If $A=\left[\begin{array}{rr}r & -2 \\ 3 & 7\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]$, then the value of x is
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If $A=\left[\begin{array}{rr}r & -2 \\ 3 & 7\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]$, then the value of $x$ is

(A) 2

(B) 3

(C) -4

(D) 4

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SOLUTION —

$\begin{array}{c}\text {Here,}|A|=\left[\begin{array}{rr}x & -2 \\3 & 7\end{array}\right]=7 x+6 \\\therefore \quad A^{-1}=\frac{1}{7 x+6}\left[\begin{array}{rr}7 & 2 \\-3 & x\end{array}\right]=\left[\begin{array}{ll}\frac{7}{34} & \frac{1}{17} \\\frac{-3}{34} & \frac{2}{17}\end{array}\right] \\\therefore \quad \frac{7}{7 x+6}=\frac{7}{34}\Rightarrow x=4\end{array}$

So, The correct option will be (D).

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