Q

If A inverse equals determinant 3 -1 1 -15 6 -5 5 -2 2 and B equals determinant 1 2 -2 -1 3 0 0 -2 1 , find (AB)^-1

Q : 7        If     $\dpi{100} A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$  and $\dpi{100} B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$, find $\dpi{100} (AB)^-^1$.

Views

We know from the identity that;

$(AB)^{-1} = B^{-1}A^{-1}$.

Then we can find easily,

Given $\dpi{100} A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$  and  $\dpi{100} B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

Then we have to basically find the $B^{-1}$ matrix.

So, Given matrix $\dpi{100} B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

$|B| = 1(3-0) -2(-1-0)-2(2-0) = 3+2-4 = 1 \neq 0$

Hence its inverse $B^{-1}$ exists;

Now, as we know that

$B^{-1} = \frac{1}{|B|} adjB$

So, calculating cofactors of B,

$B_{11} = (-1)^{1+1}(3-0) = 3$                   $B_{12} = (-1)^{1+2}(-1-0) = 1$

$B_{13} = (-1)^{1+3}(2-0) = 2$                  $B_{21} = (-1)^{2+1}(2-4) = 2$

$B_{22} = (-1)^{2+2}(1-0) = 1$                  $B_{23} = (-1)^{2+3}(-2-0) = 2$

$B_{31} = (-1)^{3+1}(0+6) = 6$                    $B_{32} = (-1)^{3+2}(0-2) = 2$

$B_{33} = (-1)^{3+3}(3+2) = 5$

$adjB = \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|} adjB = \frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

Now, We have both $A^{-1}$ as well as $B^{-1}$ ;

Putting in the relation we know; $(AB)^{-1} = B^{-1}A^{-1}$

$(AB)^{-1}=\frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$

$= \begin{bmatrix} 9-30+30 &-3+12-12 &3-10+12 \\ 3-15+10&-1+6-4 &1-5+4 \\ 6-30+25 &-2+12-10 &2-10+10 \end{bmatrix}$

$= \begin{bmatrix} 9 &-3 &5 \\ -2&1 &0 \\ 1 &0 &2\end{bmatrix}$

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