solve×+y+z=9;. 2×+5y+7z=52; and 2×+y-z=0 by using cramer s rule
Answers
Answer:
Correct option is
A
1:3:5
Here Δ=
∣
∣
∣
∣
∣
∣
∣
∣
1
2
2
1
5
1
1
7
−1
∣
∣
∣
∣
∣
∣
∣
∣
applying C
2
→C
2
−C
1
and C
3
→C
3
−C
1
∴Δ=
∣
∣
∣
∣
∣
∣
∣
∣
1
2
2
0
3
−1
0
5
−3
∣
∣
∣
∣
∣
∣
∣
∣
=1(−9+5)=−4
Δ
1
=
∣
∣
∣
∣
∣
∣
∣
∣
9
52
0
1
5
1
1
7
−1
∣
∣
∣
∣
∣
∣
∣
∣
Applying C
2
→C
2
+C
3
∴Δ
1
=
∣
∣
∣
∣
∣
∣
∣
∣
9
52
0
2
12
0
1
7
−1
∣
∣
∣
∣
∣
∣
∣
∣
=−1(108−104)=−4
Δ
2
=
∣
∣
∣
∣
∣
∣
∣
∣
1
2
2
9
52
0
1
7
−1
∣
∣
∣
∣
∣
∣
∣
∣
Applying C
1
→C
1
+2C
3
∴Δ
2
=
∣
∣
∣
∣
∣
∣
∣
∣
3
16
0
9
52
0
1
7
−1
∣
∣
∣
∣
∣
∣
∣
∣
=−1(156−144)=−12
and
Δ
3
=
∣
∣
∣
∣
∣
∣
∣
∣
1
2
2
1
5
1
9
52
0
∣
∣
∣
∣
∣
∣
∣
∣
Applying C
1
→C
1
−2C
2
∴Δ
3
=
∣
∣
∣
∣
∣
∣
∣
∣
−1
−8
0
1
5
1
9
52
0
∣
∣
∣
∣
∣
∣
∣
∣
=−1(−51+72)
=−20
Clearly Δ
1
:Δ
2
:Δ
3
=−4:−12:−20=1:3:5
Step-by-step explanation: