Question

For a matrix M = $\left[\begin{array}{cc}\frac{3}{5}& \frac{4}{5}\\ x& \frac{3}{5}\end{array}\right]$ , the transpose of the matrix is equal to the inverse of the matrix, ${\left[M\right]}^T$ = ${\left[M\right]}^{-1}$ . The value of x is given by

• A. -$\frac{4}{5}$
• B. -$\frac{3}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A -$\frac{4}{5}$

[M] = $\left[\begin{array}{cc}\frac{3}{5}& \frac{4}{5}\\ x& 3/5\end{array}\right]$

${\left[M\right]}^{-1}$ = $\frac{1}{\frac{9}{25}-\frac{4x}{5}}$ $\left[\begin{array}{cc}3/5& -4/5\\ -x& 3/5\end{array}\right]$

Given, ${\left[M\right]}^T$ = ${\left[M\right]}^{-1}$

$\left[\begin{array}{cc}3/5& x\\ 4/5& 3/5\end{array}\right]$ = $\frac{25}{(9-20x)}$ $\left[\begin{array}{cc}3/5& -4/5\\ -X& 3/5\end{array}\right]$

Comparing $a_{11}$ both sides

$\Rightarrow$ $\frac{25}{9-20x}$ = 1 $\Rightarrow$ 90-20x = 25

X = $\frac{-16}{20}$ = $\frac{-4}{5}$

X = -4/5