Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ b+c & c+a & a+b\end{array}\right|$ is
Options:
Solution:
1446 Upvotes
Verified Answer
The correct answer is:
$(a-b)(b-c)(c-a)$
We have, $\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ b+c & c+a & a+b\end{array}\right|$
On applying $C_{1} \rightarrow C_{1}-C_{2}, C_{2} \rightarrow C_{2}-C_{3}$, we get $\left|\begin{array}{ccc}0 & 0 & 1 \\ -c(a-b) & -a(b-c) & a b \\ -(a-b) & -(b-c) & a+b\end{array}\right|$
Taking common $(a-b),(b-c)$ from $C_{1}, C_{2}$ respectively, we get
$$
(a-b)(b-c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
-c & -a & a b \\
-1 & -1 & a+b
\end{array}\right|
$$
On expanding along $R_{1}$, we get
$$
\begin{aligned}
&(a-b)(b-c) \cdot 1(c-a) \\
&=(a-b)(b-c)(c-a)
\end{aligned}
$$
On applying $C_{1} \rightarrow C_{1}-C_{2}, C_{2} \rightarrow C_{2}-C_{3}$, we get $\left|\begin{array}{ccc}0 & 0 & 1 \\ -c(a-b) & -a(b-c) & a b \\ -(a-b) & -(b-c) & a+b\end{array}\right|$
Taking common $(a-b),(b-c)$ from $C_{1}, C_{2}$ respectively, we get
$$
(a-b)(b-c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
-c & -a & a b \\
-1 & -1 & a+b
\end{array}\right|
$$
On expanding along $R_{1}$, we get
$$
\begin{aligned}
&(a-b)(b-c) \cdot 1(c-a) \\
&=(a-b)(b-c)(c-a)
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.