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