Question

Let A=$\begin{bmatrix}3&7\\2&5\end{bmatrix}$and B=$\begin{bmatrix}6&8\\7&9\end{bmatrix}$,Verify that (AB)^-1=B^-1A^-1.

Answer 1

Let A=$\begin{bmatrix}3&7\\2&5\end{bmatrix}$

we have IAI=15-14=1

Now,A₁₁=5 ,A₁₂=-2, A₂₁=-7, A₂₂=3

therefore adj A=$\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$

therefore A^-1=$\frac{1}{\mid A\mid}$.adj A=$\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$

Now let B=$\begin{bmatrix}6&8\\7&9\end{bmatrix}$

we have IBI=54-56=-2

so adj B=$\begin{bmatrix}9&-8\\-7&6\end{bmatrix}$

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

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}$

=$\begin{bmatrix}-\frac{45}{2}-8&\frac{63}{2}+12\\\frac{35}{2}+6&-\frac{49}{2}-9\end{bmatrix}$

=$\begin{bmatrix}-\frac{61}{2}&\frac{87}{2}\\\frac{47}{2}&-\frac{67}{2}\end{bmatrix}$...(1)

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

=$\begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix}$

=$\begin{bmatrix}67&87\\47&61\end{bmatrix}$

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

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

=$\begin{bmatrix}-\frac{61}{2}&\frac{87}{2}\\\frac{47}{2}&-\frac{67}{2}\end{bmatrix}$  ....(2)

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

Hence, the given result is proved