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Question: Answered & Verified by Expert
If $a, b, c$, are non zero complex numbers satisfying $a^2+b^2+c^2=0$ and
$\left|\begin{array}{ccc}b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2\end{array}\right|=k a^2 b^2 c^2$, then $k$ is equal to
MathematicsDeterminantsJEE MainJEE Main 2012 (19 May Online)
Options:
  • A
    1
  • B
    3
  • C
    4
  • D
    2
Solution:
1753 Upvotes Verified Answer
The correct answer is:
4
Let $\Delta=\left|\begin{array}{ccc}b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2\end{array}\right|$
Multiply $C_1$ by $a, C_2$ by b and $C_3$ by $\mathrm{c}$ and hence divide by $a b c$.

$$
=\frac{1}{a b c}\left|\begin{array}{ccc}
a\left(b^2+c^2\right. & \left.a b^2\right) & a c^2 \\
a^2 b & b\left(c^2+a^2\right. & \left.b c^2\right) \\
a^2 c & b^2 c & c\left(a^2+b^2\right.
\end{array}\right|
$$
Take out $a, b, c$ common from $R_1, R_2$ and $R_3$ respectively.
$$
\begin{aligned}
& \therefore \quad \Delta=\frac{a b c}{a b c}\left|\begin{array}{ccc}
b^2+c^2 & b^2 & c^2 \\
a^2 & c^2+a^2 & c^2 \\
a^2 & b^2 & a^2+b^2
\end{array}\right| \\
& \text { Apply } C_1 \rightarrow C_1-C_2-C_3 \\
& \Delta=\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
-2 c^2 & c^2+a^2 & c^2 \\
-2 b^2 & b^2 & a^2+b^2
\end{array}\right| \\
& =-2\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
c^2 & c^2+a^2 & c^2 \\
b^2 & b^2 & a^2+b^2
\end{array}\right| \\
&
\end{aligned}
$$
Apply $C_2-C_1$ and $C_3-C_1$
$$
\begin{aligned}
& =-2\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
c^2 & a^2 & 0 \\
b^2 & 0 & a^2
\end{array}\right| \\
& =-2\left[-b^2\left(c^2 a^2\right)+c^2\left(-a^2 b^2\right)\right] \\
& =2 a^2 b^2 c^2+2 a^2 b^2 c^2=4 a^2 b^2 c^2 \\
& \text { But } \Delta=k a^2 b^2 c^2 \therefore k=4
\end{aligned}
$$

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