www.sakshieducation.com
MATRICES -3
EXERCISE – 3(h)
A. Solve the following system of equations.
i) By using Cramer’s rule and matrix inversion method when the coefficient
matrix is non-singular.
ii) Using Gauss-Jordan method. Also determine whether the system has a unique
solution or infinite number of solutions or no solution and find solutions if exist.
1. 5x – 6y + 4z = 15
7x + 4y – 3z = 19
2x + y + 6z = 46
∆
∆
∆
Hint : x = 1 , y = 2 , z = 3
∆
∆
∆
Sol. i) Cramer’s rule :
5 −6 4
∆= 7
2
−3
6
4
1
= 5(24 + 3) + 6(42 + 6) + 4(7 − 8)
= 135 + 288 − 4 = 419
15 −6 4
∆1 = 19
4
−3
46
1
6
= 15(24 + 3) + 6(114 + 138) + 4(19 − 184)
= 405 + 1512 − 660 = 1917 − 660 = 1257
5 15 4
∆ 2 = 7 19 −3
2 46 6
= 5(114 + 138) − 15(42 + 6) + 4(322 − 38)
= 1260 − 720 + 1136 = 1676
5 −6 15
∆3 = 7
2
4
1
19
46
= 5(184 − 19) + 6(322 − 38) + 15(7 − 8)
= 825 + 1074 − 15 = 2529 − 15 = 2514
www.sakshieducation.com
www.sakshieducation.com
∆1 1527
=
=3
419
∆
1676
∆
y= 2 =
=4
∆
419
∆
2514
=6
z= 3 =
∆
419
Solution is x = 3, y = 4, z = 6.
x=
ii) Matrix inversion method :
AdjA
Hint : A −1 =
det A
 5 −6 4 
A =  7 4 −3
 2 1 6 
A1 =
4 −3
= 24 + 3 = 27
1 6
B1 = −
C1 =
7 4
2 1
A2 = −
B2 =
7 −3
= −(42 + 6) = −48
1 6
−6 4
1
5 4
2 6
C2 = −
= 7 − 8 = −1
6
= −(−36 − 4) = 40
= 30 − 8 = 22
5 −6
= −(5 + 12) = −17
2 1
−6 4
= 18 − 16 = 2
4 −3
5 4
= −(−15 − 28) = 43
B3 = −
7 −3
5 −6
C3 =
= 20 + 42 = 62
7 4
A3 =
 A1 A 2
Adj A =  B1 B2
 C1 C2
Det A = ∆ = 419
A3   27 40 2 
B3  =  −48 22 43
C3   −1 −17 62 
www.sakshieducation.com
www.sakshieducation.com
 27
AdjA
1 
A =
=
48
DetA 419 
 −1
 27
1 
−1
x=A D=
−48
419 
 −1
−1
40 2 
22 43
−17 62 
40 2  15 
22 43 19 
−17 62   46 
 +405 + 760 + 92 
1 
=
−720 + 418 + 1978

419
 −15 − 323 + 2852 
1257   3 
1 
=
1676  =  4 

419
 2514   6 
∴ Solution is x = 3, y = 4, z = 6.
iii) Gauss-Jordan method :
 5 −6 4 15 
Augmented matrix is A =  7 4 −3 19 
 2 1 6 46 
R2 → 5R2 – 7R1, R3 → 5R3 – 2R1
15 
5 −6 4

A ~ 0 62 −43 −10 
0 17 22 200 
R1 → 31R1 + 3R2, R3 → 62R3 – 17R2
−5
435 
155 0

A ~  0 62 −43
−10 
0 2095 12570 
 0
 1 
R3 → R3 

 2095 
155 0 −5 435 
A ~  0 62 −43 −10 
 0
0
1
6 
R1 → R1 + 5R3, R2 → R2 + 43R3
155 0 0 465
A ~  0 62 0 248
 0
0 1 6 
www.sakshieducation.com
www.sakshieducation.com
 1 
 1 
R1 → R1 
 R2 → R2  
 155 
 62 
1 0 0 3 
A ~ 0 1 0 4 
0 0 1 6 
∴ Unique solution exists.
Solution is x = 3, y = 4, z = 6.
2.
I.
x+y+z=1
2x + 2y + 3z = 6
x + 4y + 9z = 3
i) Cramer’s rule
1 1 1
∆= 2 2 3
1 4 9
= 1(18 − 12) − 1(18 − 3) + 1(8 − 2)
= 6 − 15 + 6 = −3
1 1 1
∆1 = 6 2 3
3 4 9
= 1(18 − 12) − 1(54 − 9) + 1(24 − 6)
= 6 − 45 + 18 = −21
1 1 1
∆2 = 2 6 3
1 3 9
= 1(54 − 9) − 1(18 − 3) + 1(6 − 6)
= 45 − 15 = 30
1 1 1
∆3 = 2 2 6
1 4 3
= 1(6 − 24) − 1(6 − 6) + 1(8 − 2)
= −18 − 0 + 6 = −12
www.sakshieducation.com
www.sakshieducation.com
x=
∆1 −21
=
=7
∆
−3
y=
∆ 2 30
=
= −10
∆ −3
∆3 −12
=
=4
∆
−3
Solution is x = 7, y = –10, z = 4.
z=
ii) Matrix inversion method :
1 1 1 
x 


Let A =  2 2 3 , X =  y  and D =
1 4 9 
 z 
A1 =
2 3
4 9
B1 = −
C1 =
B2 =
1 4
2 3
2 3
2 2
= −(4 − 1) = −3
= 3− 2 =1
1 1
1 1
= −(9 − 4) = −5
= 9 −1 = 8
1 1
1 1
B3 = −
C3 =
4 9
1 9
= −(18 − 3) = −15
= 8−2 = 6
1 1
1 1
C2 = −
A3 =
1 9
1 4
A2 = −
= 18 − 12 = 6
2 3
2 2
1 
6
 
 3
= −(3 − 2) = −1
= 2−2 = 0
 A1 A 2
AdjA =  B1 B2
 C1 C2
DetA = ∆ = −3
A3   6 −5 1 
B3  =  −15 8 −1
C3   6 −3 0 
www.sakshieducation.com
www.sakshieducation.com
A
−1
 6 −5 1 
AdjA
1
=
= −  −15 8 −1
DetA
3
 6 −3 0 
 6 −5 1  1 
1
x = A −1D =  −15 8 −1 6 
3
 6 −3 0   3
 6 − 30 + 3 
 −21  7 
1
1

= −  −15 + 48 − 3 = −  30  =  −10 
3
3
 6 − 18 + 0 
 −12   4 
∴ Solution is x = 7, y = –10, z = 4.
iii) Gauss-Jordan method:
1 1 1 1 
Augmented matrix is A =  2 2 3 6 
1 4 9 3
R 2 → R 2 − 2R1 , R 3 → R 3 − R1
1 1 1 1 
A ~ 0 0 1 4 
0 3 8 2 
R 3 → R 3 − 8R 2 , R1 → R1 − R 2
1 1 0 −3 
A ~ 0 0 1 4 
0 3 0 −30 
1
R3 → R3  
3
1 1 0 −3 
A ~ 0 0 1 4 
0 1 0 −10 
Unique solution exists
R1 → R1 − R 3 , R 2 ↔ R 3
1 0 0 7 
A ~ 0 1 0 −10 
0 0 1 4 
∴ Solution is x = 7, y = –10, z = 4.
www.sakshieducation.com
www.sakshieducation.com
3.
x – y + 3z = 5
4x + 2y – z = 0
–x + 3y + z = 5
Sol. (i) Cramer’s rule :
1 −1 3
∆= 4
−1
2
−1
3
1
= 1(2 + 3) + 1(4 − 1) + 3(12 + 2)
= 5 + 3 + 42 = 50
5 −1 3
∆1 = 0
2
−1
5
3
1
= 5(2 + 3) + 1(0 + 5) + 3(0 − 10)
= 25 + 5 − 30 = 0
1 5 3
∆2 = 4
0 −1
−1 5
1
= 1(0 + 5) − 5(4 − 1) + 3(20 − 0)
= 5 − 15 + 60 = 50
1
∆3 = 4
−1
−1 5
2
0
3
5
= 1(10 − 0) + 1(20 − 0) + 5(12 + 2)
= 10 + 20 + 70 = 100
0
∆
x= 1 =
=0
∆ 50
∆
50
=1
y= 2 =
∆ 50
∆3 100
=
=2
∆
50
∴ Solution is x = 0, y = 1, z = 2/
z=
www.sakshieducation.com
www.sakshieducation.com
ii) Matrix inversion method :
 1 −1 3 
Let A =  4 2 −1 , X =
 −1 3 1 
A1 =
2 −1
3
B1 = −
C1 =
4
−1
−1
1
2
−1 3
C3 =
3
= 1+ 3 = 4
1
−1
−1
3
−1
3
2
−1
B3 = −
1
3
4 −1
1 −1
4
= −(−1 − 9) = 10
1
−1 1
C2 = −
A3 =
3
1
2
= −(4 − 1) = −3
= 12 + 2 = 14
−1 3
A2 = −
B2 =
= 2+3=5
1
4
x 
5 
 y  and D = 0 
 
 
5 
 z 
= −(3 − 1) = −2
= 1 − 6 = −5
= −(−1 − 12) = 13
= 2+4 = 6
 A1 A 2 A3   5 10 −5
∴ AdjA =  B1 B2 B3  =  −3 4 13 
 C1 C2 C3  14 −2 6 
det A = a1A1 + b1B1 + c1C1
= 1 ⋅ 5 − 1 ⋅ (−30 + 3 ⋅14
= 5 + 3 + 42 = 50
 5 10 −5
AdjA 1 
−1
A =
=
−3 4 13 

det A 50
14 −2 6 
www.sakshieducation.com
www.sakshieducation.com
 5 10 −5 5 
1 
X=A D=
−3 4 13  0 

50
14 −2 6  5 
 25 + 0 − 25 
 0  0
1 
1    

=
−15 + 0 + 65 =
50 = 1
50 
50    
 70 + 0 + 30 
100   2 
−1
∴ Solution is x = 0, y = 1, z = 2.
iii) Gauss Jordan method :
 1 −1 3 5 
Augmented matrix is  4 2 −1 0 
 −1 3 1 5 
By R 2 → R 2 − 4R1 , R 3 → R 3 + R1
5 
1 −1 3
A ~ 0 6 −13 −20 
0 2
4
10 
1
R3 → R3  
2
5 
1 −1 3

A ~ 0 6 −13 −20 
0 1
2
5 
R1 → R1 + R 3 , R 2 → R 2 − 6R 3
10 
1 0 5

A ~ 0 0 −25 −50 
0 1 2
5 
 −1 
R2 → R2  
 25 
1 0 5 10 
A ~ 0 0 1 2 
0 1 2 5 
R1 → R1 − 5R 2 , R 3 → R 3 − 2R 2
1 0 0 0 
A ~ 0 0 1 2 
0 1 0 1 
www.sakshieducation.com
www.sakshieducation.com
R 2 ↔ R3
1 0 0 0 
A ~ 0 1 0 1 
0 0 1 2 
Unique solution
∴ Solution is x = 0, y = 1, z = 2.
4.
2x + 6y + 11 = 0
6x + 20y – 6z + 3 = 0
6y – 18z + 1 = 0
2 6
0
Sol. ∆ = 6 20
−6
0 6 −18
= 2(−360 + 36) − 6(−108 − 0)
= −648 + 648 = 0
∴ Cramer’s rule and matrix inversion method cannot be used. ∵ ∆ = 0
ii) Gauss Jordan method :
0 −11
2 6

Augmented matrix is  6 20 −6 −3 
 0 6 −18 −1 
R 2 → R 2 − 3R1
 2 6 0 −11
A ~  0 2 −6 30 
 0 6 −18 −1 
R 3 → R 3 − 3R 2
 2 6 0 −11
A ~  0 2 −6 30 
 0 0 0 −93
ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations do not have a solution.
www.sakshieducation.com
www.sakshieducation.com
5.
2x – y + 3z = 9
x+y+z=6
x–y+z=2
Sol. i) Cramer’s rule :
2 −1 3
∆= 1 1 1
1 −1 1
= 2(1 + 1) + 1(1 − 1) + 3(−1 − 1)
= 4 + 0 − 6 = −2
9 −1 3
∆1 = 6 1 1
2 −1 1
= 9(1 + 1) + 1(6 − 2) + 3(−6 − 2)
= 18 + 4 − 24 = −2
2 9 3
∆2 = 1 6 1
1 2 1
= 2(6 − 2) − 9(1 − 1) + 3(2 − 6)
= 8 − 0 − 12 = −4
2 −1 9
∆3 = 1 1 6
1 −1 2
= 2(2 + 6) + 1(2 − 6) + 9(−1 − 1)
= 16 − 4 − 18 = −6
∆
−2
∆
−4
x= 1 =
= 1, y = 2 =
=2
∆ −2
∆ −2
∆3 −6
=
=3
∆ −2
Solution is x = 1, y = 2, z = 3.
z=
www.sakshieducation.com
www.sakshieducation.com
ii) Matrix inversion method :
 2 −1 3
Let A = 1 1 1 , X =
1 −1 1
1 1
= 1+1 = 2
A1 =
−1 1
1 1
B1 = −
1 1
x 
 y  and D =
 
 z 
9 
6
 
 2 
=0
1 1
= −1 − 1 = −2
1 −1
−1 3
= −(−1 + 3) = −2
A2 = −
−1 1
C1 =
B2 =
2 3
1 1
2
1
−1
A3 =
1
C2 = −
B3 = −
C3 =
= 2 − 3 = −1
−1
= −(−2 + 1) = 1
−1
3
= −1 − 3 = −4
1
2 3
1 1
= −(2 − 3) = 1
2 −1
= 2 +1 = 3
1 1
 A1 A 2 A3   2 −2 −4 
AdjA =  B1 B2 B3  =  0 −1 1 
 C1 C2 C3   −2 1 3 
DetA = a1A1 + b1B1 + c1C1
= 2(2) − 1 ⋅ 0 + 3(−2)
= 4 − 6 = −2
 2 −2 −4 
AdjA
1
A −1 =
= −  0 −1 1 
DetA
2
 −2 1 3 
www.sakshieducation.com
www.sakshieducation.com
 2 −2 −4  9 
1
X = A D = −  0 −1 1   6 
2
 −2 1
3   2 
−1
 18 − 12 − 8 
 −2  1 
1
1
= −  −6 + 2  = −  −4  =  2 
2
2
 −18 + 6 + 6 
 −6   3 
Solution is x = 1, y = 2, z = 3.
iii) Gauss-Jordan method :
 2 −1 3 9 
Augmented matrix A = 1 1 1 6 
1 −1 1 2 
R1 → R1 − 2R 2 , R 3 → R 3 − R1
0 −3 1 −3
A ~ 1 1 1 6 
0 −2 0 −4 
0 −3 1 −3
 1
R 3 → R 3  −  gives A ~ 1 1 1 6 
 2
0 1 0 2 
R1 → R1 + 3R 3 , R 2 → R 2 − R 3
0 0 1 3 
A ~ 1 0 1 4 
0 1 0 2 
0 0 1 3 
R 2 → R 2 − R1 gives A ~ 1 0 0 1 
0 1 0 2 
By R1 ↔ R 2 , R 2 ↔ R 2 − R 3 we get
1 0 0 1 
A ~ 0 1 0 2 
0 0 1 3 
∴ The given equations have a unique solution.
Solution is x = 1, y = 2, z = 3.
www.sakshieducation.com
www.sakshieducation.com
6.
2x – y + 8z = 13
3x + 4y + 5z = 18
5x – 2y + 7z = 20
Sol. i) Cramer’s rule :
2 −1 8
∆= 3 4 5
5 −2 7
= 2(28 + 10) + 1(21 − 25) + 8(−6 − 20)
= 76 − 4 − 208 = −136
13 −1 8
∆1 = 18 4 5
20 −2 7
= 13(28 + 10) + 1(126 − 100) + 8(−36 − 80)
= 494 + 26 − 928 = −408
2 13 8
∆ 2 = 3 18 5
2 20 7
= 2(126 − 100) − 13(21 − 25) + 8(60 − 90)
= 52 + 52 − 240 = −136
2 −1 13
∆3 = 3 4 18
5 −2 20
= 2(80 + 36) + 1(60 − 90) + 13(−6 − 20)
= 232 − 30 − 338 = −136
∆
−408
x= 1 =
=3
∆ −136
∆
−136
y= 2 =
=1
∆ −136
∆
−136
z= 3 =
=1
∆ −136
∴ Solution is x = 3, y = 1, z = 1.
www.sakshieducation.com
www.sakshieducation.com
ii) Matrix inversion method :
 2 −1 8 
x 


Let A =  3 4 5  , X =  y  and D =
 5 −2 7 
 z 
A1 =
4 5
= 28 + 10 = 38
−2 7
3 5
= −(21 − 25) = 4
5 7
B1 = −
C1 =
3 4
= −6 − 20 = −26
5 −2
A2 = −
B2 =
−1 8
= −(−7 + 16) = −9
−2 7
2 8
= (14 − 40) = −26
5 7
2
5
−1
A3 =
4
2
B3 = −
3
C2 = −
C3 =
13 
18 
 
 20 
−1
= −(−4 + 5) = −1
−2
8
= −5 − 32 = −37
5
8
= −(10 − 24) = 14
5
2 −1
= 8 + 3 = 11
3 4
 A1 A 2 A3   38 −9 −37 
AdjA =  B1 B2 B3  =  4 −26 14 
 C1 C2 C3   −26 1
11 
Det A = a1A1 + b1B1 + c1C1
= 2 ⋅ 38 + (−1)4 + 8(−26)
= 76 − 4 − 208 = −136
www.sakshieducation.com
www.sakshieducation.com
 38 −9 −37 
AdjA
1 
=−
A =
4 −26 14 
Det A
136 
 −26 1
11 
 38 −9 −37  13 
1 
−1
X=A D=−
4 −26 14  18 

280
 −26 1
11   20 
−1
 494 −162 −740 
1 
52 −468 +280 
=−

136
 −338 −18 +220 
 −408 3
1 
=−
−136  = 1

136
 −136  1
Solution is x = 3, y = 1, z = 1.
iii) Gauss Jordan method :
 2 −1
Augmented matrix is A =  3 4
 5 −2
R 2 → 2R 2 − 3R1 , R 3 → 2R 3 − 5R 2
8 13 
5 18 
7 20 
we get
13 
 2 −1 8

A ~  0 11 −14 −3 
 0 1 −26 −25
R1 → R1 + R 3 , R 2 → R 2 − 11R 3 , we get
 2 0 −18 −12 
A ~  0 0 272 272 
 0 1 −26 −25
 1 
R2 → R2 
 we get
 272 
 2 0 −18 −12 
A ~  0 0 1
1 
 0 1 −26 −25
R1 → R1 + 18R 2 , R 3 → R 3 + 26R 2 , we get
www.sakshieducation.com
www.sakshieducation.com
2 0 0 6
A ~  0 0 1 1 
 0 1 0 1 
 1 0 0 3
1
R 2 ↔ R 3 , R1   we get A ~ 0 1 0 1
2
0 0 1 1
∴ The given equations have a unique solution and solution is x = 3, y = 1, z = 1.
7.
2x – y + 3z = 8
–x + 2y + z = 4
3x + y – 4z = 0
Sol. i) Cramer’s rule :
2 −1 3
∆ = −1
3
2
1
1
−4
= 2(−8 − 1) + 1(4 − 3) + 3(−1 − 6)
= −18 + 1 − 21 = −38
8 −1 3
∆1 = 4
0
2
1
1
−4
= 8(−8 − 1) + 1(−16 − 0) + 3(4 − 0)
= −72 − 16 + 12 = −76
2 8 3
∆ 2 = −1 4 1
3 0 −4
= 2(−16 − 0) − 8(4 − 3) + 3(−0 − 12)
= −32 − 8 − 36 = −76
2 −1 8
∆3 = −1
3
2
1
4
0
= 2(0 − 4) + 1(0 − 12) + 8(−1 − 6)
= −8 − 12 − 56 = −76
www.sakshieducation.com
www.sakshieducation.com
∆1 −76
=
=2
∆ −38
∆
−76
=2
y= 2 =
∆ −38
∆
−76
=2
z= 3 =
∆ −38
∴ Solution is x = 2, y = 2, z = 2.
x=
ii) Matrix inversion method :
 2 −1 3 
Let A =  −1 2 1  , X =
 3 1 −4 
A1 =
2
1
1 −4
B1 = −
x 
 y , D =
 
 z 
8 
4
 
 0 
= −8 − 1 = −9
−1 1
= −(4 − 3) = −1
3 −4
−1 2
= −1 − 6 = −7
3 1
−1 3
A2 = −
= −(4 − 3) = −1
1 −4
C1 =
B2 =
2
3 −4
C2 = −
A3 =
= −8 − 9 = −17
2 −1
= −(2 + 3) = −5
3 1
−1 3
= −1 − 6 = −7
2 1
B3 = −
C3 =
3
2 3
= −(4 + 3) = −7
−1 2
−1
= 4 −1 = 3
−1 2
2
 A1 A 2 A3   −9 −1 −7 
AdjA =  B1 B2 B3  =  −1 −17 −7 
 C1 C2 C3   −7 −5 3 
Det A = a1A1 + b1B1 + c1C1
= 2(−9) − 1(−1) + 3(−7)
= −18 + 1 − 21 = −38
www.sakshieducation.com
www.sakshieducation.com
 −9 −1 −7 
AdjA
1 
= −  −1 −17 −7 
A =
DetA
38
 −7 −5 3 
 −9 −1 −7  8 
1 
−1
X = A D = −  −1 −17 −7   4 
38
 −7 −5 3   0 
−1
 −72 − 4 
 −76   2 
1 
1 

= −  −8 − 68  = −  −76  =  2 
38
38
 −56 − 20 
 −76   2 
Solution is x = 2, y = 2, z = 2.
iii) Gauss Jordan method :
 2 −1 3 8 
Augmented matrix is A =  −1 2 1 4 
 3 1 −4 0 
R1 → R1 + 2R2, R3 → R3 + 3R2, we get
 0 3 5 16 
A ~  −1 2 1 4 
 0 7 −1 12 
R 2 → 3R 2 − 2R1 , R 3 → 3R 3 − 7R1 , we have
16 
0 3 5

A ~  −3 0 −7 −20 
 0 0 −38 −76 
 1 
R 3 → R 3  −  , we get
 38 
 0 3 5 16 
A ~  −3 0 −7 −20 
 0 0 1
2 
R1 → R1 − 5R 2 , R 2 → R 2 + 7R 3 , we get
0 3 0 6
A ~  −3 0 0 −6 
 0 0 1 2 
1
 1
R1 → R1   , R 2 → R 2  −  , R1 → R 2 we get
3
 3
www.sakshieducation.com
www.sakshieducation.com
1 0 0 2 
A ~ 0 1 0 2 
0 0 1 2 
∴ The given equations have a unique solution and solution is x = 2, y = 2, z = 2.
8.
Solve x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Sol. i) Cramer’s rule :
1 1 1
∆ = 2 5 −7
2 1 −1
= 1(−5 − 7) − 1(−2 − 14) + 1(2 − 10)
= −12 + 16 − 8 = −4
9 1 1
∆1 = 52 5 7
0 1 −1
= 9(−5 − 7) − 1(−52 − 0) + 1(52 − 0)
= −108 + 52 + 52 = −4
1 9 1
∆ 2 = 2 52 7
2 0 −1
= 1(−52 − 0) − 9(−2 − 14) + 1(0 − 104)
= −52 + 144 − 104 = −20
1 1 9
∆3 = 2 5 52
2 1 0
= 1(0 − 52) − 1(0 − 104) + 9(2 − 10)
= −52 + 104 − 72 = −20
∆
−4
x= 1 =
=1
∆ −4
∆
−12
y= 2 =
=3
∆
−4
∆
−20
=5
z= 3 =
∆
−4
Solution is x = 1, y = 3, z = 5
www.sakshieducation.com
www.sakshieducation.com
ii) Matrix inversion method :
1 1 1 
x 


Let A =  2 5 7  , X =  y  , D =
 2 1 −1
 z 
A1 =
5 7
= −5 − 7 = −12
1 −1
B1 = −
C1 =
2 7
= −(−2 − 14) = 16
2 −1
2 5
= 2 − 10 = −8
2 1
A2 = −
B2 =
9
52 
 
 0 
1 1
= −(−1 − 1) = 2
1 −1
1 1
= −1 − 2 = −3
2 −1
1 1
= −(1 − 2) = 1
2 1
1 1
= 7−5 = 2
A3 =
5 7
C2 = −
B3 = −
C3 =
1 1
= −(7 − 2) = −5
2 7
1 1
= 5−2 = 3
2 5
 A1 A 2 A3   −12 2 2 
AdjA =  B1 B2 B3  =  16 −3 −5
 C1 C2 C3   −8 1 3 
Det A = a1A1 + b1B1 + c1C1
= 1(−12) + 1(16) + 1(−8)
= −12 + 16 − 8 = −4
 −12 +2 2 
AdjA
1
−1
A =
= −  16 −3 −5
DetA
4
 −8 1 3 
www.sakshieducation.com
www.sakshieducation.com
 −12 +2 2   9 
1
X = A D = −  16 −3 −5 52 
4
 −8 1 3   0 
 −108 + 104 
 −4  1 
1
1

= −  144 − 156  = −  −12  = 3
4
4
 −72 + 52 
 −20  5
−1
∴ Solution is x = 1, y = 3, z = 5.
iii) Gauss Jordan method :
1 1 1 9 
Augmented matrix A =  2 5 7 52 
 2 1 −1 0 
R 2 → R 2 − 2R1 , R 3 → R 3 − R 2
9 
1 1 1

A ~ 0 3 5 34 
0 −4 −8 −52 
R1 → 3R1 − R 2 , R 3 → 3R 3 + 4R 2
 3 0 −2 −7 
A ~ 0 3 5 34 
0 0 −4 −20 
 1
R 3 → R 3  −  , we obtain
 4
 3 0 −2 −7 
A ~ 0 3 5 34 
0 0 1
5 
R1 → R1 + 2R 3 , R 2 → R 2 − 5R 3 , we get
 3 0 0 3
A ~ 0 3 0 9 
0 0 1 5
1
1
R1 → R1   , R 2 → R 2   we have
3
3
1 0 0 1 
A ~ 0 1 0 3
0 0 1 5
∴ The given equations have a unique solution and solution is x = 1, y = 3, z = 5.
www.sakshieducation.com