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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 440
Question 440 of 472
Medium
MCQ
The values of \[\alpha\], for which \[\left|\[
1
3
2
α
+
3
2
1
1
3
α
+
1
3
2
α
+
3
3
α
+
1
0
\begin{array}{|ccccccc|}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}
1
1
2
α
+
3
2
3
3
1
3
α
+
1
α
+
2
3
α
+
3
1
0
\]\right|=0\], lie in the interval
A
\[(-2,1)\]
B
\[\left(-\frac{3}{2}, \frac{3}{2}\right)\]
C
\[(-3,0)\]
D
\[(0,3)\]
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Solutions · 1
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Official solution
\[\left|\[
1
3
2
α
+
3
2
1
1
3
α
+
1
3
2
α
+
3
3
α
+
1
0
\begin{array}{|ccccccc|}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}
1
1
2
α
+
3
2
3
3
1
3
α
+
1
α
+
2
3
α
+
3
1
0
\]\right|=0\] \[
⇒
(
2
α
+
3
)
{
7
α
6
}
−
(
3
α
+
1
)
{
−
7
6
}
=
0
⇒
(
2
α
+
3
)
⋅
7
α
6
+
(
3
α
+
1
)
⋅
7
6
=
0
⇒
2
α
2
+
3
α
+
3
α
+
1
=
0
⇒
2
α
2
+
6
α
+
1
=
0
⇒
α
=
−
3
+
7
2
,
−
3
−
7
2
\begin{aligned} & \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\ & \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\ & \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\ & \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\ & \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2} \end{aligned}
⇒
(
2
α
+
3
)
{
6
7
α
}
−
(
3
α
+
1
)
{
6
−
7
}
=
0
⇒
(
2
α
+
3
)
⋅
6
7
α
+
(
3
α
+
1
)
⋅
6
7
=
0
⇒
2
α
2
+
3
α
+
3
α
+
1
=
0
⇒
2
α
2
+
6
α
+
1
=
0
⇒
α
=
2
−
3
+
7
,
2
−
3
−
7
\] Hence option (C) is correct.
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