
Now
$
|A|=\Delta=\left|\begin{array}{ll}
3 & 2 \\
2 & 3
\end{array}\right|=9-4=5 \neq 0
$
$
\begin{aligned}
\Delta_a & =\left|\begin{array}{ll}
12 & 2 \\
13 & 3
\end{array}\right|=36-26=10 \\
\Delta_y & =\left|\begin{array}{ll}
3 & 12 \\
2 & 13
\end{array}\right|=39-24=15 \\
a & =\frac{\Delta_a}{\Delta}=\frac{10}{5}=2 ; \\
y & =\frac{\Delta_y}{\Delta}=\frac{15}{5}=-3 \quad \therefore a=2, y=3 \text { but } \frac{1}{x}=a \\
\Rightarrow \quad x & =\frac{1}{a}=\frac{1}{2} \text { and } y=3 \quad \therefore x=\frac{1}{2} ; y=3 .
\end{aligned}
$
(iii) $3 x+3 y-z=11,2 x-y+2 z=9,4 x+3 y+2 z=25$
The matrix form is
$
\begin{aligned}
& \left(\begin{array}{rrr}
3 & 3 & -1 \\
2 & -1 & 2 \\
4 & 3 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
11 \\
9 \\
25
\end{array}\right) \\
\Delta & =|\mathrm{A}|=\left|\begin{array}{rrr}
3 & 3 & -1 \\
2 & -1 & 2 \\
4 & 3 & 2
\end{array}\right| \\
& \text { (i.e) } \mathrm{AX}=\mathrm{B} \\
& =3(-2-6)-3(4-8)-1(6+4)=3(-8)-3(-4)-1(10) \\
& =-24+12-10=-22 \\
\Delta_x & =\left|\begin{array}{ccc}
11 & 3 & -1 \\
9 & -1 & 2 \\
25 & 3 & 2
\end{array}\right|=11(-2-6)-3(18-50)-1(27+25) \\
& =11(-8)-3(-32)-1(52)=-88+96-52=-140+96=-44 \\
\Delta_y & =\left|\begin{array}{lll}
3 & 11 & -1 \\
2 & 9 & 2
\end{array}\right|=3(18-50)-11(4-8)-1(50-36)
\end{aligned}
$

$\begin{aligned}
& =3(-32)-11(-4)-1(14)=-96+44-14=-66 \\
\Delta_z & =\left|\begin{array}{ccc}
3 & 3 & 11 \\
2 & -1 & 9 \\
4 & 3 & 25
\end{array}\right|=3(-25-27)-3(50-36)+11(6+4) \\
& =3(-52)-3(14)+11(10)=-156-42+110=-88 \\
\Delta & =-22 ; \Delta x=-44 ; \Delta y=-66 ; \Delta z=-88 \\
x & =\frac{\Delta_x}{\Delta}=\frac{-44}{-22}=2 \\
y & =\frac{\Delta_y}{\Delta}=\frac{-66}{-22}=3 \\
z & =\frac{\Delta_z}{\Delta}=\frac{-88}{-22}=4 \\
x & =2 ; y=3 ; z=4
\end{aligned}$

(iv)
$
\begin{aligned}
\frac{3}{x}-\frac{4}{y}-\frac{2}{z}-1 & =0, \frac{1}{x}+\frac{2}{y}+\frac{1}{z}-2=0, \frac{2}{x}-\frac{5}{y}-\frac{4}{z}+1=0 \\
\frac{3}{x}-\frac{4}{y}-\frac{2}{z} & =1 \\
\frac{1}{x}+\frac{2}{y}+\frac{1}{z} & =2 \\
\frac{2}{x}-\frac{5}{x}-\frac{4}{z} & =-1 \\
\frac{1}{x} & =a, \frac{1}{y}=b \text { and } \frac{1}{z}=c .
\end{aligned}
$
The above equations become
$
\begin{gathered}
3 a-4 b-2 c=1 \\
a+2 b+c=2 \\
2 a-5 b-4 c=-1
\end{gathered}
$
The matrix form of the above equations is
$
\begin{aligned}
\left(\begin{array}{rrr}
3 & -4 & -2 \\
1 & 2 & 1 \\
2 & -5 & -4
\end{array}\right)\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right) & =\left(\begin{array}{r}
1 \\
2 \\
-1
\end{array}\right) \\
|A| & =\Delta=\left|\begin{array}{rrr}
3 & -4 & -2 \\
1 & 2 & 1 \\
2 & -5 & -4
\end{array}\right|=3(-8+5)+4(-4-2)-2(-5-4) \\
& =3(-3)+4(-6)-2(-9)=-9-24+18=-15 \\
\Delta_a & =\left|\begin{array}{ccc}
1 & -4 & -2 \\
2 & 2 & 1 \\
-1 & -5 & -4
\end{array}\right|=1(-8+5)+4(-8+1)-2(-10+2) \\
& =1(-3)+4(-7)-2(-8)=-3-28+16=-15 \\
\Delta_b & =\left|\begin{array}{ccc}
3 & 1 & -2 \\
1 & 2 & 1 \\
2 & -1 & -4
\end{array}\right|=3(-8+1)-1(-4-2)-2(-1-4) \\
& =3(-7)-1(-6)-2(-5)=-21+6+10=-5 \\
\Delta_c & =\left|\begin{array}{ccc}
3 & -4 & 1 \\
1 & 2 & 2 \\
2 & -5 & -1
\end{array}\right|=3(-2+10)+4(-1-4)+1(-5-4) \\
& =3(8)+4(-5)+1(-9)=24-20-9=-5
\end{aligned}
$

Here
$
\begin{aligned}
& \Delta=-15, \Delta_a=-15, \Delta_b=-5, \Delta_c=-5 \\
& a=\frac{\Delta_a}{\Delta}=\frac{-15}{-15}=1 \\
& b=\frac{\Delta_b}{\Delta}=\frac{-5}{-15}=\frac{1}{3} \text { and } \\
& c=\frac{\Delta_c}{\Delta}=\frac{-5}{-15}=\frac{1}{3}
\end{aligned}
$
Thus $a=1, b=\frac{1}{3}, c=\frac{1}{3}$
But
$
\begin{array}{ll}
\frac{1}{x}=a & \Rightarrow x=\frac{1}{a}=\frac{1}{1}=1 \\
\frac{1}{y}=b & \Rightarrow y=\frac{1}{b}=\frac{1}{1 / 3}=3 \\
\frac{1}{z}=c & \Rightarrow z=\frac{1}{c}=\frac{1}{1 / 3}=3
\end{array}
$
and
$
\therefore \quad x=1, y=3, z=3
$
Question 2.
In a competitive examination, one mark is awarded for every correct answer while $\frac{1}{4}$ mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer's rule to solve the problem).
Solution:
No. of Questions answered $=100$
Let the No. of questions answered correctly be $x$ and the No. of questions answered wrongly be $y$
Here, $x+y=100$ and $x-\frac{1}{4} y=80$
(i.e) $x+y=100$ and $4 x-y=320$
The matrix form is 

$
\begin{aligned}
\left(\begin{array}{rr}
1 & 1 \\
4 & -1
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right) & =\left(\begin{array}{l}
100 \\
320
\end{array}\right) \\
\Delta & =|\mathrm{A}|=\left|\begin{array}{rr}
1 & 1 \\
4 & -1
\end{array}\right|=-1-4=-5 \\
\Delta_x & =\left|\begin{array}{rr}
100 & 1 \\
320 & -1
\end{array}\right|=-100-320=-420 \\
\Delta_y & =\left|\begin{array}{ll}
1 & 100 \\
4 & 320
\end{array}\right|=320-400=-80 \\
x & =\frac{\Delta_x}{\Delta}=\frac{-420}{-5}=84 \text { and } \\
y & =\frac{\Delta_y}{\Delta}=\frac{-80}{-5}=16
\end{aligned}
$
No. of questions answered correctly $x=84$.

Question 3.
A chemist has one solution which is $50 \%$ acid and another solution which is $25 \%$ acid. How much each should be mixed to make 10 litres of a $40 \%$ acid solution? (Use Cramer's rule to solve the problem).
Solution:
Let the no. of litres in $50 \%$ acid used be $x$ litres and the no. of litres in $25 \%$ acid used be $y$ litres
Here $x+y=10$ and $\frac{50}{100} x+\frac{25}{100} y=\frac{40}{100} \times 10$
$\begin{array}{lcc}\text { (i.e) } & x+y=10 & \text { and } \frac{x}{2}+\frac{y}{4}=4 \\ \Rightarrow & 2 x+y=16\end{array}$
Now the given equations are $x+y=10$ and $2 x+y=16$
The equations in matrix form is
$
\begin{aligned}
& \begin{array}{r}
\left(\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=\left(\begin{array}{l}
10 \\
16
\end{array}\right) \\
\Delta=|A|=\left|\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right|=1-2=-1
\end{array} \\
& \Delta_x=\left|\begin{array}{ll}
10 & 1 \\
16 & 1
\end{array}\right|=10-16=-6 \\
& \Delta_y=\left|\begin{array}{ll}
1 & 10 \\
2 & 16
\end{array}\right|=16-20=-4 \\
& x=\frac{\Delta_x}{\Delta}=\frac{-6}{-1}=6 \text { and } y=\frac{\Delta_y}{\Delta}=\frac{-4}{-1}=4 \\
&
\end{aligned}
$
(i.e) we have to $\operatorname{mix} 6$ litres in $50 \%$ acid and 4 litres in $25 \%$ acid.

Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer's rule to solve the problem).
Solution:
Time is taken for pump A to fill the tank be $\mathrm{x}$ minutes and time taken for pump B to fill the tank be $y$ minutes
Given $\quad \frac{1}{x}+\frac{1}{y}=\frac{1}{10}$ and $\frac{1}{x}-\frac{1}{y}=\frac{1}{30}$
To solve the above equations: put $\frac{1}{x}=a$ and $\frac{1}{y}=b$
Now $\quad a+b=\frac{1}{10}$ and $a-b=\frac{1}{30}$
The matrix form of the above equations is
$
\left(\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right)\left(\begin{array}{l}
a \\
b
\end{array}\right)=\left(\begin{array}{c}
\frac{1}{10} \\
\frac{1}{30}
\end{array}\right) \quad \text { (i.e) } \mathrm{AX}=\mathrm{B}
$
$
\begin{aligned}
|A| & =\Delta=\left|\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right|=-1-1=-2 \\
\Delta_a & =\left|\begin{array}{cc}
\frac{1}{10} & 1 \\
\frac{1}{30} & -1
\end{array}\right|=\frac{-1}{10}-\frac{1}{30}=\frac{-3-1}{30}=\frac{-4}{30} \\
\Delta_b & =\left|\begin{array}{cc}
10 \\
1 & \frac{1}{30}
\end{array}\right|=\frac{1}{30}-\frac{1}{10}=\frac{1-3}{30}=\frac{-2}{30}
\end{aligned}
$
$
\begin{aligned}
& a=\frac{\Delta_a}{\Delta}=\frac{\frac{-4}{30}}{-2}=\frac{2}{30}=\frac{1}{15} \\
& b=\frac{\Delta_b}{\Delta}=\frac{\frac{-2}{-2}}{-2}=\frac{1}{30} \\
& a=\frac{1}{x} \quad \Rightarrow x=\frac{1}{a}=\frac{1}{\frac{1}{15}}=15
\end{aligned}
$

and
$
b=\frac{1}{y} \quad \Rightarrow y=\frac{1}{b}=\frac{1}{\frac{1}{30}}=30
$
(i.e) Pump A can fill the tank in 15 minutes and pump B can fill the tank in 30 minutes.

Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is ₹ 150 . The cost of the two dosai, two idlies and four vadais is ₹ 200 . The cost of five dosai, four idlies and two vadais is $₹ 250$. The family has $₹ 350$ in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
Solution:
Let the cost of 1 dosai be $₹ x$ the cost of 1 idli be ₹ $y$ and the cost of 1 vadai be $₹ z$
Here $2 x+3 y+2 z=150$
$2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=200$
$5 \mathrm{x}+4 \mathrm{y}+2 \mathrm{z}=250$
Writing the above equations in matrix form
$
\left(\begin{array}{lll}
2 & 3 & 2 \\
2 & 2 & 4 \\
5 & 4 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
150 \\
200 \\
250
\end{array}\right)
$
(i.e) $\mathrm{AX}=\mathrm{B}$
Now
$
\begin{aligned}
& \Delta=|A|=\left|\begin{array}{lll}
2 & 3 & 2 \\
2 & 2 & 4 \\
5 & 4 & 2
\end{array}\right|=2(4-16)-3(4-20)+2(8-10) \\
& =2(-12)-3(-16)+2(-2)=-24+48-4=20 \\
& \Delta_x=\left|\begin{array}{lll}
150 & 3 & 2 \\
200 & 2 & 4 \\
250 & 4 & 2
\end{array}\right|=150(4-16)-200(6-8)+250(12-4) \\
& =150(-12)-200(-2)+250(8)=-1800+400+2000=600 \\
& \Delta_y=\left|\begin{array}{lll}
2 & 150 & 2 \\
2 & 200 & 4 \\
5 & 250 & 2
\end{array}\right|=2(400-1000)-2(300-500)+5(600-400) \\
& =2(-600)-2(-200)+5(200)=-1200+400+1000=200 \\
& \Delta_z=\left|\begin{array}{lll}
2 & 3 & 150 \\
2 & 2 & 200 \\
5 & 4 & 250
\end{array}\right|=2(500-800)-2(750-600)+5(600-300) \\
& =2(-300)-2(150)+5(300)=600-300+1500=600 \\
& x=\frac{\Delta_x}{\Delta}=\frac{600}{20}=30 \\
& y=\frac{\Delta_y}{\Delta}=\frac{200}{20}=10 \text { and } \\
& z=\frac{\Delta_z}{\Delta}=\frac{600}{20}=30 \\
&
\end{aligned}
$

So, cost of 1 dosai $=x=₹ 30$
cost of $1 \mathrm{idli}=\mathrm{y}=₹ 10$ and
cost of 1 vadai $=z=₹ 30$
Now cost of 3 dosai and 6 idlis and 6 vadais $=3 \times 30+6 \times 10+6 \times 30=90+60+180=₹ 330$ and they are having ₹ 350 . So they will be able to manage to pay the bill.