Question

If $A=\left[\begin{array}{cc}x& -2\\ 3& 7\end{array}\right]$ 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. $4$
• E. $-2$

Answer 1

Answer: (D)

Given matrix is $A=\left[\begin{array}{cc}x& -2\\ 3& 7\end{array}\right]$
$\therefore$ $|A|=7x+6$
$\therefore$ $A^{-1}=\left[\begin{array}{cc}\frac{7}{7x+6}& \frac{2}{7x+6}\\ \frac{-3}{7x+6}& \frac{x}{7x+6}\end{array}\right]$
Comparing it with given inverse matrix $A^{-1}=\left[\begin{array}{cc}\frac{7}{34}& \frac{1}{17}\\ \frac{-3}{24}& \frac{2}{17}\end{array}\right],$ we get $\frac{7}{7x+6}=\frac{7}{34}$
$\Rightarrow$ $x=4$