If inverse of $\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$ does not…

If inverse of $\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$ does not exist, then $x=$
  1. -3
  2. 2
  3. 3
  4. 0

Solution

$\begin{aligned} & \text { Here }\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]=0 \\ & \therefore(6-28)-2(-24-14)+x(16+20)=0 \\ & \therefore-22+76+18 x=\Rightarrow x=-3\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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