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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 433
Question 433 of 472
Medium
MCQ
Let the system of linear equations
x
+
y
+
k
z
=
2
x+y+kz=2
x
+
y
+
k
z
=
2
2
x
+
3
y
−
z
=
1
2x+3y-z=1
2
x
+
3
y
−
z
=
1
3
x
+
4
y
+
2
z
=
k
3x+4y+2z=k
3
x
+
4
y
+
2
z
=
k
have infinitely many solutions. Then the system
(
k
+
1
)
x
+
(
2
k
−
1
)
y
=
7
(k+1)x+(2k-1)y=7
(
k
+
1
)
x
+
(
2
k
−
1
)
y
=
7
(
2
k
+
1
)
x
+
(
k
+
5
)
y
=
10
(2k+1)x+(k+5)y=10
(
2
k
+
1
)
x
+
(
k
+
5
)
y
=
10
has :
A
unique solution satisfying \[x-y=1\]
B
infinitely many solutions
C
no solution
D
unique solution satisfying \[x+y=1\]
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Official solution
x
+
y
+
k
z
=
2
x + y + kz = 2
x
+
y
+
k
z
=
2
............(i)
2
x
+
3
y
−
z
=
1
2x + 3y - z = 1
2
x
+
3
y
−
z
=
1
..........(ii)
3
x
+
4
y
+
2
z
=
k
3x + 4y + 2z = k
3
x
+
4
y
+
2
z
=
k
......(iii) (1) + (2)
3
x
+
4
y
+
z
(
k
−
1
)
=
3
3x + 4y + z(k - 1) = 3
3
x
+
4
y
+
z
(
k
−
1
)
=
3
Comparing with (3)
k
=
3
k = 3
k
=
3
Now,
4
x
+
5
y
=
7
4x + 5y = 7
4
x
+
5
y
=
7
\[ \Rightarrow 3x + 3y = 3\]
7
x
+
8
y
=
10
7x + 8y = 10
7
x
+
8
y
=
10
as \[{4 \over 7} \ne {5 \over 8}\] \[\therefore\] Unique solution satisfying
x
+
y
=
1
x + y = 1
x
+
y
=
1
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