Exercise 1.5-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.
Examine the consistency of the following system of equations. If it is consistent then solve the same.
$
\begin{aligned}
& \text { (i) } 4 x+3 y+6 z=25 \\
& x+5 y+7 z=13 \\
& 2 x+9 y+z=1
\end{aligned}
$
Solution:
The augmented matrix is
$
\begin{aligned}
{[A, B] } & =\left[\begin{array}{rrrr}
4 & 3 & 6 & 25 \\
1 & 5 & 7 & 13 \\
2 & 9 & 1 & 1
\end{array}\right] \sim\left[\begin{array}{rrrr}
4 & 3 & 6 & 25 \\
0 & 1 & 13 & 25 \\
0 & 15 & -4 & -23
\end{array}\right] \begin{array}{l}
\mathrm{R}_2 \rightarrow 2 \mathrm{R}_2-\mathrm{R}_3 \\
\mathrm{R}_3 \rightarrow 2 \mathrm{R}_3-\mathrm{R}_1
\end{array} \\
& \sim\left[\begin{array}{rrrr}
4 & 3 & 6 & 25 \\
0 & 1 & 13 & 25 \\
0 & 0 & -199 & -398
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-15 \mathrm{R}_2
\end{aligned}
$
The last equivalent matrix is in the echelon form. It has three non-zero rows.
$
\therefore \quad \rho(\mathrm{A}, \mathrm{B})=3 \text {; }
$
Also
$
A \sim\left[\begin{array}{rrr}
4 & 3 & 6 \\
0 & 1 & 13 \\
0 & 0 & -199
\end{array}\right] ; \text { i.e. } \rho(A)=3 \text {. }
$
Thus
$
\rho(A)=\rho(A, B)=3=\text { number of unknowns. }
$

$\therefore$ The given system is consistent and has a unique solution
$4 x+3 y+6 z=25$
From (3),
$
\begin{aligned}
y+13 z & =25 \\
-199 z & =-398 \\
z & =\frac{-398}{-199}=2
\end{aligned}
$
Substituting $z=2$ in (2), we get
$
y+26=25 \quad \Rightarrow y=25-26=-1
$
Substituting $z=2, y=-1$ in (1), we get
$
\begin{aligned}
4 x-3+12 & =25 & & \Rightarrow 4 x+9=25 \\
4 x & =25-9=16 & & \Rightarrow x=\frac{16}{4}=4
\end{aligned}
$
$\therefore$ The unique solution is $\mathrm{x}=4, \mathrm{y}=-1, \mathrm{z}=2$.
Question 2.
Verify whether the given system of equations is consistent. If it is consistent, solve them. $2 x+5 y+7 z=$ $52, x+7+z=9,2 x+y-z=0$

Solution:
The given system of equations is equivalent to the single matrix equation.
$
\begin{aligned}
& {\left[\begin{array}{rrr}
2 & 5 & 7 \\
1 & 1 & 1 \\
2 & 1 & -1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
52 \\
9 \\
0
\end{array}\right]} \\
& \mathrm{AX}=\mathrm{B}
\end{aligned}
$
The augmented matrix is
$
\begin{aligned}
{[A, B] } & =\left[\begin{array}{cccc}
2 & 5 & 7 & 52 \\
1 & 1 & 1 & 9 \\
2 & 1 & -1 & 0
\end{array}\right] \\
& \sim\left[\begin{array}{cccc}
1 & 1 & 1 & 9 \\
2 & 5 & 7 & 52 \\
2 & 1 & -1 & 0
\end{array}\right] R_1 \leftrightarrow R_2 \\
& \sim\left[\begin{array}{cccc}
1 & 1 & 1 & 9 \\
0 & 3 & 5 & 34 \\
0 & -1 & -3 & -18
\end{array}\right] R_2 \rightarrow R_2-2 R_1 \\
& \sim\left[\begin{array}{cccc}
1 & 1 & 1 & 9 \\
0 & -1 & -3 & -18 \\
0 & 3 & 5 & 34
\end{array}\right] R_3-2 R_1 \\
& \sim\left[\begin{array}{cccc}
1 & 1 & 1 & 9 \\
0 & -1 & -3 & -18 \\
0 & 0 & -4 & -20
\end{array}\right] R_3 \rightarrow R_3+3 R_2
\end{aligned}
$
The last equivalent matrix is in the echelon form. It has three non-zero rows.
$
\therefore \rho(\mathrm{A}, \mathrm{B})=3
$
Also
$
A \sim\left[\begin{array}{ccc}
1 & 1 & 1 \\
0 & -1 & -3 \\
0 & 0 & -4
\end{array}\right]
$
Since there are three non-zero rows, $\rho(\mathrm{A})=3$
$
\therefore \quad \rho(A)=\rho[A, B]=3=\text { number of unknowns. }
$
The given system is consistent and has a unique solution.
To find the solution, we see that the given system of equations is equivalent to the matrix equation.

$
\begin{aligned}
& {\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & -1 & -3 \\
0 & 0 & -4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
9 \\
-18 \\
-20
\end{array}\right]} \\
& \mathrm{x}+\mathrm{y}+\mathrm{z}=9 \ldots \ldots(1) \\
& -\mathrm{y}-3 \mathrm{z}=-18 \ldots \ldots \ldots(2) \\
& -4 \mathrm{z}=-20 \ldots \ldots \ldots(3) \\
& (3) \Rightarrow \mathrm{z}=5,(2) \Rightarrow \mathrm{y}=18-3 \mathrm{z}=13,(1) \Rightarrow \mathrm{x}=9-\mathrm{y}-\mathrm{z} \Rightarrow \mathrm{x}=9-3-5=1 \\
& \therefore \text { Solution is } \mathrm{x}=1, \\
& \mathrm{y}=3, \\
& \mathrm{z}=5
\end{aligned}
$
Question 3.
Examine the consistency of the equations.
$
\begin{aligned}
& 2 x-3 y+3 z=5 \\
& 3 x+y-3 z=13 \\
& 2 x+19 y-47 z=32
\end{aligned}
$
Solution:
The given system of equations can be written in the form of a matrix equation as
$
\left[\begin{array}{rrr}
2 & -3 & 7 \\
3 & 1 & -3 \\
2 & 19 & -47
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
5 \\
13 \\
32
\end{array}\right]
$
The augmented matrix is

$
\begin{aligned}
& {[A, B]=\left[\begin{array}{rrrr}
2 & -3 & 7 & 5 \\
3 & 1 & -3 & 13 \\
2 & 19 & -47 & 32
\end{array}\right] \sim\left[\begin{array}{rrrr}
1 & -\frac{3}{2} & \frac{7}{2} & \frac{5}{2} \\
3 & 1 & -3 & 13 \\
2 & 19 & -47 & 32
\end{array}\right] R_1 \rightarrow \frac{1}{2} R_1} \\
& \sim\left[\begin{array}{cccc}
1 & -\frac{3}{2} & \frac{7}{2} & \frac{5}{2} \\
0 & \frac{11}{2} & -\frac{27}{2} & \frac{11}{2} \\
0 & 22 & -54 & 27
\end{array}\right] \begin{array}{l}
R_2 \rightarrow R_2-3 R_1 \\
R_3 \rightarrow R_3-2 R_1
\end{array} \\
& \sim\left[\begin{array}{cccc}
1 & -\frac{3}{2} & \frac{7}{2} & \frac{5}{2} \\
0 & \frac{11}{2} & -\frac{27}{2} & \frac{11}{2} \\
0 & 0 & 0 & 5
\end{array}\right] R_3 \rightarrow R_3-4 R_2 \\
&
\end{aligned}
$
The last equivalent matrix is in the echelon form. It has three non-zero rows.
$
\begin{aligned}
\therefore \quad \rho[\mathrm{A}, \mathrm{B}] & =3 \text { and } \rho(\mathrm{A})=2 \\
\rho(\mathrm{A}) & \neq \rho[\mathrm{A}, \mathrm{B}]
\end{aligned}
$
$\therefore$ The given system is inconsistent and hence has no solution.
Question 4.
Show that the equations: $x+y+z=6, x+2 y+3 z=14, x+4 y+7 z=30$ are consistent and solve them.

Solution:
The matrix equation corresponding to the given system is.
$
\begin{aligned}
{\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 7
\end{array}\right]\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] } & =\left[\begin{array}{c}
6 \\
14 \\
30
\end{array}\right] \\
\mathrm{AX} & =\mathrm{B}
\end{aligned}
$
The augmented matrix is

$
\begin{aligned}
{[\mathrm{A}, \mathrm{B}] } & =\left[\begin{array}{cccc}
1 & 1 & 1 & 6 \\
1 & 2 & 3 & 14 \\
1 & 4 & 7 & 30
\end{array}\right] \sim\left[\begin{array}{cccc}
1 & 1 & 1 & 6 \\
0 & 1 & 2 & 8 \\
0 & 3 & 6 & 24
\end{array}\right] \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1
\end{array} \\
& \sim\left[\begin{array}{llll}
1 & 1 & 1 & 6 \\
0 & 1 & 2 & 8 \\
0 & 0 & 0 & 0
\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_2
\end{aligned}
$
In the last equivalent matrix, there are two non-zero rows.
$
\begin{aligned}
\therefore \quad \rho(\mathrm{A}, \mathrm{B}) & =2 \text { and } \rho(\mathrm{A})=2 \\
\rho(\mathrm{A}) & =\rho(\mathrm{A}, \mathrm{B})
\end{aligned}
$
$\therefore$ The given system is consistent. But the value of the common rank is less than the number of unknowns. The given system has an infinite number of solutions. The given system is equivalent to the matrix equation
$
\begin{aligned}
& {\left[\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
6 \\
8 \\
0
\end{array}\right]} \\
& x+y+z=6 \ldots \ldots(1) \\
& y+2 z=8 \ldots(2)
\end{aligned}
$
(2) $\Rightarrow y=8-2 z ;(1) \Rightarrow x=6-y-z=6-(8-2 z)-z=z-2$
Taking $\mathrm{z}=\mathrm{k}$, we get $\mathrm{x}=\mathrm{k}-2, \mathrm{y}=8-2 \mathrm{k} ; \mathrm{k} \in \mathrm{R}$
Putting $\mathrm{k}=1$, we have one solution as $\mathrm{x}=-1, \mathrm{y}=6, \mathrm{z}=1$. Thus by giving different values for $\mathrm{k}$ we get different solutions. Hence the given system has infinite number of solutions.