Search any question & find its solution
Question:
Answered & Verified by Expert
$A=\left|\begin{array}{ccc}2 a & 3 r & x \\ 4 b & 6 s & 2 y \\ -2 c & -3 t & -z\end{array}\right|=\lambda\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$, then what is the value of $\lambda ?$
Options:
Solution:
2293 Upvotes
Verified Answer
The correct answer is:
$-12$
Given, $\left|\begin{array}{ccc}2 a & 3 r & x \\ 4 b & 6 s & 2 y \\ -2 c & -3 t & -z\end{array}\right|=\lambda\left|\begin{array}{ccc}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
Taking 2 common from $C_{1}$ and 3 from $C_{2}$ in LHS
$\therefore 2 \times 3\left|\begin{array}{ccc}a & r & x \\ 2 b & 2 s & 2 y \\ -c & -t & -z\end{array}\right|=\lambda\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
Taking 2 common from $R_{2}$ and $-1$ from $R_{3}$ in LHS
$\therefore-12\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|=\lambda\left|\begin{array}{ccc}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
$\Rightarrow \lambda=-12$
Taking 2 common from $C_{1}$ and 3 from $C_{2}$ in LHS
$\therefore 2 \times 3\left|\begin{array}{ccc}a & r & x \\ 2 b & 2 s & 2 y \\ -c & -t & -z\end{array}\right|=\lambda\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
Taking 2 common from $R_{2}$ and $-1$ from $R_{3}$ in LHS
$\therefore-12\left|\begin{array}{lll}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|=\lambda\left|\begin{array}{ccc}a & r & x \\ b & s & y \\ c & t & z\end{array}\right|$
$\Rightarrow \lambda=-12$
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.