ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.6

Continuous practice using ML Aggarwal Class 12 Solutions ISC Chapter 4 Determinants Ex 4.6 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.6

Using cramer’s rule, solve the following (1 to 7) systems of linear equations:
1. (i) 3x + y = 5
x + 2y = 3
(ii) 2x – y = 17
3x + 5y = 6
Solution:
(i) The given system is equivalent to
AX = B
where A = \(\left[\begin{array}{ll}
3 & 1 \\
1 & 2
\end{array}\right]\) ;
X = \(=\left[\begin{array}{l}
x \\
y
\end{array}\right]\) ;
B = \(\left[\begin{array}{l}
5 \\
3
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
3 & 1 \\
1 & 2
\end{array}\right|\)
= 6 – 1
= 5 ≠ 0
∴ given system has unique solution.
|A₁| = \(\left|\begin{array}{ll}
5 & 1 \\
3 & 2
\end{array}\right|\)
= 10 – 3 = 7
|A₂| = \(\left|\begin{array}{ll}
3 & 5 \\
1 & 3
\end{array}\right|\)
= 9 – 5 = 4
Then by cramer’s rule, we have
x = \(\frac{\left|\mathrm{A}_1\right|}{|\mathrm{A}|}\)
= \(\frac{7}{5}\)
and y = \(\frac{\left|\mathrm{A}_2\right|}{|\mathrm{A}|}\)
= \(\frac{4}{5}\)

(ii) For given system of equations, we have
D = \(\left|\begin{array}{rr}
2 & -1 \\
3 & 5
\end{array}\right|\)
= 10 + 3
= 13 ≠ 0
∴ given system of eqns. have unique solution.
D₁ = \(\left|\begin{array}{rr}
17 & -1 \\
6 & 5
\end{array}\right|\)
= 85 +6
= 91
and D₂ = \(\left|\begin{array}{rr}
2 & 17 \\
3 & 6
\end{array}\right|\)
= 12 – 51
= – 39
Then by Cramer’s rule, we have
x = \(\frac{D_1}{D}\)
= \(\frac{91}{13}\)
= 7;
y = \(\frac{D_2}{D}\)
= [/latex]-\frac{39}{13}[/latex]
= – 3
Hence x = 7, y = – 3 be the required solution.

Question 2.
(i) 2x – 7y – 13 = 0 ; 5x + 6y – 9 = 0
(ii) \(\frac{2}{x}+\frac{3}{y}\) = 2 ; \(\frac{5}{x}+\frac{8}{7}=\frac{31}{6}\)
Solution:
(i) The given system of equations be:
2x – 7y = 13
5x + 6y = 9
Here D = \(\left|\begin{array}{rr}
2 & -7 \\
5 & 6
\end{array}\right|\)
= 12 + 35
= 47 ≠ 0
∴ Thus given system has unique solution.
D₁ = \(\left|\begin{array}{rr}
13 & -7 \\
9 & 6
\end{array}\right|\)
= 78 +63
= 141
D₂ = \(\left|\begin{array}{rr}
2 & 13 \\
5 & 9
\end{array}\right|\)
= 18 – 65
= – 47
∴ x = \(\frac{D_1}{D}=\frac{141}{47}\)
= 3 ;
y = \(\frac{D_2}{D}=-\frac{47}{47}\)
= – 1
Hence the required solution of given system be x = 3 and y = – 1.

(ii) putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v in given system of eqn’s we have,
2u + 3v = 2
5u + 8v = \(\frac{31}{6}\)
Here, D = \(\left|\begin{array}{ll}
2 & 3 \\
5 & 8
\end{array}\right|\)
= 16 – 15
= 1 ≠ 0
∴ given system of equations has unique solution.
D₁ = \(\left|\begin{array}{rr}
2 & 3 \\
\frac{31}{6} & 8
\end{array}\right|\)
= 16 – \(\frac{31}{2}\)
= \(\frac{1}{2}\)
D₂ = \(\left|\begin{array}{rr}
2 & 2 \\
5 & \frac{31}{6}
\end{array}\right|\)
= \(\frac{31}{3}\) – 10
= \(\frac{1}{3}\)
Thus by cramer’s rule, we have
u = \(\frac{D_1}{D_2}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\)
and v = \(\frac{D_2}{D}=\frac{1}{3}\)
⇒ \(\frac{1}{x}=\frac{1}{2}\)
⇒ x = 2
and \(\frac{1}{y}=\frac{1}{3}\)
⇒ y = 3
Hence the required solution of given system be x = 2 and y = 3.

Question 3.
3x + ay = 4
2x + ay = 2, a ≠ 0.
Solution:
The given system of equations is,
3x + ay = 4 ………….(1)
and 2x + ay = 2 ………..(2)
Here |A| = \(\left|\begin{array}{ll}
3 & a \\
2 & a
\end{array}\right|\)
= 3a – 2a = a
∴ The given system of equations has unique solution and it is given by using cramer’s rule.
x = \(\frac{|\mathrm{A}|}{|\mathrm{A}|}\),
y = \(\frac{\left|\mathrm{A}_2\right|}{|\mathrm{A}|}\) …………..(3)
Where, |A₁| = \(\left|\begin{array}{ll}
4 & a \\
2 & a
\end{array}\right|\)
= 4a – 2a = 2a
and |A₂| = \(\left|\begin{array}{ll}
3 & 4 \\
2 & 2
\end{array}\right|\)
= 6 – 8 = – 2
∴ From (3) ;
x = \(\frac{2 a}{a}\) = 2 ;
y = \(\frac{-2}{a}\)

Question 4.
(i) 5x – 7y + z = 11
6x – 8y -z = 11
3x + 2y – 6z = 7
(ii) x + y + z + 1 = 0
x + 2y + 3z + 4 = 0
x + 3y + 4z + 6 = 0
Solution:
(i) The given system of eqns. is,
5x – 7y + z = 11 ………….(1)
6x – 8y – z = 15 ………….(2)
and 3x + 2y – 6z = 7 ………….(3)
The given system of eqns is equivalent to
AX = B
Where A = \(\left[\begin{array}{rrr}
5 & -7 & 1 \\
6 & -8 & -1 \\
3 & 2 & -6
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{r}
11 \\
15 \\
7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rrr}
5 & -7 & 1 \\
6 & -8 & -1 \\
3 & 2 & -6
\end{array}\right|\)
Expanding along R₁ ; we get
= 5 (48 + 2) + 7 (- 36 + 3) + (12 + 24)
= 250 – 231 + 36
= 55 ≠ 0
∴ given system of eqns. has unique solution and by Cramer’s rule
x = \(\frac{\left|\mathrm{A}_1\right|}{|\mathrm{A}|}\) ;
y = \(\frac{\left|\mathrm{A}_2\right|}{|\mathrm{A}|}\) ;
z = \(\frac{\left|\mathrm{A}_3\right|}{|\mathrm{A}|}\) …………..(2)
Hence, |A₁| = \(\left|\begin{array}{rrr}
11 & -7 & 1 \\
15 & -8 & -1 \\
7 & 2 & -6
\end{array}\right|\)
= 11 (48 + 2) + 7 (- 90 + 7) + 1 (30 + 56)
= 550 – 581 + 86
= 55

|A₂| = \(\left|\begin{array}{rrr}
5 & 11 & 1 \\
6 & 15 & -1 \\
3 & 7 & -6
\end{array}\right|\)
= 5 (- 90 + 7) – 11 (- 36 + 3) + 1 (42 – 45)
= – 415 + 363 – 3
= – 418 + 363
= – 55

|A₃| = \(\left|\begin{array}{rrr}
5 & -7 & 11 \\
6 & -8 & 15 \\
3 & 2 & 7
\end{array}\right|\)
= 5 (- 56 – 30) + 7 (42 – 45) + 11 (12 + 24)
= – 430 – 21 + 396
= – 55

∴ From (2) ;
x = \(\frac{55}{55}\) = 1 ;
y = \(\frac{-55}{55}\) = – 1
and z = \(\frac{-55}{55}\) = – 1
Thus x = 1 ; y = – 1 and z = – 1

(ii) The given system of eqns is equivalent to AX = B
where A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & 4
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{l}
-1 \\
-4 \\
-6
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & 4
\end{array}\right|\)
= 1 (- 1) – 1 (1) + 1 (1)
= – 1 – 1 + 1
= – 1 ≠ 0
∴ The given system of equations has unique solution and it is given by using cramer’s rule.
x = \(\frac{\left|A_1\right|}{|A|}\) ;
y = \(\frac{\left|A_2\right|}{|A|}\)
and z = \(\frac{\left|A_3\right|}{|A|}\) ………..(1)
where, |A₁| = \(\left|\begin{array}{lll}
-1 & 1 & 1 \\
-4 & 2 & 3 \\
-6 & 3 & 4
\end{array}\right|\)
= – 1 (- 1) – 1 (2) + 1 (0)
= 1 – 2
= – 1
|A₂| = \(\left|\begin{array}{lll}
1 & -1 & 1 \\
1 & -4 & 3 \\
1 & -6 & 4
\end{array}\right|\)
= 1 (2) + 1 (1) + 1 (- 2)
= 2 + 1 – 2 = 1
|A₃| = \(\left|\begin{array}{lll}
1 & 1 & -1 \\
1 & 2 & -4 \\
1 & 3 & -6
\end{array}\right|\)
= 1 (0) – 1 (- 2) – 1 (1)
= 2 – 1 = 1
∴ From (1) ; we have
x = \(\frac{-1}{-1}\) = 1 ;
y = \(\frac{1}{-1}\) = – 1;
z = \(\frac{1}{-1}\) = – 1.

Question 5.
(i) 4x – 2y + 9z + 2 = 0
3x + 4y + z – 5 = 0
x – 3y + 2z – 8 = 0

(ii) x + y = 2
2x – z = 1
2y – 3z = 1
Solution:
Given system of eqn’s can be written as :
4x – 2y + 9z = – 2
3x + 4y + z = 5
x – 3y + 2z = 8
Here D = \(\left|\begin{array}{rrr}
4 & -2 & 9 \\
3 & 4 & 1 \\
1 & -3 & 2
\end{array}\right|\) ;
Expanding along R₁
= 4 (8 + 3) + 2 (6 – 1) + 9 (- 9 – 4)
= 44 + 10 – 117
= – 63 ≠ 0
Thus given system of eqns has unique solution.
D₁ = \(\left|\begin{array}{rrr}
-2 & -2 & 9 \\
5 & 4 & 1 \\
8 & -3 & 2
\end{array}\right|\) ;
expanding along R₁
= – 2 (8 + 3) + 2 (10 – 8) + 9 (- 15 – 32)
= – 22 + 4 – 423
= – 441
D₂ = \(\left|\begin{array}{rrr}
4 & -2 & 9 \\
3 & 5 & 1 \\
1 & 8 & 2
\end{array}\right|\) ;
Expanding along R₁
= 4 (10 – 8) + 2 (6 – 1) + 9 (24 – 5)
= 8 + 10 + 171
= 189
and D₃ = \(\left|\begin{array}{rrr}
4 & -2 & -2 \\
3 & 4 & 5 \\
1 & -3 & 8
\end{array}\right|\) ;
Expanding along R₁
= 4 (32 + 15) + 2 (24 – 5) – 2 (- 9 – 4)
= 188 + 38 + 262
= 252
Thus the Cramer’s rule, we have
x = \(\frac{D_1}{D}\)
= \(\frac{-441}{-63}\)
= 7 ;
y = \(\frac{D_2}{D}\)
= \(\frac{189}{-63}\)
= – 3 ;
z = \(\frac{D_3}{D}\)
= \(\frac{252}{-63}\)
= – 4
Hence the required solution of given system be
x = 7 ;
y = – 3
and z = – 4

(ii) The given system of equations is:
x + y = 2
2x – z = 1
2y – 3z = 1
Here D = \(\left|\begin{array}{rrr}
1 & 1 & 0 \\
2 & 0 & -1 \\
0 & 2 & -3
\end{array}\right|\) ;
Expanding along R₁
= 1 (0 + 2) – 1 (- 6 – 0)
= 2 + 6
= 8 ≠ 0
Thus given system of eqns. has unique solution.
D₁ = \(\left|\begin{array}{rrr}
2 & 1 & 0 \\
1 & 0 & -1 \\
1 & 2 & -3
\end{array}\right|\)
= 2 (0 + 2) – 1 (- 3 + 1)
= 4 + 2
= 6
D₂ = \(\left|\begin{array}{rrr}
1 & 2 & 0 \\
2 & 1 & -1 \\
0 & 1 & -3
\end{array}\right|\)
= 1 (- 3 + 1) – 2 (- 6 – 0)
= – 2 + 12
= 0
D₃ = \(\left|\begin{array}{lll}
1 & 1 & 2 \\
2 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)
= 1 (0 – 2) – 1 (2 – 0) 2(4 – 0)
= – 2 – 2 + 8
= 4
Thus, by cramer’s rule, we have
x = \(\frac{D_1}{D}\)
= \(\frac{6}{8}=\frac{3}{4}\) ;

y = \(\frac{\mathrm{D}_2}{\mathrm{D}}\)
= \(\frac{10}{8}=\frac{5}{4}\) ;

and z = \(\frac{D_3}{D}\)
= \(\frac{4}{8}=\frac{1}{2}\)
Hence the required solution of given system be
x = \(\frac{3}{4}\) ;
y = \(\frac{5}{4}\) and
z = \(\frac{1}{2}\).

Question 6.
(i) 2x – 3y = 1
x + 3z = 11
x + 2y + z = 7

(ii) \(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}\) = 4
\(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}\) = 1
\(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}\) = 2
Solution:
(i) The given system of eqns. are
2x – 3y = 1
x + 3z = 11
x + 2y + z = 7
Here, D = \(\left|\begin{array}{rrr}
2 & -3 & 0 \\
1 & 0 & 3 \\
1 & 2 & 1
\end{array}\right|\) ;
Expanding along R₁ ;
= 2 (0 – 6) + 3 (1 – 3) + 0
= – 12 – 6
= – 18 ≠ 0
Thus, given system of equations has unique solution.
D₁ = \(\left|\begin{array}{rrr}
1 & -3 & 0 \\
11 & 0 & 3 \\
7 & 2 & 1
\end{array}\right|\)
= 1 (0 – 6) + 3 (11 – 21) + 0
= – 6 – 30
= – 36
D₂ = \(\left|\begin{array}{rrr}
2 & 1 & 0 \\
1 & 11 & 3 \\
1 & 7 & 1
\end{array}\right|\)
= 2 (11 – 21) + 1 (1 – 3) + 0
= – 20 + 2
= – 18
D₃ = \(\left|\begin{array}{rrr}
2 & -3 & 1 \\
1 & 0 & 11 \\
1 & 2 & 7
\end{array}\right|\)
= 2 (0 – 22) + 3 (7 – 11) + 1 (2 – 0)
= – 44 – 12 + 2
= – 54
Then by Cramer’s rule, we have
x = \(\frac{D_1}{D}\)
= \( \frac{-36}{-18}\)
= 2;
y = \(\frac{D_2}{D}\)
= \(\frac{-18}{-18}\)
= 1;
and z = \(\frac{D_3}{D}\)
= \( \frac{-54}{-18}\)
= 3
thus, the required solution of given system be
x = 2 ; y = 1 and z = 3.

(ii) putting \(\frac{1}{x}\) = u ;
\(\frac{1}{y}\) = v ;
\(\frac{1}{z}\) = w
in given system of equations ; we get
Here D = \(\left|\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right|\) ;
expanding along R₁
= 2 (120 – 45) – 3 (- 80 – 30) + 10 (36 +36)
= + 150 + 330 + 720
= 1200 ≠ 0
Thus, given system has unique solution.

D₁ = \(\left|\begin{array}{rrr}
4 & 3 & 10 \\
1 & -6 & 5 \\
2 & 9 & -20
\end{array}\right|\)
= 4 (120 – 45) – 3 (- 20 – 10) + 10 (9 +12)
= 300 + 90 + 210
= 600
D₂ = \(\left|\begin{array}{rrr}
2 & 4 & 10 \\
4 & 1 & 5 \\
6 & 2 & -20
\end{array}\right|\)
= 2 (- 20 – 10) – 4 (- 80 – 30) + 10 (8 – 6)
= – 60 + 440 + 20
= 400
D₃ = \(\left|\begin{array}{rrr}
2 & 3 & 4 \\
4 & -6 & 1 \\
6 & 9 & 2
\end{array}\right|\)
= 2 (- 12 – 9) – 3 (8 – 6) + 4 (36 + 36)
= – 42 – 6 + 288
= 240
Then by cramer’s rule, we have
u = \(\frac{D_1}{D}=\frac{600}{1200}=\frac{1}{2}\)
⇒ \(\frac{1}{x}=\frac{1}{2}\)
⇒ x = 2
v = \(\frac{D_2}{D}=\frac{400}{1200}=\frac{1}{3}\)
⇒ \(\frac{1}{y}=\frac{1}{3}\)
⇒ y = 3
and w = \(\frac{D_3}{D}=\frac{240}{1200}=\frac{1}{5}\)
⇒ \(\frac{1}{z}=\frac{1}{5}\)
⇒ z = 5.
Hence cthe required solution of given system be
x = 2 ; y = 3 and z = 5.

Question 7.
x + y + z = 1
ax + by + cz = k
a²x + b²y + c²z = k where a, b, c are all different.
Solution:
The given system of eqns. is equivalent to AX = B
where A = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2
\end{array}\right|\);
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{c}
1 \\
k \\
k^2
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2
\end{array}\right|\);
operate C₂ → C₂ – C₁
C₃ → C₃ – C₁
= \(\left|\begin{array}{ccc}
1 & 0 & 0 \\
a & b-a & c-a \\
a^2 & b^2-a^2 & c^2-a^2
\end{array}\right|\) ;
Taking (b – a) common from C₂ and (c – a) common from C₃

Question 8.
The sum of three numbers is 20. If we multiply the first number by 2 and add the second number to the result and subtract the third number, we get 23. By adding second and third numbers to three times the first number, we get 46. Represent the above problem algebraically and use Cramer’s rule to find the numbers from these equations.
Solution:
The given problem can be formulated mathematically as under:
x + y + z = 20
2x + y – z = 23
3x + y + z = 46
Here D = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
2 & 1 & -1 \\
3 & 1 & 1
\end{array}\right|\);
expanding along R₁
= 1(1 + 1) – 1 (2 + 3) + 1 (2 – 3)
= 2 – 5 – 1
= – 4 ≠ 0
Thus given system of eqn’s has unique solution.
D₁ = \(\left|\begin{array}{rrr}
20 & 1 & 1 \\
23 & 1 & -1 \\
46 & 1 & 1
\end{array}\right|\)
= 20 (1 + 1) – 1 (23 + 46) + 1 (23 – 46)
= 40 – 69 – 23
= – 52
D₂ = \(\left|\begin{array}{rrr}
1 & 20 & 1 \\
2 & 23 & -1 \\
3 & 46 & 1
\end{array}\right|\)
= 1 (23 + 46) – 20 (2 + 3) + 1 (92 – 69)
= 69 – 100 + 23
= – 8
D₃ = \(\left|\begin{array}{lll}
1 & 1 & 20 \\
2 & 1 & 23 \\
3 & 1 & 46
\end{array}\right|\)
= 1 (46 – 23) – 1 (92 – 69) + 20 (2 – 3)
= 23 – 23 – 20
= – 20
Thus, by cramer’s rule, we get
x = \(\frac{D_1}{D}\)
=\frac{-52}{-4}[/latex]
= 13 ;
y = \(\frac{D_2}{D}\)
= \(\frac{-8}{-4}\)
= 2 ;
z = \(\frac{D_3}{D}\)
= \(\frac{-20}{-4}\)
= 5
Thus the required three numbers are 13, 2 and 5 respectively.

Question 9.
Which of the following equations are consistent? If consistent, solve them.
(i) 3x + y = 5
6x + 2y = 11

(ii) x + 2y = 3
4x + 8y = 12
Solution:
(i) The given system of eqns is;
3x + y = 5
6x + 2y = 11
Here, D = \(\left|\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right|\)
= 6 – 6 = 0
D = \(\left|\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right|\)
= 10 – 11
= – 1
Thus by Cramer’s rule given system has no solution and given system is inconsistent.

(ii) Given system of eqn’s is
x + 2y = 3
4x + 8y = 12
Here, D = \(\left|\begin{array}{ll}
1 & 2 \\
4 & 8
\end{array}\right|\)
= 8 – 8 = 0
D₁ = \(\left|\begin{array}{rr}
3 & 2 \\
12 & 8
\end{array}\right|\)
= 24 – 24 = 0
D₂ = \(\left|\begin{array}{rr}
1 & 3 \\
4 & 12
\end{array}\right|\)
= 12 – 12 = 0
Thus D = D₁ = D₂ = 0
Then by Cramer’s rule. the given system is consistent and has infinitely many solutions.

For Solution :
putting y = k in first equation of given system, we have
x = 3 – 2k. where k be any number
Clearly x = 3 – 2k and y = k satisfies the 2nd eqn. of given system.
Hence, the given system is consistent and has infinitely many solutions given by
x = 3 – 2k ; y = k; where k be any number.

Question 10.
Which of the following equations are consistent ? If consistent, solve them.
(i) x – y + 3z = 6
x + 3y – 3z = – 4
5x + 3y + 3z = 14

(ii) 4x – 2y + 6z = 8
2x – y + z = 4
2x – y + z = 13.

(iii) 2x + y – 2z = 4
x – 2y + z = – 2
5x – 5y + z = – 2
Solution:
(i) The given system of eqns is
x – y + 3z = 6 ………….(1)
x + 3y – 3z = – 4 …………(2)
and 5x + 3y – 3z = 14 ………….(30
Here |A| = \(\left|\begin{array}{rrr}
1 & -1 & 3 \\
1 & 3 & -3 \\
5 & 3 & 3
\end{array}\right|\)
= 1 (9 + 9) + 1 (3 + 15) – 3 (3 – 15)
= 18 + 18 – 36 = 0
Also |A₁| = \(\)
= 6 (18) + 1 (30) + 3 (- 54)
= 108 + 30 – 162
= – 24 ≠ 0
∴ The given system of eqns has no solution and they are inconsistent.

(ii) Given system of eqn’s is
4x – 2y + 6z = 8
2x – y + z = 4
2x – y + z = 13
Here D = \(\left|\begin{array}{rrr}
4 & -2 & 6 \\
2 & -1 & 3 \\
2 & -1 & 3
\end{array}\right|\) = 0
[∵ R₂ and R₃ are identical]

D₁ = \(\left|\begin{array}{rrr}
8 & -2 & 6 \\
4 & -1 & 3 \\
13 & -1 & 3
\end{array}\right|\) ;
Expanding along R₁
= 8 (- 3 + 3) + 2 (12 – 39) + 6 (- 4 + 13)
= 0 – 54 + 54
= 0

D₂ = \(\left|\begin{array}{rrr}
4 & 8 & 6 \\
2 & 4 & 3 \\
2 & 13 & 3
\end{array}\right|\) ;
Expanding along R₁
= 4 (12 – 39) – 8 (6 – 6) + 6 (26 – 8)
= – 108 + 0 + 108
= 0

D₃ = \(\left|\begin{array}{rrr}
4 & -2 & 8 \\
2 & -1 & 4 \\
2 & -1 & 13
\end{array}\right|\) ;
Expanding along R₁
= 4 (- 13 + 4) + 2 (26 – 8) + 8 (- 2 + 2)
= – 36 + 36 + 0
= 0
Here, D = D₁ = D₂ = D₃ = 0
Thus, by Cramer’s rule, given system of eqn’s has infinitely many solutions and system of eqn’s is consistent.

For solution :
putting z = k in eqn. (1) and eqn. (3) of given system, we have
4x – 2y = 8 – 6k
⇒ 2x – y = 4 – 3k
and 2x – y + 3k = 13
⇒ 2x – y = 13 – 3k
∴ 4 – 3k = 13 – 3k
⇒ 4=13
which is false.
Thus given system is inconsistent and has no solution.

(iii) The given system of eqns is ;
2x + y – 2z = 4
x – 2y + z = – 2
5x – 5y + z = – 2
Here D = \(\left|\begin{array}{rrr}
2 & 1 & -2 \\
1 & -2 & 1 \\
5 & -5 & 1
\end{array}\right|\)
= 2 (- 2 + 5) – 1 (1 – 5) – 2 (- 5 + 10)
= 6 + 4 – 10 = 0
D₁ = \(\left|\begin{array}{rrr}
4 & 1 & -2 \\
-2 & -2 & 1 \\
-2 & -5 & 1
\end{array}\right|\)
= 4 (- 2 + 5) – 1 (- 2 + 2) – 2 (- 5 + 10)
= 12 + 0 – 12 = 0

D₂ = \(\left|\begin{array}{rrr}
2 & 4 & -2 \\
1 & -2 & 1 \\
5 & -2 & 1
\end{array}\right|\)
= 2 (- 2 + 2) – 4 (1 – 5) – 2 (- 2 + 10)
= 0 + 16 – 16 = 0

D₃ = \(\left|\begin{array}{rrr}
2 & 1 & 4 \\
1 & -2 & -2 \\
5 & -5 & -2
\end{array}\right|\)
= 2 (4 – 10) – 1 (- 2 + 10) + 4 (- 5 + 10)
= – 12 – 8 + 20 = 0
Thus D = D₁ = D₂ = D₃ = 0
Then by Cramers rule, the given system is consistent and has infinitely many solutions.

For solution:
putting z = k in first two equations of given system, we have
2x + y = 4 + 2k
x – 2y = – 2 – k
Here |A| = \(\left|\begin{array}{rr}
2 & 1 \\
1 & -2
\end{array}\right|\)
= – 4 – 1
= – 5 ≠ 0
∴ given system has unique solution.
Here A₁ = \(\left|\begin{array}{rr}
4+2 k & 1 \\
-2-k & -2
\end{array}\right|\)
= – 2 (4 + 2k) – (- 2 – k)
= – 8 – 4k + 2 + k
= – 6 – 3k
and A₂ = \(\left|\begin{array}{cc}
2 & 4+2 k \\
1 & -2-k
\end{array}\right|\)
= 2 (- 2 – k) – (4 + 2k)
= – 4 – 2k – 4 – 2k
= – 8 – 4k
Thus, by Cramer’s rule, we have
x = \(\frac{\mathrm{A}_1}{\mathrm{~A}}\)
= \(\frac{-6-3 k}{-5}\)
= \(\frac{3(k+2)}{5}\)
and y = \(\frac{A_2}{A}\)
= \(\frac{-8-4 k}{-5}\)
= \(\frac{4(k+2)}{5}\)
and z = k
These values of x, y and z satisfies the 3rd eqn. of given system.
Hence the required solution of given system be
x = \(\frac{3}{5}\) (k + 2) ;
y = \(\frac{4}{5}\) (k + 2); = k,
where k be any number.

Question 11.
Which of the following systems has non trivial solutions ? If so, find these solutions.
(i) 5x + 5y + 2z = 0
2x + 5y + 4z = 0
4x + 5y + 2z = 0
(ii) 2x – 3y – z = 0
x + 3y – 2z =0
x – 3y = 0
Solution:
(i) The given homogeneous system of eqns. is
5x + 5y + 2z = 0
2x + 5y + 4z = 0
4x + 5y + 2z = 0
Here, D = \(\left|\begin{array}{lll}
5 & 5 & 2 \\
2 & 5 & 4 \\
4 & 5 & 2
\end{array}\right|\)
= 5 (10 – 20) – 5 (4 – 16) + 2 (10 – 20)
= – 50 + 60 – 20
= – 10 ≠ 0
Thus, given system has trivial solution.
i.e., x = 0 = y = z.

(ii) The given system of eqns is,
2x – 3y – z = 0 …………..(1)
x + 3y – 2z =0 …………..(2)
x – 3y = 0 …………..(3)
Here, |A| = \(\left|\begin{array}{ccc}
2 & -3 & -1 \\
1 & 3 & -2 \\
1 & -3 & 0
\end{array}\right|\)
= 2 (- 6) + 3 (0 + 2) – 1 (- 3 – 3)
= – 12 + 6 + 6
= 0
∴The given system of eqns. has infinite number of solutions.
Let z = k
From (1),
2x – 3y = k …………..(3)
and From (2),
x + 3y = 2k …………(4)
Here, |D| = \(\left|\begin{array}{rr}
2 & -3 \\
1 & 3
\end{array}\right|\)
= 6 + 3
= 9 ≠ 0
Also, |D₁| = \(\left|\begin{array}{rr}
k & -3 \\
2 k & 3
\end{array}\right|\)
= 3k + 6k
= 9k
and |D₂| = \(\left|\begin{array}{rr}
2 & k \\
1 & 2 k
\end{array}\right|\)
= 4k – k
= 3k
∴ by Cramer’s rule,
x = \(\frac{\left|D_1\right|}{|D|}\)
= \(\frac{9 k}{9}\) = k ;
and y = \(\frac{\left|\mathrm{D}_2\right|}{|\mathrm{D}|}\)
= \(\frac{3 k}{9}\)
= \(\frac{k}{3}\)
Putting in eqn. (3) ; we get
L.H.S. = k – 3 \(\left(\frac{k}{3}\right)\)
= k – k = 0
= R.H.S.
Hence x = k, y = \(\frac{k}{3}\) and z = k is a solution of given system for all values of k ∈ R.

Question 12.
If the system of equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solution, then find the possible values of k.
Solution:
Given system of equations is;
x – ky – z = 0
kx – y – z = 0
x + y – z = 0
The given homogeneous system of equations has non-trivial solution then
D = \(\left|\begin{array}{rrr}
1 & -k & -1 \\
k & -1 & -1 \\
1 & 1 & -1
\end{array}\right|\) = 0
⇒ 1 (1 + 1) + k (- k + 1) – 1 (k + 1) = 0
⇒ 2 – k² + k – k – 1 = O
⇒ k² = 1
⇒ k = ± 1.

Question 13.
Find the real value(s) of a for which the system of equations x + ay = 0, y + az = 0, z + ax = 0 has infinitely many solutions.
Solution:
The homogeneous system of equations is;
x + ay = 0
y + az = 0
z + ax = 0
The given homogeneous system of eqn’s has infinitely many solutions.
∴ |D| = 0
⇒ \(\left|\begin{array}{lll}
1 & a & 0 \\
0 & 1 & a \\
a & 0 & 1
\end{array}\right|\) = 0
Expanding along r₁ ; we have
1 (1 – 0) – a (0 – a²) = 0
⇒ 1 + a³ = 0
⇒ (a + 1) (a² – a + 1) = 0
a + 1 = 0 or
a² – a + 1 = 0
it does not give real values of a.
i.e., a = – 1.

Question 14.
If the equations x = cy + bz, y = az + cx, z = bx + ay are consistent, prove that a² + b² + c² + 2abc = 1.
Solution:
Given homogeneous system of eqn’s is equivalent to AX = O
where A = \(\left[\begin{array}{rrr}
-1 & c & b \\
c & -1 & a \\
b & a & -1
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
O = \(\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]\)
The given system of equations have non-trivial or infinitely many solution or consistent.
Then |A| = 0
⇒ \(\left|\begin{array}{rrr}
-1 & c & b \\
c & -1 & a \\
b & a & -1
\end{array}\right|\) = 0
expanding along R₁ ; we have
– 1 (1 – a²) – c (- c – ab) + b (ac + b) = 0
⇒ – 1 + a² + c² + b² + 2abc = 0
⇒ a² + b ² + c² + 2abc = 0