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3 and 4 .Determinants and Matrices
hard
Let $P =\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$ where $\alpha \in R .$ Suppose $Q =\left[ q _{ ij }\right]$ is a matrix satisfying $PQ = kI _{3}$ for some non-zero $k \in R .$ If $q_{23}=-\frac{k}{8}$ and $|Q|=\frac{k^{2}}{2}$, then $\alpha^{2}+ k ^{2}$ is equal to...........
A
$17$
B
$21$
C
$13$
D
$19$
(JEE MAIN-2021)
Solution
$PQ = kI$
$| P | \cdot| Q |= k ^{3}$
$\Rightarrow| P |=2 k \neq 0 \Rightarrow P$ is an invertible matrix
$\because PQ = kI$
$\therefore Q=k P^{-1} I$
$\therefore Q=\frac{\text { adj.P }}{2}$
$\because q _{23}=-\frac{ k }{8}$
$\therefore \frac{-(3 \alpha+4)}{2}=-\frac{ k }{8} \Rightarrow k =4$
$\therefore| P |=2 k \Rightarrow k =10+6 \alpha \ldots( i )$
Put value of $k$ in (i).. we get $\alpha=-1$
Standard 12
Mathematics
