Answer
$\dfrac{2}{3}$
Work Step by Step
General formula to calculate the determinate of a matrix is:
$D=\begin{vmatrix}p&q\\r&s\end{vmatrix}=ps-qr$
As per question, we have
Thus, $D=\begin{vmatrix}\frac{2}{3}&\frac{1}{3}\\\frac{-1}{2}&\frac{3}{4}\end{vmatrix}$
or, $D=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3}$
