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Mathematics Question on Determinants

Let A=[37\25]\begin{bmatrix}3&7\\\2&5\end{bmatrix}and B=[68\79]\begin{bmatrix}6&8\\\7&9\end{bmatrix},Verify that (AB)-1=B-1A-1.

Answer

Let A=[37\25]\begin{bmatrix}3&7\\\2&5\end{bmatrix}

we have IAI=15-14=1

Now,A11=5 ,A12=-2, A21=-7, A22=3

therefore adj A=[5−7−23]\begin{bmatrix}5&-7\\\\-2&3\end{bmatrix}

therefore A-1=1∣A∣\frac{1}{\mid A\mid}.adj A=[5−7−23]\begin{bmatrix}5&-7\\\\-2&3\end{bmatrix}

Now let B=[68\79]\begin{bmatrix}6&8\\\7&9\end{bmatrix}

we have IBI=54-56=-2

so adj B=[9−8−76]\begin{bmatrix}9&-8\\\\-7&6\end{bmatrix}

therefore B-1=1∣B∣\frac{1}{\mid B\mid} adj B=-\frac{1}{27}$$\begin{bmatrix}9&-8\\\\-7&6\end{bmatrix}
=[−97472−3]\begin{bmatrix}-\frac{9}{7}&4\\\\\frac{7}{2}&-3\end{bmatrix}

Now,B-1 A-1=\begin{bmatrix}-\frac{9}{7}&4\\\\\frac{7}{2}&-3\end{bmatrix}$$\begin{bmatrix}5&-7\\\\-2&3\end{bmatrix}

=[−452−8632+12352+6−492−9]\begin{bmatrix}-\frac{45}{2}-8&\frac{63}{2}+12\\\\\frac{35}{2}+6&-\frac{49}{2}-9\end{bmatrix}

=[−612872472−672]\begin{bmatrix}-\frac{61}{2}&\frac{87}{2}\\\\\frac{47}{2}&-\frac{67}{2}\end{bmatrix}...(1)

Then ,AB=\begin{bmatrix}3&7\\\2&5\end{bmatrix}$$\begin{bmatrix}6&8\\\7&9\end{bmatrix}

=[18+4924+63\12+3516+45]\begin{bmatrix}18+49&24+63\\\12+35&16+45\end{bmatrix}

=[6787\4761]\begin{bmatrix}67&87\\\47&61\end{bmatrix}

therefore we have IABI=67x61-87x47=4087-4089=-2
Also adj(AB)=[61−87−4767]\begin{bmatrix}61&-87\\\\-47&67\end{bmatrix}

therefore (AB)-1=1∣AB∣\frac{1}{\mid AB\mid}adj AB=-\frac{1}{2}$$\begin{bmatrix}61&-87\\\\-47&67\end{bmatrix}

=[−612872472−672]\begin{bmatrix}-\frac{61}{2}&\frac{87}{2}\\\\\frac{47}{2}&-\frac{67}{2}\end{bmatrix} ....(2)

From (1) and (2), we have:
(AB)-1 = B-1 A-1

Hence, the given result is proved