\text { If } A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right], \text { verify that }(A B)^{-1}=B^{-1} A^{-1} \text {. }

Solution:

Here, A=\left(\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right) and B=\left(\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right)
Now find
A B=\left(\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right)\left(\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right)=\left(\begin{array}{ll} 18+16 & 21+18 \\ 42+40 & 49+45 \end{array}\right)=\left(\begin{array}{ll} 34 & 39 \\ 82 & 94 \end{array}\right)
We now need to find the inverses of some matrices as below:

(i) Given matrix as A=\left(\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right).
Let’s now calculate the determinant of the given matrix first.
|A|=\left|\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right|=15-14=1 \neq 0
So, the given matrix has inverse.
Find: the adjoint of the given matrix.

Step: 1 Find the minor matrix of A.
M_{A}=\left(\begin{array}{ll} |5| & |7| \\ |2| & |3| \end{array}\right)=\left(\begin{array}{ll} 5 & 7 \\ 2 & 3 \end{array}\right)

Step: 2 Find the co-factor matrix of A.
C_{A}=\left(\begin{array}{ll} (-1)^{1+1}(5) & (-1)^{1+2}(7) \\ (-1)^{2+1}(2) & (-1)^{2+2}(3) \end{array}\right)=\left(\begin{array}{cc} 5 & -7 \\ -2 & 3 \end{array}\right)

Step: 3 By transpose of C_{A} we will have \operatorname{adj} A.
\operatorname{adj} A=C_{A}^{T}=\left(\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right)^{T}=\left(\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right)
Finally the inverse of the matrix is
A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{1}\left(\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right)=\left(\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right)

(ii) Given matrix as B=\left(\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right).
Let’s now calculate the determinant of the given matrix first.
|B|=\left|\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right|=54-56=-2 \neq 0
So, the given matrix has inverse.
Find: the adjoint of the given matrix.

Step: 1 Find the minor matrix of B.
M_{B}=\left(\begin{array}{ll} |9| & |8| \\ |7| & |6| \end{array}\right)=\left(\begin{array}{ll} 9 & 8 \\ 7 & 6 \end{array}\right)

Step: 2 Find the co-factor matrix of B.
C_{B}=\left(\begin{array}{ll} (-1)^{1+1}(9) & (-1)^{1+2}(8) \\ (-1)^{2+1}(7) & (-1)^{2+2}(6) \end{array}\right)=\left(\begin{array}{cc} 9 & -8 \\ -7 & 6 \end{array}\right)

Step: 3 By transpose of C_{B} we will have \operatorname{adj} B.
\operatorname{adj} B=C_{B}^{T}=\left(\begin{array}{cc}9 & -8 \\ -7 & 6\end{array}\right)^{T}=\left(\begin{array}{cc}9 & -7 \\ -8 & 6\end{array}\right)
Finally the inverse of the matrix is
B^{-1}=\frac{1}{|B|} \text { adj } B=\frac{1}{-2}\left(\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right)=\frac{-1}{2}\left(\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right)

(iii) Given matrix as A B=\left(\begin{array}{ll}34 & 39 \\ 82 & 94\end{array}\right).
Let’s now calculate the determinant of the given matrix first.
|A B|=\left|\begin{array}{cc} 34 & 39 \\ 82 & 94 \end{array}\right|=3196-3198=-2 \neq 0
So, the given matrix has inverse.
Find: the adjoint of the given matrix.

Step: 1 Find the minor matrix of A B.
M_{A B}=\left(\begin{array}{ll} |94| & |82| \\ |39| & |34| \end{array}\right)=\left(\begin{array}{ll} 94 & 82 \\ 39 & 34 \end{array}\right)

Step: 2 Find the co-factor matrix of A B.
C_{A B}=\left(\begin{array}{ll} (-1)^{1+1}(94) & (-1)^{1+2}(82) \\ (-1)^{2+1}(39) & (-1)^{2+2}(34) \end{array}\right)=\left(\begin{array}{cc} 94 & -82 \\ -39 & 34 \end{array}\right)

Step: 3 By transpose of C_{A B} we will have a d j A B.
\operatorname{adj} A B=C_{A B}^{T}=\left(\begin{array}{cc}94 & -82 \\ -39 & 34\end{array}\right)^{T}=\left(\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right)
Finally the inverse of the matrix is
(A B)^{-1}=\frac{1}{|A B|} \operatorname{adj} A B=\frac{1}{-2}\left(\begin{array}{cc} 94 & -39 \\ -82 & 34 \end{array}\right)=\frac{-1}{2}\left(\begin{array}{cc} 94 & -39 \\ -82 & 34 \end{array}\right)

(iv) Now on finally multiplying the inverse of B with inverse of A we obtain
\begin{array}{l} B^{-1} A^{-1}=\frac{-1}{2}\left(\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right)\left(\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right)= \\ -\frac{1}{2}\left(\begin{array}{cc} 45+49 & -18-21 \\ -40-42 & 16+18 \end{array}\right)=\frac{-1}{2}\left(\begin{array}{cc} 94 & -39 \\ -82 & 34 \end{array}\right) \end{array}
Hence, from the results of (iii) and (iv) we get (A B)^{-1}=B^{-1} A^{-1}.