Search any question & find its solution
Question:
Answered & Verified by Expert
$\left|\begin{array}{lll}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|=$
Options:
Solution:
2602 Upvotes
Verified Answer
The correct answer is:
0
$\left|\begin{array}{ccc}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|=\left|\begin{array}{ccc}0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & c^2-a b\end{array}\right|$ by $\left\{\begin{array}{l}R_1 \rightarrow R_1-R_2 \\ R_2 \rightarrow R_2-R_3\end{array}\right.$
$(a-b)(b-c)\left|\begin{array}{lll}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^2-a b\end{array}\right|=0 \quad, \quad\left\{R_1=R_2\right\}$
$(a-b)(b-c)\left|\begin{array}{lll}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^2-a b\end{array}\right|=0 \quad, \quad\left\{R_1=R_2\right\}$
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.