જો $A=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]$ અને $B=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]$ હોય,તો ચકાસો કે $(AB)^{-1}=B^{-1} A^{-1}$.

(A) સૌ પ્રથમ,આપણે $AB$ નો ગુણાકાર શોધીએ:
$AB = \left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right] \left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right] = \left[\begin{array}{cc}2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3)\end{array}\right] = \left[\begin{array}{cc}-1 & 5 \\ 5 & -14\end{array}\right]$.
ત્યારબાદ,આપણે નિશ્ચાયક $|AB| = (-1)(-14) - (5)(5) = 14 - 25 = -11$ શોધીએ.
કારણ કે $|AB| \neq 0$,તેથી $(AB)^{-1}$ અસ્તિત્વ ધરાવે છે.
$(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{-11} \left[\begin{array}{cc}-14 & -5 \\ -5 & -1\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}14 & 5 \\ 5 & 1\end{array}\right]$.
હવે,આપણે $A^{-1}$ અને $B^{-1}$ શોધીએ:
$|A| = 2(-4) - 3(1) = -8 - 3 = -11$.
$A^{-1} = \frac{1}{-11} \left[\begin{array}{cc}-4 & -3 \\ -1 & 2\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}4 & 3 \\ 1 & -2\end{array}\right]$.
$|B| = 1(3) - (-2)(-1) = 3 - 2 = 1$.
$B^{-1} = \frac{1}{1} \left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right] = \left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right]$.
છેલ્લે,$B^{-1} A^{-1}$ ની ગણતરી કરીએ:
$B^{-1} A^{-1} = \left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right] \left( \frac{1}{11} \left[\begin{array}{cc}4 & 3 \\ 1 & -2\end{array}\right] \right) = \frac{1}{11} \left[\begin{array}{cc}3(4)+2(1) & 3(3)+2(-2) \\ 1(4)+1(1) & 1(3)+1(-2)\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}14 & 5 \\ 5 & 1\end{array}\right]$.
આમ,$(AB)^{-1} = B^{-1} A^{-1}$ સાબિત થાય છે.