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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
I. One Mark Questions
Choose the correct statement.
Question 1.
If $\mathrm{A}$ is a matrix of order $\mathrm{m} \times \mathrm{n}$, then
(a) $\rho(\mathrm{A})=\mathrm{m}$
(b) $\rho(\mathrm{A})=\mathrm{n}$
(c) $\rho(A)=\min$ of $\{\mathrm{m}, \mathrm{n}\}$
(d) $\rho(\mathrm{A})<\mathrm{m}$
Answer:
(c) $\rho(\mathrm{A})=\min$ of $\{\mathrm{m}, \mathrm{n}\}$
Question 2.
If  is a transition probability matrix, then the value of a is
(a) $0.5$
(b) $0.7$
(c) 0
(d) $0.6$
Answer:
(b) $0.7$
Hint:
$
\begin{aligned}
& 0.3+\alpha=1 \\
& \Rightarrow \alpha=1-0.3=0.7
\end{aligned}
$
Question 3.
Let $\mathrm{AX}=\mathrm{B}$ be a system of $\mathrm{n}$ non-homogeneous linear equation. Then which of the following is correct?

(a) $|\mathrm{A}|=0$
(b) $\mathrm{A}=\mathrm{B}$
(c) $|\mathrm{A}| \neq 0$
(d) $\rho($ A $)<$ n
Answer:
(c) $|\mathrm{A}| \neq 0$
Question 4.
Choose the incorrect pair.
(a) $\left(\begin{array}{lll}3 & 2 & 4\end{array}\right)$
Column matrix
(b) $\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
Unit matrix
(c) $\left(\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right)$
(d) nonsingular matrix A
$|\mathrm{A}| \neq 0$
Answer:
(a) $(32$ 4) - Column matrix
Question 5.
If $\mathrm{A}$ is matrix $[\mathrm{A}, \mathrm{B}]$ is the augmented matrix then which of the following is true?
(a) $\rho([\mathrm{A}, \mathrm{B}])=\rho(\mathrm{A})$
(b) $\rho([\mathrm{A}, \mathrm{B}]) \geq \rho(\mathrm{A})$
(c) $\rho([\mathrm{A}, \mathrm{B}])=\rho(\mathrm{A})>\mathrm{n}$
(d) $\rho([\mathrm{A}, \mathrm{B}])<\rho($ A $)$
Answer:
(b) $\rho([\mathrm{A}, \mathrm{B}]) \geq \rho(\mathrm{A})$

Question 6.
Choose the echelon matrix.
(a) $\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 0 & 0\end{array}\right]$
(b) $\left[\begin{array}{ll}1 & 2 \\ 0 & 1 \\ 0 & 0\end{array}\right]$
(c) $\left[\begin{array}{ll}1 & 2 \\ 0 & 1 \\ 0 & 1\end{array}\right]$
(d) $\left[\begin{array}{ll}1 & 0 \\ 0 & 0 \\ 0 & 1\end{array}\right]$
Answer:
(b) $\left[\begin{array}{ll}1 & 2 \\ 0 & 1 \\ 0 & 0\end{array}\right]$
Question 7.
Let $\mathrm{A}$ be a non-singular matrix of order $(3 \times 3)$. Then $|\operatorname{adj} \mathrm{A}|$ is equal to
(a) $|\mathrm{A}|$
(b) $|\mathrm{A}|^2$
(c) $|\mathrm{A}|^3$
(d) $3|\mathrm{~A}|$
Answer:
(b) $|\mathrm{A}|^2$

Question 8.
For what value of $\mathrm{k}$ does the matrix? $\mathrm{A}=\left[\begin{array}{cc}k & -1 \\ 3 & 2\end{array}\right]$ does not have inverse?
(a) 1
(b) $\frac{-3}{2}$
(c) 0
(d) $-1$
Answer:
(b) $\frac{-3}{2}$
Hint:
For $\mathrm{A}=\left[\begin{array}{rr}k & -1 \\ 3 & 2\end{array}\right]$, if $|\mathrm{A}|=0$, then inverse does not exists
$
\left|\begin{array}{rr}
k & -1 \\
3 & 2
\end{array}\right|=0 \text { gives } 2 k+3=0 \text { or } k=\frac{-3}{2}
$
Question 9.
Rank of the matrix $A=\left(\begin{array}{rrrr}1 & -1 & 2 & 0 \\ 0 & 2 & 0 & 3\end{array}\right)$ is
(a) 1
(b) 0
(c) 2
(d) $-1$
Answer:
(c) 2
Hint:
The second order minors, $\left|\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right|,\left|\begin{array}{cc}-1 & 2 \\ 2 & 0\end{array}\right|,\left|\begin{array}{cc}2 & 0 \\ 0 & 3\end{array}\right|$ are not zero. So rank is 2

Question 10.
Choose the correct answer.
(a) A system of linear equations always has a unique solution.
(b) A system of linear equations can have more than one solution.
(c) A system of linear equations need not be consistent.
(d) All of the above
Answer:
(d) All of the above
Question 11.
For the matrix equation $\left(\begin{array}{cc}1 & 2 \\ 3 & -1\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}7 \\ 3\end{array}\right)$ the augmented matrix is
(a) $\left(\begin{array}{ccc}1 & 2 & 7 \\ 3 & -1 & 3\end{array}\right)$
(b) $\left(\begin{array}{ll}1 & 7 \\ 3 & 3\end{array}\right)$
(c) $\left(\begin{array}{ll}7 & 1 \\ 3 & 3\end{array}\right)$
(d) $\left(\begin{array}{rrr}1 & 7 & 2 \\ 3 & 3 & -1\end{array}\right)$
Answer:
(a) $\left(\begin{array}{ccc}1 & 2 & 7 \\ 3 & -1 & 3\end{array}\right)$
Question 12.
Find the correct statement. Cramer's rule can be used to solve,
(a) $\mathrm{n}$ equations in $\mathrm{n}$ unknowns
(b) When the determinant of the coefficient matrix is non-zero.
(c) The number of unknowns need not be equal to the number of equations.
(d) For infinite solutions
Answer:
(a) $\mathrm{n}$ equations in $\mathrm{n}$ unknowns \& (b) When the determinant of the coefficient matrix is non-zero.

Question 13.
Which of the following is a transition probability matrix?
(a) $\left(\begin{array}{cc}0.3 & 0.6 \\ 0.15 & 0.85\end{array}\right)$
(b) $\left(\begin{array}{ccc}0.9 & 0.075 & 0.025 \\ 0.15 & 0.8 & 0.05 \\ 0.25 & 0.25 & 0.5\end{array}\right)$
(c) $\left(\begin{array}{ll}1 & 0.2 \\ 0.3 & 0.25\end{array}\right)$
(d) $\left(\begin{array}{cc}0.35 & 0.6 \\ 0.2 & 0.7\end{array}\right)$
(i) only a
(ii) only $b$
(iii) both $\mathrm{c}$ and $\mathrm{d}$
(iv) both $\mathrm{a}$ and $\mathrm{b}$
Answer:
(iv) both $a$ and $b$
Hint:
The sum of all the probabilities should be equal to one
Question 14.
Choose the correct statements.
(a) the rank of an identity matrix is zero.
(b) the rank of an adjoint matrix is more than the rank of a matrix
(c) the rank of a diagonal matrix is equal to the number of rows
(d) the rank of an echelon matrix is greater than the matrix
Answer:
(c) the rank of a diagonal matrix is equal to the number of rows

Question 15.

Answer:
$
1-\mathrm{d}, 2-\mathrm{c}, 3-\mathrm{a}, 4-\mathrm{b}
$
II. 2 Mark Questions
Question 1.
Show that the inverse of $\mathrm{A}=\left(\begin{array}{cc}-6 & 9 \\ 4 & -6\end{array}\right)$ does not exist
Solution:
Show $|\mathrm{A}|=0$
Question 2.
Find the rank of $\mathrm{A}=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7\end{array}\right)$

Solution:
$
\begin{aligned}
& \text { Given } A=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 5 & 7
\end{array}\right) \\
& \text { Consider }\left(\begin{array}{ccc}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 5 & 7
\end{array}\right) \\
& =1(21-20)-2(14-12)+3(10-9) \\
& =1-4+3 \\
& =0
\end{aligned}
$
Since third order minor equals zero, $\rho(\mathrm{A})<3$
Consider $\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=3-4=-1 \neq 0$
There is a minor of order 2 which is not zero. Hence $\rho(\mathrm{A})=2$
Question 3.
Find the rank of $\mathrm{A}=\left[\begin{array}{rrrr}4 & 5 & -6 & -1 \\ 7 & -3 & 0 & 8\end{array}\right]$
Solution:
Given $\mathrm{A}=\left[\begin{array}{rrrr}4 & 5 & -6 & -1 \\ 7 & -3 & 0 & 8\end{array}\right]$
Consider the minor $\left|\begin{array}{cc}5 & -6 \\ -3 & 0\end{array}\right|=-18 \neq 0$
Since a minor of order 2 is not zero, the rank is 2

Question 4.
Solve the system by Cramer's rule, $4 x-3 y=11; 6 x+5 y=7$
Solution:
$
4 x-3 y=11 ; 6 x+5 y=7
$
The matrix equation is $\left(\begin{array}{rr}4 & -3 \\ 6 & -5\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}11 \\ 7\end{array}\right)$
$
\text { A } \mathrm{X}=\mathrm{B}
$
Now $\Delta=\left|\begin{array}{rr}4 & -3 \\ 6 & 5\end{array}\right|=20+18=38 \neq 0$
$\therefore$ we can apply Cramer's rule
$
\begin{aligned}
& \Delta_x=\left|\begin{array}{rr}
11 & -3 \\
7 & 5
\end{array}\right|=55+21=76 \\
& \Delta_y=\left|\begin{array}{rr}
4 & 11 \\
6 & 7
\end{array}\right|=28-66=-38 \\
& x=\frac{\Delta x}{\Delta}=\frac{76}{38}=2, \quad y=\frac{\Delta y}{\Delta}=\frac{-38}{38}=-1
\end{aligned}
$
Solution is $(2,-1)$

Question 5.
Solve the system by the rank method.
$
3 \mathrm{x}+5 \mathrm{y}=-7 ; \mathrm{x}+4 \mathrm{y}=-14
$
Solution:
$
3 x+5 y=-7 ; x+4 y=-14
$
The matrix equation of the system is
$
\left(\begin{array}{ll}
3 & 5 \\
1 & 4
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=\left(\begin{array}{c}
-7 \\
-14
\end{array}\right)
$
A $\quad \mathrm{X}=\mathrm{B}$
The Augmented matrix $[\mathrm{A}, \mathrm{B}]$ is
$
\begin{array}{rlr} 
& \left(\begin{array}{ccc}
3 & 5 & -7 \\
1 & 4 & -14
\end{array}\right) & \\
\sim & \left(\begin{array}{ccc}
1 & 4 & -14 \\
3 & 5 & -7
\end{array}\right) & \mathrm{R}_1 \leftrightarrow \mathrm{R}_2 \\
& \sim\left(\begin{array}{ccc}
1 & 4 & -14 \\
0 & -7 & 35
\end{array}\right) & \mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1
\end{array}
$
The number of non-zero rows is 2
So $\rho(A)=\rho([A, B])=2$
Now the new matrix equation is
$
\begin{aligned}
& -7 y=35 \Rightarrow y=-5 \\
& x+4 y=-14 \Rightarrow x=-14-4(-5)=6
\end{aligned}
$
Solution is $(6,-5)$
Question 6.
Find the solution of the system $5 x+y=-13 ; 3 x-2 y=0$

Solution:
$
5 x+y=-13 ; 3 x-2 y=0
$
We solve the system by Cramer's rule

The matrix equation is $\left(\begin{array}{cc}5 & 1 \\ 3 & -2\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-13 \\ 0\end{array}\right)$
$
\Delta=\left|\begin{array}{cc}
5 & 1 \\
3 & -2
\end{array}\right|=-10-3=-13 \neq 0
$
So the system has a unique solution
$
\begin{array}{ll}
\Delta_x=\left|\begin{array}{cc}
-13 & 1 \\
0 & -2
\end{array}\right|=26 & \Delta_y=\left|\begin{array}{rr}
5 & -13 \\
3 & 0
\end{array}\right|=39 \\
\text { So } x=\frac{\Delta_x}{\Delta}=\frac{26}{-13}=-2 & y=\frac{\Delta_y}{\Delta}=\frac{39}{-13}=-3
\end{array}
$
Hence solution is $(-2,-3)$
III. 3 and 5 Mark Questions
Question 1.
Find $\lambda$ so that the matrix $\left(\begin{array}{ccc}8 & 9 & 7 \\ 7 & 8 & 6 \\ 9 & 10 & \lambda\end{array}\right)$ is a singular matrix.
Solution:
Let $A=\left(\begin{array}{ccc}8 & 9 & 7 \\ 7 & 8 & 6 \\ 9 & 10 & \lambda\end{array}\right)$
Given $A$ is singular matrix $\Rightarrow|A|=0$
$
\begin{aligned}
& \left|\begin{array}{ccc}
8 & 9 & 7 \\
7 & 8 & 6 \\
9 & 10 & \lambda
\end{array}\right|=0 \\
& 8(8 \lambda-60)-9(7 \lambda-54)+7(70-72)=0 \\
& 64 \lambda-480-63 \lambda+486-14=0 \\
& \lambda+6-14=0 \Rightarrow \lambda=8
\end{aligned}
$

Question 2.
Find $\mathrm{k}$ so that the matrix $\left(\begin{array}{ccc}1 & 2 & k \\ 3 & 4 & 5 \\ 7 & 10 & 12\end{array}\right)$ is a non-singular matrix.
Solution:
Let $\mathrm{A}=\left(\begin{array}{ccc}1 & 2 & k \\ 3 & 4 & 5 \\ 7 & 10 & 12\end{array}\right)$ Give $\mathrm{A}$ is non-singular. That is $|\mathrm{A}| \neq 0$
$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & 2 & k \\
3 & 4 & 5 \\
7 & 10 & 12
\end{array}\right| \quad \neq 0 \\
& 1(48-50)-2(36-35)+k(30-28) \neq 0 \\
& -2-2+2 k \neq 0 \\
& 2 \mathrm{k} \neq 4 \Rightarrow k \neq 2
\end{aligned}
$
Question 3.
Solve by Cramer' rule, $x-y+z=2 ; 2 x-y=0 ; 2 y-z=1$
Solution:
$
\mathrm{x}-\mathrm{y}+\mathrm{z}=2 ; 2 \mathrm{x}-\mathrm{y}=0 ; 2 \mathrm{y}-\mathrm{z}=1
$
The matrix equation corresponding to the system is
$
\left(\begin{array}{ccc}
1 & -1 & 1 \\
2 & -1 & 0 \\
0 & 2 & -1
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
2 \\
0 \\
1
\end{array}\right)
$

The system has a unique solution
$
\begin{aligned}
& \Delta_x=\left|\begin{array}{ccc}
2 & -1 & 1 \\
0 & -1 & 0 \\
1 & 2 & -1
\end{array}\right|=2(1)+1(0)+1(1)=3 \\
& \Delta_y=\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & 0 & 0 \\
0 & 1 & -1
\end{array}\right|=1(0)-2(-2)+1(2)=6 \\
& \Delta_z=\left|\begin{array}{ccc}
1 & -1 & 2 \\
2 & -1 & 0 \\
0 & 2 & 1
\end{array}\right|=1(-1)+1(2)+2(4)=9 \\
& x=\frac{\Delta_x}{\Delta}=\frac{3}{3}=1 \quad y=\frac{\Delta_y}{\Delta}=\frac{6}{3}=2 \quad z=\frac{\Delta_z}{\Delta}=\frac{9}{3}=3 \\
&
\end{aligned}
$
Therefore the solution is $(1,2,3)$

Question 4.
Solve by rank method $x+2 y-3 z=-4 ; 2 x+3 y+2 z=2 ; 3 x-3 y-4 z=11$
Solution:
$
x+2 y-3 z=-4 ; 2 x+3 y+2 z=2 ; 3 x-3 y-4 z=11
$
The matrix equation is
$
\begin{aligned}
\left(\begin{array}{ccc}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\left(\begin{array}{c}
-4 \\
2 \\
11
\end{array}\right) \\
\mathrm{A} \quad \mathrm{X} & =\mathrm{B}
\end{aligned}
$
Augmented matrix [A, B]
$
\begin{aligned}
& \left(\begin{array}{cccr}
1 & 2 & -3 & -4 \\
2 & 3 & 2 & 2 \\
3 & -3 & -4 & 11
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 2 & -3 & -4 \\
0 & -1 & 8 & 10 \\
0 & -9 & 5 & 23
\end{array}\right) \quad \begin{array}{l}
\quad 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 & -3 & -4 \\
0 & -1 & 8 & 10 \\
0 & 0 & -67 & -67
\end{array}\right) \quad R_3 \rightarrow R_3-9 R_2 \\
&
\end{aligned}
$

The last equivalent matrix is in echelon form. $\rho(A)=\rho([\mathrm{A}, \mathrm{B}])=3=$ Number of unknowns The new matrix equation is given by $\left(\begin{array}{ccc}1 & 2 & -3 \\ 0 & -1 & 8 \\ 0 & 0 & -67\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}-4 \\ 10 \\ -67\end{array}\right)$
$
\begin{aligned}
& x+2 y-3 z=-4 \ldots \ldots \text { (1) } \\
& -y+8 z=10 \ldots \ldots \text { (2) } \\
& -67 z=-67 \ldots \ldots \ldots \text { (3) } \\
& \text { (3) } \Rightarrow z=1 \\
& \text { (2) } \Rightarrow-y+8=10 \Rightarrow-y=2 \Rightarrow y=-2 \\
& \text { (1) } \Rightarrow x=-4-2(-2)+3(1) \Rightarrow x=3
\end{aligned}
$
Hence the solution is $(x, y, z)=(3,-2,1)$

Question 5.
The sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add the second and third to it, we get 46. By using the rank method find the numbers.
Solution:
Let the three number be $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ respectively.
According to the problem,
$
\begin{aligned}
& x+y+z=20 \\
& 2 x+y-z=23 \\
& 3 x+y+z=46
\end{aligned}
$
The matrix equation is given by

$
\begin{aligned}
& x+y+z=20 \ldots \ldots . \text { (1) } \\
& -y-3 z=-17 \ldots \ldots . \text { (2) } \\
& 4 z=20 \ldots \ldots \text { (3) } \\
& \text { (3) } \Rightarrow z=5 \\
& \text { (2) } \Rightarrow-y=-17+3(5)=-2 \Rightarrow y=2 \\
& \text { (1) } \Rightarrow x=20-2-5=13
\end{aligned}
$
Hence the three numbers are $(13,2,5)$
Question 6.
Weekly expenditure in an office for three weeks is given as follows. Calculate the salary for each type of staff using Cramer's rule.

Solution:
Let the salary for the three types of staff $A, B$, and $C$ be $₹ x, ₹ y$, and $₹ z$ respectively.
According to the problem we have,
$
\begin{aligned}
& 4 x+2 y+3 z=4900 \\
& 3 x+3 y+2 z=4500 \\
& 4 x+3 y+4 z=5800
\end{aligned}
$
Now $\begin{aligned} \Delta=\left|\begin{array}{lll}3 & 3 & 2 \\ 4 & 3 & 4\end{array}\right|=4(12-6)-2(12-8)+3(9-12) \\ =24-8-9=7 \neq 0\end{aligned}$

Hence a unique solution exists and we can use Cramer's rule to solve the system.
$
\begin{aligned}
\Delta_x=\left|\begin{array}{lll}
4900 & 2 & 3 \\
4500 & 3 & 2 \\
5800 & 3 & 4
\end{array}\right| & =4900(6)-2(18000-11600)+3(13500-17400) \\
& =29400-2(6400)+3(-3900) \\
& =4900
\end{aligned}
$

Question 7.
Show that the equations $x-3 y-8 z=-10 ; 3 x+y-4 z=0; 2 x+5 y+6 z=13$ are consistent and have infinite sets of solutions.

The last equivalent matrix is in echelon form. It has two non-zero rows. $\rho(\mathrm{A})=\rho([\mathrm{A}, \mathrm{B}])=2<$ number of unknowns The system is consistent and has infinitely many solutions The changed matrix equation is given by $\left(\begin{array}{ccc}1 & -3 & -8 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}-10 \\ 3 \\ 0\end{array}\right)$
$
\begin{aligned}
& x-3 y-8 z=-10 \ldots \ldots(1) \\
& y+2 z=3 \ldots \ldots(2) \\
& (2) \Rightarrow y=3-2 z \\
& (1) \Rightarrow x=-10+3 y+8 z \\
& \Rightarrow x=-10+3(3-2 z)+8 z \\
& \Rightarrow x=-10+9-6 z+8 z \\
& \Rightarrow x=2 z-1
\end{aligned}
$
Let us take $\mathrm{z}=\mathrm{k}, \mathrm{k} \in \mathrm{R}$. We get $\mathrm{y}=3-2 \mathrm{k}$ and $\mathrm{x}=2 \mathrm{k}-1$.
By giving different values for $\mathrm{k}$, we get different solutions.

Question 8.
Find the rank of the matrix by reducing to echelon form $A=\left(\begin{array}{cccc}1 & 1 & 1 & 1 \\ 1 & 3 & -2 & 1 \\ 2 & 0 & -3 & 2\end{array}\right)$
Solution:
$
\begin{aligned}
& A=\left(\begin{array}{cccc}
1 & 1 & 1 & 1 \\
1 & 3 & -2 & 1 \\
2 & 0 & -3 & 2
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & 2 & -3 & 0 \\
0 & -2 & -5 & 0
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_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 & 1 \\
0 & 2 & -3 & 0 \\
0 & 0 & -8 & 0
\end{array}\right) \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_2 \\
&
\end{aligned}
$
The matrix is now in echelon form. Consider the minor of order 3 ,
$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & 2 & -3 \\
0 & 0 & -8
\end{array}\right|=1(-16)-1(0)+1(0)=-16 \neq 0
$
Hence rank of $\mathrm{A}=3$

Question 9.
Show that the equations $x-3 y+4 z=3 ; 2 x-5 y+7 z=6 ; 3 x-8 y+11 z=1$ are inconsistent.

Solution:
$
x-3 y+4 z=3 ; 2 x-5 y+7 z=6 ; 3 x-8 y+11 z=1
$
The matrix equation is given by
$
\begin{aligned}
& \left(\begin{array}{ccc}
1 & -3 & 4 \\
2 & -5 & 7 \\
3 & -8 & 11
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
3 \\
6 \\
1
\end{array}\right) \\
& \text { A } \quad \mathrm{X}=\mathrm{B} \\
&
\end{aligned}
$

The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and $[\mathrm{A}]$ has 2 non-zero rows.
$
\begin{aligned}
& \rho([\mathrm{A}, \mathrm{B}])=3 ; \rho(\mathrm{A})=2 \\
& \rho(\mathrm{A}) \neq \rho([\mathrm{A}, \mathrm{B}])
\end{aligned}
$
The system is inconsistent and has no solution.
Question 10.
Find $k$ if the equations $2 x+3 y-z=5 ; 3 x-y+4 z=2 ; x+7 y-6 z=k$ are consistent.
Solution:
Given that the system $2 x+3 y-z=5 ; 3 x-y+4 z=2, x+7 y-6 z=k$ is consistent.

The matrix equation is $\left(\begin{array}{ccc}2 & 3 & -1 \\ 3 & -1 & 4 \\ 1 & 7 & -6\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}5 \\ 2 \\ k\end{array}\right)$
$
\text { A } \quad X=B
$
Augmented matrix [A, B]
$
\begin{aligned}
& \left(\begin{array}{cccc}
2 & 3 & -1 & 5 \\
3 & -1 & 4 & 2 \\
1 & 7 & -6 & k
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 7 & -6 & k \\
2 & 3 & -1 & 5 \\
3 & -1 & 4 & 2
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_1 \leftrightarrow \mathrm{R}_3 \\
\mathrm{R}_2 \leftrightarrow \mathrm{R}_3
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 7 & -6 & k \\
0 & -11 & -11 & 5-2 k \\
0 & -22 & -22 & 2-3 k
\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 & 7 & -6 & k \\
0 & -11 & -11 & 5-2 k \\
0 & 0 & 0 & -8+k
\end{array}\right) \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_2 \\
&
\end{aligned}
$
$
\rho(\mathrm{A})=2 \rho([\mathrm{A}, \mathrm{B}])=2 \text { or } 3
$
For the equations to be consistent, $\rho([A, B])=\rho(A)=2$
$
-8+\mathrm{k}=0 \Rightarrow \mathrm{k}=8
$
Question 11.
Find $k$ if the equations $x+y+z=3 ; x+3 y+2 z=6 ; x+5 y+3 z=k$ are inconsistent.

Solution:
Given that the system is inconsistent

The matrix equation is $\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 5 & 3\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}3 \\ 6 \\ k\end{array}\right)$
$\mathrm{A} \quad \mathrm{X}=\mathrm{B}$
Augmented matrix $[\mathrm{A}, \mathrm{B}]$
$
\begin{aligned}
& \left(\begin{array}{llll}
1 & 1 & 1 & 3 \\
1 & 3 & 2 & 6 \\
1 & 5 & 3 & k
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 3 \\
0 & 2 & 1 & 3 \\
0 & 4 & 2 & k-3
\end{array}\right) \quad \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}{cccc}
1 & 1 & 1 & 3 \\
0 & 2 & 1 & 3 \\
0 & 0 & 0 & k-9
\end{array}\right) \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_2
\end{aligned}
$
For the equations to be inconsistent, $\rho[(\mathrm{A}, \mathrm{B})] \neq \rho(\mathrm{A})$ Here $\rho(A)=2$ So $\rho([A, B]) \neq 2$
That is the last row of the echelon matrix should not be zero
$
\Rightarrow \mathrm{k} \neq 9
$
Question 12.
Two newspapers A and B are published in a city. Their present market shares are $15 \%$ for A and $85 \%$ for $\mathrm{B}$. Of those who bought A the previous year, $65 \%$ continued to buy it again while $35 \%$ switched over to B. Of those who bought B the previous year, $55 \%$ buy it again and $45 \%$ switch over to A. Find their market shares after two years.
Solution:
Let the present shares of $A$ and $B$ be denoted by $(0.150 .85)$
The transition probability matrix is given by

Question 13.
The pattern of sunny and rainy days on a planet has two states. Every sunny day is followed by another sunny day with a probability of $0.8$. Every rainy day is followed by another rainy day with a probability of $0.6$. Today is sunny on the planet. What is the chance of rain the day after tomorrow?
Solution:
Let $\mathrm{S}$ and $\mathrm{R}$ denote sunny and rainy days on the planet The transition probability matrix is given by

We have to find the probability of rain after two days. We have to find $\mathrm{T}^2$

Hence if today is sunny on the planet, the chance of rain the day after tomorrow is $0.28$.