- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
medium
જો $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, તો $A$ મેળવો.
A
$12$
B
$24$
C
$-12$
D
$-24$
(IIT-1982) (JEE MAIN-2015)
Solution
(b) Trick : Put $x = 1$, then we have
$\left| {\,\begin{array}{*{20}{c}}2&2&{ – 1}\\4&3&0\\6&1&1\end{array}\,} \right| = A – 12 \Rightarrow \left| {\,\begin{array}{*{20}{c}}0&2&{ – 1}\\1&3&0\\5&1&1\end{array}\,} \right| = A – 12$
{Apply ${C_1} \to {C_1} – {C_2}$}
$ \Rightarrow $ $ – 2 + ( – 1)\,( – 14) = A – 12 \Rightarrow A = 24$.
Standard 12
Mathematics
