Exercise 1.1 - Chapter 1 Applications of Matrices and Determinants 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Applications of Matrices and Determinants
$\mathbf{E x} 1.1$
Question 1.
Find the rank of each of the following matrices.
Solution:
(i) Let $\mathrm{A}=\left(\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right)$
Order of $\mathrm{A}$ is $2 \times 2$.
$\rho(\mathrm{A}) \leq 2$
Consider the second order minor
$
\left|\begin{array}{lr}
5 & 6 \\
7 & 8
\end{array}\right|=40-42=-2 \neq 0
$
There is a minor of order 2 , which is not zero.
$\rho(A)=2$
(ii) Let $\mathrm{A}=\left(\begin{array}{ll}1 & -1 \\ 3 & -6\end{array}\right)$
Order of $\mathrm{A}$ is $2 \times 2$
$\rho(\mathrm{A}) \leq 2$
Consider the second order minor
$
\begin{aligned}
& \left|\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right|=-6+3=-3 \neq 0 \\
& \rho(A)=2
\end{aligned}
$
(iii) Let $A=\left(\begin{array}{ll}1 & 4 \\ 2 & 8\end{array}\right)$
Since $A$ is of order $2 \times 2, \rho(A) \leq 2$
Now $\left|\begin{array}{ll}1 & 4 \\ 2 & 8\end{array}\right|=8-8=0$
Since second order minor vanishes $\rho(A) \neq 2$
But first order minors, $|1|,|4|,|2|,|8|$ are non zero.
$\rho(A)=1$
(iv) Let $\mathrm{A}=\left(\begin{array}{ccc}2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1\end{array}\right)$
Order of $\mathrm{A}$ is $3 \times 3$
$\rho(A) \leq 3$
Consider the third order minor

$
\begin{aligned}
& \left|\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right| \\
& =2(1+5)+1(3+5)+1(3-1) \\
& =2(6)+8+2 \\
& =22 \neq 0
\end{aligned}
$
There is a minor of order 3 , which is non zero.
$
\rho(A)=3
$
(v) Let $A=\left(\begin{array}{ccc}-1 & 2 & -2 \\ 4 & -3 & 4 \\ -2 & 4 & -4\end{array}\right)$
Since order of $\mathrm{A}$ is $3 \times 3, \rho(\mathrm{A}) \leq 3$
Now,
$
\begin{aligned}
& \left|\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right| \\
& =-1(12-16)-2(-16+8)-2(16-6) \\
& =4+16-20 \\
& =0
\end{aligned}
$
Since the third order minor vanishes, $\rho(A) \neq 3$
$
\text { Consider }\left|\begin{array}{cc}
-1 & 2 \\
4 & -3
\end{array}\right|=3-8=-5 \neq 0
$
There is a minor order 2 , which is not zero $\rho(A)=2$
(vi) Let $A=\left(\begin{array}{cccc}1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 3 & -7\end{array}\right)$
Let us transform the matrix A to an echelon form by using elementary transformations.
$
\begin{aligned}
& A=\left(\begin{array}{cccc}
1 & 2 & -1 & 3 \\
2 & 4 & 1 & -2 \\
3 & 6 & 3 & -7
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 2 & -1 & 3 \\
0 & 0 & 3 & -8 \\
0 & 0 & 6 & -16
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 2 & -1 & 3 \\
0 & 0 & 3 & -8 \\
0 & 0 & 0 & 0
\end{array}\right) \quad R_3 \rightarrow R_3-2 R_2 \\
&
\end{aligned}
$

The above matrix is in echelon form.
The number of non zero rows is $2 \Rightarrow \rho(A)=2$
(vii) Let $\mathrm{A}=\left(\begin{array}{cccc}3 & 1 & -5 & -1 \\ 1 & -2 & 1 & -5 \\ 1 & 5 & -7 & 2\end{array}\right)$
Order of $\mathrm{A}$ is $3 \times 4 \quad \therefore \rho(\mathrm{A}) \leq 3$
Consider the third order minors,

(viii) Let $A=\left(\begin{array}{cccc}1 & -2 & 3 & 4 \\ -2 & 4 & -1 & -3 \\ -1 & 2 & 7 & 6\end{array}\right)$
Let us transform the matric A to an echelon form by elementary transformations
$
\begin{aligned}
& A=\left(\begin{array}{cccr}
1 & -2 & 3 & 4 \\
-2 & 4 & -1 & -3 \\
-1 & 2 & 7 & 6
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & -2 & 3 & 4 \\
0 & 0 & 5 & 5 \\
0 & 0 & 0 & 0
\end{array}\right) \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_2 \\
&
\end{aligned}
$
The number of non-zero rows is 2
$
\therefore \rho(\mathrm{A})=2
$

Question 2.
If $\mathrm{A}=\left(\begin{array}{ccc}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{ccc}1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1\end{array}\right)$, then find the rank of $\mathrm{AB}$ and the rank of
BA.
Solution:
$
\text { Given } \mathrm{A}=\left(\begin{array}{ccc}
1 & 1 & -1 \\
2 & -3 & 4 \\
3 & -2 & 3
\end{array}\right) \text { and } \mathrm{B}=\left(\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -6 \\
5 & 1 & -1
\end{array}\right)
$

$\begin{aligned}
& \mathrm{AB}=\left[\begin{array}{lll}
1-2-5 & -2+4-1 & 3-6+1 \\
2+6+20 & -4-12+4 & 6+18-4 \\
3+4+15 & -6-8+3 & 9+12-3
\end{array}\right] \\
& A B=\left(\begin{array}{ccc}
-6 & 1 & -2 \\
28 & -12 & 20 \\
22 & -11 & 18
\end{array}\right) \\
&
\end{aligned}$

Question 3.
Solve the following system of equations by rank method.
$
x+y+z=9,2 x+5 y+7 z=52,2 x+y-z=0
$
Solution:
The given equations are $x+y+z=9,2 x+5 y+7 z=52,2 x+y-z=0$ The matrix equation corresponding to the given system is
$$
\begin{aligned}
\left(\begin{array}{ccc}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\left(\begin{array}{c}
9 \\
52 \\
0
\end{array}\right) \\
\mathrm{A} & =\mathrm{B}
\end{aligned}
$
Augmented matrix
$
\begin{aligned}
& {[A, B]=\left(\begin{array}{cccc}
1 & 1 & 1 & 9 \\
2 & 5 & 7 & 52 \\
2 & 1 & -1 & 0
\end{array}\right)} \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 9 \\
0 & 3 & 5 & 34 \\
0 & -1 & -3 & -18
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 9 \\
0 & 3 & 5 & 34 \\
0 & 0 & -4 & -20
\end{array}\right) \quad R_3 \rightarrow 3 R_3+R_2 \\
&
\end{aligned}
$
$
\text { Now } A \sim\left(\begin{array}{lll}
1 & 1 & 1 \\
0 & 3 & 5 \\
0 & 0 & 4
\end{array}\right) \quad \Rightarrow \rho(A)=3
$
Since augmented matrix $[A, B] \sim\left(\begin{array}{cccc}1 & 1 & 1 & 9 \\ 0 & 3 & 5 & 34 \\ 0 & 0 & -4 & -20\end{array}\right)$ has three non-zero rows, $\rho([A, B])=3$ That is, $\rho(\mathrm{A})=\rho([\mathrm{A}, \mathrm{B}])=3=$ number of unknowns.
So the given system is consistent and has unique solution.
To find the solution, we rewrite the echelon form into the matrix form.

$
\begin{aligned}
&\left(\begin{array}{ccc}
1 & 1 & 1 \\
0 & 3 & 5 \\
0 & 0 & -4
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{r}
9 \\
34 \\
-20
\end{array}\right) \\
& x+y+z=9 \quad(1) \\
& 3 y+5 z=34 \quad(2) \\
&-4 z=-20(3) \\
&(3) \Rightarrow z=5 \\
&(2) \Rightarrow 3 y=34-25=9 \\
& y=3 \\
&(1) \Rightarrow \quad x=9-3-5 \\
& x=1
\end{aligned}
$
$\therefore x=1, y=3, z=5$ is the unique solution of the given equations.

Question 4.
Show that the equations $5 x+3 y+7 z=4,3 x+26 y+2 z=9,7 x+2 y+10 z=5$ are consistent and solve them by rank method.
Solution:
The given equations are,
$
\begin{aligned}
& 5 x+3 y+7=4 \\
& 3 x+26 y+2 z=9 \\
& 7 x+2 y+10 z=5
\end{aligned}
$
The matrix equation corresponding to the given system is
$
\begin{aligned}
\left(\begin{array}{ccc}
5 & 3 & 7 \\
3 & 26 & 2 \\
7 & 2 & 10
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\left(\begin{array}{l}
4 \\
9 \\
5
\end{array}\right) \\
\mathrm{A} \quad \mathrm{X} & =\mathrm{B}
\end{aligned}
$
Augmented matrix [A,B]

$
\sim\left(\begin{array}{cccc}
1 & \frac{3}{5} & \frac{7}{5} & \frac{4}{5} \\
0 & \frac{121}{5} & \frac{-11}{5} & \frac{33}{5} \\
0 & 0 & 0 & 0
\end{array}\right) \quad \mathrm{R}_3 \rightarrow 11 \mathrm{R}_3+\mathrm{R}_2
$
The equivalent matrix is in echelon form. It has two non-zero rows. $\therefore \rho(\mathrm{A})=\rho([\mathrm{A}, \mathrm{B}])=2<$ number of unknowns.
So the equations are consistent and have infinitely many solutions
$
\left(\begin{array}{ccc}
1 & \frac{3}{5} & \frac{7}{5} \\
0 & \frac{121}{5} & \frac{-11}{5} \\
0 & 0 & 0
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
\frac{4}{5} \\
\frac{33}{5} \\
0
\end{array}\right)
$

$
\begin{aligned}
& \Rightarrow \quad x+\frac{3}{5} y+\frac{7}{5} z=\frac{4}{5} \text {. } \\
& \frac{121}{5} y-\frac{11}{5} z=\frac{33}{5} \\
& \text { (2) } \Rightarrow \frac{121}{5} y=\frac{11}{5} z+\frac{33}{5} \text { (or) } \frac{11}{5} y=\frac{1}{5} z+\frac{3}{5} \\
& 11 y=z+3 \\
& y=\frac{z+3}{11} \\
& \text { (1) } \Rightarrow \quad x=\frac{4}{5}-\frac{3}{5} y-\frac{7}{5} z \\
& x=\frac{4}{5}-\frac{3}{5}\left(\frac{z+3}{11}\right)-\frac{7}{5} z \\
& x=\frac{4}{5}-\frac{3 z}{55}-\frac{9}{55}-\frac{7}{5} z \Rightarrow x=\frac{-16 z}{11}+\frac{7}{11} \\
&
\end{aligned}
$
Let us take $\mathrm{z}=\mathrm{k}, \mathrm{k} \in \mathrm{R}$. We get, $y=\frac{k+3}{11}, x=\frac{-16}{11} k+\frac{7}{11}$
By giving different values for $\mathrm{k}$, we get different solutions. Thus the solutions of the given system are given by $x=\frac{1}{11}(7-16 k) ; y=\frac{1}{11}(3+k) ; z=k$

Question 5.
Show that the following system of equations have unique solution: $x+y+z=3, x+2 y+3 z=4, x+4 y+9 z=6$ by rank method.
Solution:
The given system of equations can be written in matrix equation,
$
\left(\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 9
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
3 \\
4 \\
6
\end{array}\right)
$
$\mathrm{A} \quad \mathrm{X}=\mathrm{B}$
Augmented matrix [A,B]
$
\begin{aligned}
& \left(\begin{array}{llll}
1 & 1 & 1 & 3 \\
1 & 2 & 3 & 4 \\
1 & 4 & 9 & 6
\end{array}\right) \\
& \sim\left(\begin{array}{llll}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 1 \\
0 & 3 & 8 & 3
\end{array}\right) \underset{R_3 \rightarrow R_3-R_1}{R_2 \rightarrow R_2-R_1} \\
& \sim\left(\begin{array}{llll}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 1 \\
0 & 0 & 2 & 0
\end{array}\right) \mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_2 \\
&
\end{aligned}
$
The last matrix is in echelon form. It has 3 non-zero rows, $\rho(A)=\rho([A, B])=3=$ number of unknowns.
The given system is consistent and has a unique solution.
To find the solution, we write the echelon form into matrix form.
$
\begin{aligned}
& \left(\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
3 \\
1 \\
0
\end{array}\right) \\
& \mathrm{x}+\mathrm{y}+\mathrm{z}=3 \ldots \ldots(1) \\
& \mathrm{y}+2 \mathrm{z}=1 \ldots \ldots(2) \\
& 2 \mathrm{z}=0 \ldots \ldots(3) \\
& (3) \Rightarrow \mathrm{z}=0 \\
& (2) \Rightarrow \mathrm{y}=1 \\
& (1) \Rightarrow \mathrm{x}=2
\end{aligned}
$
So the unique solution is $\mathrm{x}=2, \mathrm{y}=1, \mathrm{z}=0$

Question 6.
For what values of the parameter $\mathrm{X}$, will the following equations fail to have unique solution: $3 \mathrm{x}$ $-y+\lambda z=1,2 x+y+z=2, x+2 y-\lambda z=-1$ by rank method.
Solution:
The given system can be written in matrix equation form as given below:
$
\begin{aligned}
& \left(\begin{array}{ccc}
3 & -1 & \lambda \\
2 & 1 & 1 \\
1 & 2 & -\lambda
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
1 \\
2 \\
-1
\end{array}\right) \\
& \text { A } \quad X=B \\
& \text { Augmented matrix [A,B] } \\
& \left(\begin{array}{cccc}
3 & -1 & \lambda & 1 \\
2 & 1 & 1 & 2 \\
1 & 2 & -\lambda & -1
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 2 & -\lambda & -1 \\
2 & 1 & 1 & 2 \\
3 & -1 & \lambda & 1
\end{array}\right) \quad R_1 \leftrightarrow R_3 \\
& \sim\left(\begin{array}{cccc}
1 & 2 & -\lambda & -1 \\
0 & -3 & 1+2 \lambda & 4 \\
0 & -7 & 4 \lambda & 4
\end{array}\right) \quad \begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-3 R_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 2 & -\lambda & -1 \\
0 & -3 & 1+2 \lambda & 4 \\
0 & 0 & \frac{-2}{3} \lambda-\frac{7}{3} & \frac{-16}{3}
\end{array}\right) \\
& \mathrm{R}_3 \rightarrow \mathrm{R}_3-7 / 3 \mathrm{R}_2 \\
&
\end{aligned}
$
For the system to be inconsistent (or) not to have unique solution, $\rho([\mathrm{A}, \mathrm{B}]) \neq \rho(\mathrm{A})$

$
\begin{aligned}
& \operatorname{So} \rho(\mathrm{A}) \neq 3 \Rightarrow \frac{-2}{3} \lambda-\frac{7}{3}=0 \\
& -2 \lambda=7 \\
& \lambda=\frac{-7}{2} \\
&
\end{aligned}
$
So when $\lambda=\frac{-7}{2}$, the equations fail to have unique solution.
Note: The system cannot have an infinite number of solutions, since $\rho([A, B])=3=$ a number of unknowns.
Question 7.
The price of three commodities, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are and $\mathrm{z}$ respectively Mr. Anand purchases 6 units of $Z$ and sells 2 units of $X$ and 3 units of $Y$. Mr.Amar purchases a unit of $Y$ and sells 3 units of $X$
and 2 units of $Z$. Mr. Amit purchases a unit of $X$ and sells 3 units of $Y$ and a unit of $Z$. In the process they earn ₹ $5,000 /-$, ₹ $2,000 /$ - and ₹ $5,500 /$ - respectively. Find the prices per unit of three commodities by rank method.
Solution:

The price of three commodities $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ are given as $\mathrm{x}, \mathrm{y}, \mathrm{z}$. We form the following system of equations from the given conditions.
Anand $\rightarrow 2 x+3 y-6 z=5000$
Amar $\rightarrow 3 \mathrm{x}-\mathrm{y}+2 \mathrm{z}=2000$
Amit $\rightarrow-x+3 y+z=5500$
The matrix equation is given by

$\begin{aligned}
& \left(\begin{array}{ccc}
2 & 3 & -6 \\
3 & -1 & 2 \\
-1 & 3 & 1
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
5000 \\
2000 \\
5500
\end{array}\right) \\
& \text { A } \quad \mathrm{X}=\mathrm{B} \\
& \text { Augmented matrix [A,B] } \\
& \left(\begin{array}{cccc}
2 & 3 & -6 & 5000 \\
3 & -1 & 2 & 2000 \\
-1 & 3 & 1 & 5500
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
-1 & 3 & 1 & 5500 \\
3 & -1 & 2 & 2000 \\
2 & 3 & -6 & 5000
\end{array}\right) \quad \mathrm{R}_1 \leftrightarrow \mathrm{R}_3 \\
& \sim\left(\begin{array}{cccc}
-1 & 3 & 1 & 5500 \\
0 & 8 & 5 & 18500 \\
0 & 9 & -4 & 16000
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2+3 \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3+2 \mathrm{R}_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
-1 & 3 & 1 & 5500 \\
0 & 72 & 45 & 166500 \\
0 & 72 & -32 & 128000
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow 9 \mathrm{R}_2 \\
\mathrm{R}_3 \rightarrow 8 \mathrm{R}_3
\end{array} \\
&
\end{aligned}$

Question 8.
An amount of $₹ 5,000 /$ - is to be deposited in three different bonds bearing $6 \%, 7 \%$ and $8 \%$ per year respectively. Total annual income is ₹ $358 /-$. If the income from the first two investments is ₹ $70 /$ - more than the income from the third, then find the amount of investment in each bond by the rank method.
Solution:
Let the amount of investment in the three different bonds be Rs. $\mathrm{x}$, Rs. $\mathrm{y}$ and Rs. $\mathrm{z}$ respectively. We get the following equations according to the given conditions,
$
x+y+z=5000
$

$
\begin{aligned}
& \frac{6}{100} x+\frac{7}{100} y+\frac{8}{100} z=358 \text { (or) } 6 x+7 y+8 z=35800 \\
& \frac{6}{100} x+\frac{7}{100} y=70+\frac{8}{100} z \text { (or) } 6 x+7 y-8 z=7000
\end{aligned}
$
This can be written as $\left(\begin{array}{ccc}1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}5000 \\ 35800 \\ 7000\end{array}\right)$
$
\begin{aligned}
& \text { A } \quad \mathrm{X}=\mathrm{B} \\
& \text { Augmented matrix [A,B] } \\
& \left(\begin{array}{cccc}
1 & 1 & 1 & 5000 \\
6 & 7 & 8 & 35800 \\
6 & 7 & -8 & 7000
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 5000 \\
6 & 7 & 8 & 35800 \\
0 & 0 & -16 & -28800
\end{array}\right) \quad R_3 \rightarrow R_3-R_2 \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 5000 \\
0 & 1 & 2 & 5800 \\
0 & 0 & -16 & -28800
\end{array}\right) \quad R_2 \rightarrow R_2-6 R_1 \\
&
\end{aligned}
$
The above equivalent matrix is in echelon form with 3 non-zero rows. So $\rho(A)=\rho([\mathrm{A}, \mathrm{B}])=3=$ number of unknowns. the system has a unique solution. The matrix equation is given by
$
\begin{aligned}
& \left(\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & -16
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
5000 \\
5800 \\
-28800
\end{array}\right) \\
& \mathrm{x}+\mathrm{y}+\mathrm{z}=5000 \ldots(1) \\
& \mathrm{y}+2 \mathrm{z}=5800 \ldots(2) \\
& -16 \mathrm{z}=-28800 \ldots(3) \\
& (3) \Rightarrow \mathrm{z}=1800 \\
& (2) \Rightarrow \mathrm{y}=5800-2(1800)=2200 \\
& (1) \Rightarrow \mathrm{x}=5000-2200-1800=1000
\end{aligned}
$
The amount invested in the three bonds are ₹ 1000 , ₹ 2200 and ₹ 1800 .

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