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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     A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}  and B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}, find (AB)^-^1.

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We know from the identity that;

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

Then we can find easily, 

Given A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}  and  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 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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