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Standard XII
Mathematics
Question
The system of equations
3
x
−
y
+
4
z
=
2
;
x
+
2
y
−
z
=
3
;
4
x
−
y
+
λ
z
=
−
1
has no solution. The value of
λ
is
1
3
5
6
A
6
B
1
C
3
D
5
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Solution
Verified by Toppr
We have
3
x
−
4
+
4
z
=
2
x
+
2
y
−
z
=
3
4
x
−
y
+
λ
z
=
−
1
Now,
Δ
=
⎡
⎢
⎣
3
−
1
4
1
2
−
1
4
−
1
λ
⎤
⎥
⎦
=
3
(
2
λ
−
1
)
+
(
λ
+
4
)
+
4
(
−
1
−
8
)
=
7
λ
+
1
−
36
=
7
λ
−
35
Δ
1
=
⎡
⎢
⎣
2
−
1
4
3
2
−
1
−
1
−
1
λ
⎤
⎥
⎦
=
2
(
2
λ
−
1
)
+
(
3
λ
−
1
)
+
4
(
−
3
+
2
)
=
7
λ
+
5
And,
Δ
2
=
⎡
⎢
⎣
3
2
4
1
3
−
1
4
−
1
λ
⎤
⎥
⎦
=
7
λ
−
63
For
λ
=
5
No solution
For
n
solution
Δ
=
0
Δ
1
,
Δ
2
,
Δ
3
≠
0
Hence, the option
(
C
)
is the correct answer.
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