Q : 12        Let   \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}   and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix} .  Verify that  \small (AB)^-^1=B^{-1}A^{-1}.

Answers (1)

We have \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix} and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}.

then calculating;

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}

Finding the inverse of AB.

Calculating the cofactors fo AB:

AB_{11}=(-1)^{1+1}(61) = 61        AB_{12}=(-1)^{1+2}(47) = -47

AB_{21}=(-1)^{2+1}(87) = -87     AB_{22}=(-1)^{2+2}(67) = 67

Then we have adj(AB):

adj(AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse: 

(AB)^{-1}=\frac{1}{|AB|}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}                                                      .....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1   and |B| = 54- 56 = -2

adjA = \begin{bmatrix} 5 &-7 \\ -2 & 3 \end{bmatrix}     and    adjB = \begin{bmatrix} 9 &-8 \\ -7 & 6 \end{bmatrix}

therefore we have 

A^{-1} = \frac{1}{|A|}adjA= \frac{1}{1} \begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}    and  B^{-1} = \frac{1}{|B|}adjB= \frac{1}{-2} \begin{bmatrix} 9&-8 \\ -7& 6 \end{bmatrix}= \begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}

Now calculating B^{-1}A^{-1}.

B^{-1}A^{-1} =\begin{bmatrix} \frac{-9}{2} & 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}                ........................(2)

From (1) and (2) we get 

\small (AB)^-^1=B^{-1}A^{-1}

Hence proved.

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