Exercise 1.5 - 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 $1.5$
Question 1.
Solve the following systems of linear equations by Gaussian elimination method:
(i) $2 x-2 y+3 z=2, x+2 y-z=3,3 x-y+2 z=1$
(ii) $2 x+4 y+6 z=22,3 x+8 y+5 z=27,-x+y+2 z=2$
Solution:
(i) $2 x-2 y+3 z=2, x+2 y-z=3$ and $3 x-y+2 z=1$
The matrix form of the above equations is $\left(\begin{array}{rrr}2 & -2 & 3 \\ 1 & 2 & -1 \\ 3 & -1 & 2\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}2 \\ 3 \\ 1\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B}$
The augment matrix $(\mathrm{A}, \mathrm{B})$ is
$
\begin{aligned}
(A, B) & =\left[\begin{array}{rrrr}
2 & -2 & 3 & 2 \\
1 & 2 & -1 & 3 \\
3 & -1 & 2 & 1
\end{array}\right] \sim\left[\begin{array}{rrrr}
1 & 2 & -1 & 3 \\
2 & -2 & 3 & 2 \\
3 & -1 & 2 & 1
\end{array}\right] R_1 \leftrightarrow R_2 \\
& \sim\left(\begin{array}{rrrr}
1 & 2 & -1 & 3 \\
0 & -6 & 5 & -4 \\
0 & -7 & 5 & -8
\end{array}\right) R_2 \rightarrow R_2-2 R_1 ; R_3 \rightarrow R_3-3 R_1 \\
& \sim\left(\begin{array}{rrrr}
1 & 2 & -1 & 3 \\
0 & -6 & 5 & -4 \\
0 & 0 & -5 & -20
\end{array}\right) R_3 \rightarrow 6 R_3-7 R_2
\end{aligned}
$
The above matrix is in echelon form.
Now writing the equivalent equations
$
\begin{aligned}
& \left(\begin{array}{rrr}
1 & 2 & -1 \\
0 & -6 & 5 \\
0 & 0 & -5
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{r}
3 \\
-4 \\
-20
\end{array}\right) \\
& \text { (i.e) } \quad x+2 y-z=3 \\
& -6 y+5 z=-4 \\
& -5 z=-20 \\
& \text { from }(3) \Rightarrow \quad z=\frac{-20}{-5}=4 \\
&
\end{aligned}
$

Substituting $z=4$ in (2) we get
$
\begin{aligned}
& -6 y+20=-4 \\
& \Rightarrow-6 y=-4-20=-24 \\
& \Rightarrow y=4
\end{aligned}
$
Substituting $z=4$ and $y=4$ in (1) we get
$
\begin{aligned}
& \mathrm{x}+8-4=3 \\
& \Rightarrow \mathrm{x}+4=3 \\
& \Rightarrow \mathrm{x}=3-4=-1
\end{aligned}
$
So, $x=-1 ; y=4 ; z=4$
(ii) $2 x+4 y+6 z=22$
$3 x+8 y+5 z=27$
$
-x+y+2 z=2 \ldots \ldots \text { (3) }
$
Divide equation (1) by 2 we get
$
\begin{aligned}
& x+2 y+3 z=11 \\
& 3 x+8 y+5 z=27 \\
& -x+y+2 z=2 \ldots
\end{aligned}
$
The matrix form of the above equations is $\left(\begin{array}{ccc}1 & 2 & 3 \\ 3 & 8 & 5 \\ -1 & 1 & 2\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}11 \\ 27 \\ 2\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B}$
The augment matrix $(A, B)$ is
$
\begin{aligned}
(A, B) & =\left[\begin{array}{rrrr}
1 & 2 & 3 & 11 \\
3 & 8 & 5 & 27 \\
-1 & 1 & 2 & 2
\end{array}\right] \\
& \sim\left[\begin{array}{rrrr}
1 & 2 & 3 & 11 \\
0 & 2 & -4 & -6 \\
0 & 3 & 5 & 13
\end{array}\right] \mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1 ; \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_1 \\
& \sim\left[\begin{array}{rrrr}
1 & 2 & 3 & 11 \\
0 & 1 & -2 & -3 \\
0 & 3 & 5 & 13
\end{array}\right] \mathrm{R}_2 \rightarrow \frac{\mathrm{R}_2}{2} \\
& \sim\left[\begin{array}{rrrr}
1 & 2 & 3 & 11 \\
0 & 1 & -2 & -3 \\
0 & 0 & 11 & 22
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_2
\end{aligned}
$
The above matrix is in echelon form.
Now writing the equivalent equations.

$
\begin{aligned}
& \left(\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & -2 \\
0 & 0 & 11
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
11 \\
-3 \\
22
\end{array}\right) \\
& \Rightarrow \quad x+2 y+3 z=11 \\
& y-2 z=-3 \\
& 11 z=22 \\
& \text { From (3) } \Rightarrow \quad z=\frac{22}{11}=2 \\
&
\end{aligned}
$
Substituting $z=2$ in (2) we get
$
\begin{aligned}
& y-4=-3 \\
& \Rightarrow y=-3+4=1
\end{aligned}
$
Substituting $z=2, y=1$ in (1) we get
$
\begin{aligned}
& \mathrm{x}+2(1)+3(2)=11 \\
& \Rightarrow \mathrm{x}+2+6=11 \\
& \Rightarrow \mathrm{x}+8=11 \\
& \Rightarrow \mathrm{x}=11-8=3 \\
& \mathrm{x}=3, \mathrm{y}=1, \mathrm{z}=2
\end{aligned}
$
Question 2.
If $a x^2+b x+c$ is divided by $x+3, x-5$, and $x-1$, the remainders are 21,61 and 9 respectively. Find $a, b$ and c. (Use Gaussian elimination method.)
Solution:
$\mathrm{P}(\mathrm{x})=a x^2+b \mathrm{x}+\mathrm{c}$. When $\mathrm{P}(\mathrm{x})$ is divided by $\mathrm{x}+3, \mathrm{x}-5$ and $\mathrm{x}-1$.
The remainders are respectively $\mathrm{P}(-3), \mathrm{P}(5)$ and $\mathrm{P}(1)$.
We are given that $\mathrm{P}(-3)=21 ; \mathrm{P}(5)=61 ; \mathrm{P}(1)=9$
Now $\mathrm{P}(-3)=21$
$
\begin{aligned}
& \Rightarrow \mathrm{a}(-3)^2+\mathrm{b}(-3)+\mathrm{c}=21 \\
& \Rightarrow 9 \mathrm{a}-3 \mathrm{~b}+\mathrm{c}=21 \ldots \ldots . \\
& \mathrm{P}(5)=61 \\
& \Rightarrow \mathrm{a}(5)^2+\mathrm{b}(5)+\mathrm{c}=61 \\
& \Rightarrow 25 \mathrm{a}+5 \mathrm{~b}+\mathrm{c}=61 \ldots \ldots . \\
& \mathrm{P}(1)=9
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \mathrm{a}(1)^2+\mathrm{b}(1)+\mathrm{c}=9 \\
& \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=9
\end{aligned}
$
Now the matrix form of the above three equations is $\left(\begin{array}{ccc}9 & -3 & 1 \\ 25 & 5 & 1 \\ 1 & 1 & 1\end{array}\right)\left(\begin{array}{l}a \\ b \\ c\end{array}\right)=\left(\begin{array}{c}21 \\ 61 \\ 9\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B}$
The augmented matrix (A, B) is
$
\begin{aligned}
{[A, B] } & =\left(\begin{array}{rrrr}
9 & -3 & 1 & 21 \\
25 & 5 & 1 & 61 \\
1 & 1 & 1 & 9
\end{array}\right) \\
& \sim\left(\begin{array}{rrrr}
1 & 1 & 1 & 9 \\
25 & 5 & 1 & 61 \\
9 & -3 & 1 & 21
\end{array}\right) R_1 \leftrightarrow R_3 \\
\left(\begin{array}{rrrrr}
1 & 1 & 1 & 9 \\
25 & 5 & 1 & 61 \\
9 & -3 & 1 & 21
\end{array}\right) & \sim\left(\begin{array}{rrrr}
1 & 1 & 1 & 9 \\
0 & -20 & -24 & -164 \\
0 & -12 & -8 & -60
\end{array}\right) R_3 \rightarrow R_3-9 R_1 \\
& \sim R_2-25 R_1 \\
& \sim\left(\begin{array}{rrrr}
1 & 1 & 1 & 9 \\
0 & -20 & -24 & -164 \\
0 & 12 & 8 & 60
\end{array}\right) R_3 \rightarrow-R_3 \\
& \sim\left(\begin{array}{rrrr}
1 & 1 & 1 & 9 \\
0 & -20 & -24 & -164 \\
0 & 0 & -32 & -192
\end{array}\right) R_3 \rightarrow 5 R_3+3 R_2
\end{aligned}
$
The above matrix is in echelon form now writing the equivalent equations.
$
\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & -20 & -24 \\
0 & 0 & -32
\end{array}\right]\left[\begin{array}{l}
a \\
b \\
c
\end{array}\right]=\left[\begin{array}{r}
9 \\
-164 \\
-192
\end{array}\right]
$
(i.e) $a+b+c=9$
$-20 b-24 c=-164$
$-32 c=-192$
From (3) $\Rightarrow \quad c=\frac{-192}{-32}=6$
Substituting c $=6$ in (2) we get
$
\begin{aligned}
& -20 \mathrm{~b}-24(6)=-164 \\
& \Rightarrow-20 \mathrm{~b}=-164+144=-20 \\
& \Rightarrow \mathrm{b}=1
\end{aligned}
$

Substituting $\mathrm{c}=6, \mathrm{~b}=1$ in (1) we get
$
\begin{aligned}
& \mathrm{a}+1+6=9 \\
& \Rightarrow \mathrm{a}=9-7=2
\end{aligned}
$
So $\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=6$
Question 3.
An amount of ₹ 65,000 is invested in three bonds at the rates of $6 \%, 8 \%$ and $10 \%$ per annum respectively. The total annual income is $₹ 5,000$. The income from the third bond is $₹ 800$ more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Solution:
Let the amount invested in $6 \%$ bond be ₹ $x$ and the amount invested in $8 \%$ bond be ₹ y
and the amount invested in $10 \%$ bond be ₹ $z$
Now $\mathrm{x}+\mathrm{y}+\mathrm{z}=65000$
$
\frac{6}{100} x+\frac{8}{100} y+\frac{10}{100} z=5000
$
(i.e) $6 x+8 y+10 z=500000$
$(\div$ by 2 ) $3 x+4 y+5 z=250000 \ldots \ldots$. (2)
Also given that $\frac{10}{100} z-\frac{8}{100} y=800$
(i.e) $-8 y+10 z=80000$
$(\div$ by 2$)-4 y+5 z=40000$
Now the matrix form of the above three equations is $\left(\begin{array}{ccc}1 & 1 & 1 \\ 3 & 4 & 5 \\ 0 & -4 & 5\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}65000 \\ 250000 \\ 40000\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B}$
Now, the augmented matrix (A, B) is
$
\begin{aligned}
{[A, B] } & =\left(\begin{array}{cccc}
1 & 1 & 1 & 65000 \\
3 & 4 & 5 & 250000 \\
0 & -4 & 5 & 40000
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 65000 \\
0 & 1 & 2 & 55000 \\
0 & -4 & 5 & 40000
\end{array}\right) \begin{array}{l}
\mathrm{R}_1 \rightarrow \mathrm{R}_1 \\
\mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 65000 \\
0 & 1 & 2 & 55000 \\
0 & 0 & 13 & 260000
\end{array}\right) \mathrm{R}_3 \rightarrow \mathrm{R}_3+4 \mathrm{R}_2
\end{aligned}
$
Now the above matrix is in echelon form. Writing the equivalent equations.
$
\begin{aligned}
& \left(\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 13
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{r}
65000 \\
55000 \\
260000
\end{array}\right) \\
& \text { (i.e) } \mathrm{x}+\mathrm{y}+\mathrm{z}=65000 \\
& \Rightarrow \mathrm{y}+2 \mathrm{z}=55000 \\
& \Rightarrow 13 \mathrm{z}=260000
\end{aligned}
$

$
\Rightarrow z=₹ 20000
$
Substituting $z=₹ 20000$ in (2) we get
$
\begin{aligned}
& \mathrm{y}+40000=55000 \\
& \Rightarrow \mathrm{y}=55000-40000=₹ 15000
\end{aligned}
$
Substituting $z=₹ 20000, y=₹ 15000$ in (1) we get
$
\begin{aligned}
& \mathrm{x}+15000+20000=65000 \\
& \Rightarrow \mathrm{x}=65000-35000=₹ 30000
\end{aligned}
$
So the amount invested in
$6 \%$ bond $\mathrm{x}=₹ 30000$
$8 \%$ bond $y=₹ 15000$
and $10 \%$ bond $z=₹ 20000$
Question 4.
A boy is walking along the path $y=a x^2+b x+c$ through the points $(-6,8),(-2,-12)$, and $(3,8)$. He wants to meet his friend at $\mathrm{P}(7,60)$. Will he meet his friend? (Use Gaussian elimination method.)
Solution:
We are given $\mathrm{y}=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}$
Also we are given $(-6,8),(-2,-12)$ and $(3,8)$ are points on the path.
(i) $(-6,8)$ is a point on $y=a x^2+b x+c$
at $x=-6, y=8$
$
\begin{aligned}
& \text { (i.e) } \mathrm{a}(36)+\mathrm{b}(-6)+\mathrm{c}=8 \\
& \Rightarrow 36 \mathrm{a}-6 \mathrm{~b}+\mathrm{c}=8
\end{aligned}
$
(ii) $(-2,-12)$ is a point on $y=a x^2+b x+c$
at $\mathrm{x}=-2, \mathrm{y}=-12$
$
\begin{aligned}
& \Rightarrow \mathrm{a}(-2)^2+\mathrm{b}(-2)+\mathrm{c}=-12 \\
& \Rightarrow 4 \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=-12 \ldots . \text { (2) }
\end{aligned}
$
(iii) $(3,8)$ is a point on $y=a x^2+b x+c$
at $\mathrm{x}=3, \mathrm{y}=8$
$
\begin{aligned}
& \Rightarrow \mathrm{a}(3)^2+6(3)+\mathrm{c}=8 \\
& \Rightarrow 9 \mathrm{a}+3 \mathrm{~b}+\mathrm{c}=8
\end{aligned}
$
The matrix form of the above three equations is $\left(\begin{array}{ccc}36 & -6 & 1 \\ 4 & -2 & 1 \\ 9 & 3 & 1\end{array}\right)\left(\begin{array}{l}a \\ b \\ c\end{array}\right)=\left(\begin{array}{c}8 \\ -12 \\ 8\end{array}\right)$
(i.e) $\mathrm{AX}=\mathrm{B}$
The augmented matrix $(\mathrm{A}, \mathrm{B})$ is

$
\begin{aligned}
&(A, B)=\left[\begin{array}{rrrr}
36 & -6 & 1 & 8 \\
4 & -2 & 1 & -12 \\
9 & 3 & 1 & 8
\end{array}\right] \\
& \sim\left[\begin{array}{rrrr}
36 & -6 & 1 & 8 \\
0 & -12 & 8 & -116 \\
0 & 18 & 3 & 24
\end{array}\right] \mathrm{R}_2 \rightarrow 9 \mathrm{R}_2-\mathrm{R}_1 \\
& \mathrm{R}_3 \rightarrow 4 \mathrm{R}_3-\mathrm{R}_1 \\
& \sim\left(\begin{array}{rrrr}
36 & -6 & 1 & 8 \\
0 & -12 & 8 & -116 \\
0 & 0 & 30 & -300
\end{array}\right) \mathrm{R}_3 \rightarrow 2 \mathrm{R}_3+3 \mathrm{R}_2
\end{aligned}
$
The above matrix is in echelon form. Now writing the equivalent equations we get
$
\begin{aligned}
& \left(\begin{array}{ccc}
36 & -6 & 1 \\
0 & -12 & 8 \\
0 & 0 & 30
\end{array}\right)\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)=\left(\begin{array}{c}
8 \\
-116 \\
-300
\end{array}\right) \\
& \text { (i.e) } 36 \mathrm{a}-6 \mathrm{~b}+\mathrm{c}=8 \\
& \Rightarrow-12 \mathrm{~b}+8 \mathrm{c}=-116 \\
& \Rightarrow 30 \mathrm{c}=-300 \\
& \Rightarrow \mathrm{c}=-10
\end{aligned}
$
Substituting $c=-10$ in (2) we get
$
\begin{aligned}
& -12 b+8(-10)=-116 \\
& \Rightarrow-12 b=-116+80=-36 \\
& \Rightarrow b=3
\end{aligned}
$
Substituting $\mathrm{c}=-10, \mathrm{~b}=3$ in (1) we get
$
\begin{aligned}
& 36 a-6(3)+(-10)=8 \\
& \Rightarrow 36 a-18-10=8 \\
& \Rightarrow 36 a=8+18+10=36 \\
& \Rightarrow a=1 \\
& a=1, b=3 \text { and } c=-10 \\
& y=(1) x^2+(3) x+(-10) \\
& y=x^2+3 x-10
\end{aligned}
$
Now at $x=7, y=(7)^2+3(7)-10=49+21-10=60$
$(7,60)$ is a point on the path so he will meet his friend.