
Now
X 5 A
21
B 52
1
5
0 255
224 25
217 210
2
4
3
5
23
1
21
2
4
3
5
.
x
y
z
2
4
3
5
5
2
23
24
2
4
3
5
Hence the required solution is x 5 2, y 523, and z 524.
3.13 Determinant Method (Cramer’s Rule)
Let us consider the system of equations:
a
1
x 1 b
1
y 5 c
1
a
2
x 1 b
2
y 5 c
2
.
Multiplying the first equation by b
2
, second by b
1
, and then subtracting the sec-
ond from the first, we get ða
1
b
2
2 a
2
b
1
Þx 5 b
2
c
1
2 b
1
c
2
.
‘x 5
b
2
c
1
2 b
1
c
2
a
1
b
2
2 a
2
b
1
5
c
1
b
1
c
2
b
2
a
1
b
1
a
2
b
2
Similarly, we obtain y 5
a
1
c
1
a
2
c
2
a
1
b
1
a
2
b
2
, provided that
a
1
b
1
a
2
b
2
6¼ 0.
Alternatively, we can write the above-mentioned formulae as show n below:
If
D 5
a
1
b
1
a
2
b
2
; D
1
5
c
1
b
1
c
2
b
2
; and D
2
5
a
1
c
1
a
2
c
2