questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 400
Question 400 of 472
Medium
MCQ
If the following system of linear equations 2x + y + z = 5 x - y + z = 3 x + y + az = b has no solution, then :
A
\[a = - {1 \over 3},b \ne {7 \over 3}\]
B
\[a \ne {1 \over 3},b = {7 \over 3}\]
C
\[a \ne - {1 \over 3},b = {7 \over 3}\]
D
\[a = {1 \over 3},b \ne {7 \over 3}\]
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
Here \[D = \left| {
(
2
1
1
1
−
1
1
1
1
a
)
\begin{pmatrix} 2 & 1 & 1 \\ 1 & { - 1} & 1 \\ 1 & 1 & a \\ \end{pmatrix}
2
1
1
1
−
1
1
1
1
a
} \right|
(
=
2
(
a
−
1
)
−
1
(
a
−
1
)
+
1
+
1
=
1
−
3
a
)
\begin{pmatrix} { = 2(a - 1) - 1(a - 1) + 1 + 1} \\ { = 1 - 3a} \\ \end{pmatrix}
(
=
2
(
a
−
1
)
−
1
(
a
−
1
)
+
1
+
1
=
1
−
3
a
)
\] \[{D_3} = \left| {
(
2
1
5
1
−
1
3
1
1
b
)
\begin{pmatrix} 2 & 1 & 5 \\ 1 & { - 1} & 3 \\ 1 & 1 & b \\ \end{pmatrix}
2
1
1
1
−
1
1
5
3
b
} \right|
(
=
2
(
−
b
−
3
)
−
1
(
b
−
3
)
+
5
(
1
+
1
)
=
7
−
3
b
)
\begin{pmatrix} { = 2( - b - 3) - 1(b - 3) + 5(1 + 1)} \\ { = 7 - 3b} \\ \end{pmatrix}
(
=
2
(
−
b
−
3
)
−
1
(
b
−
3
)
+
5
(
1
+
1
)
=
7
−
3
b
)
\] for \[a = {1 \over 3},b \ne {7 \over 3}\], system has no solutions.
Answer the question to unlock
Previous
Next