Question
If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$ find $( A B)^{-1}$
Since, (AB)-1 = B-1A-1 [Reversal law] .....(i)
Now $\left| B \right| = \left| {\begin{array}{*{20}{c}} 1&2&{ - 2} \\ { - 1}&3&0 \\ 0&{ - 2}&1 \end{array}} \right|$
= 1(3 - 0) - 2(-1 - 0) + (-2)(2 - 0) = 3 + 2 - 4 = 1 = $\ne $ 0
Therefore, B-1 exists.
$\therefore $ B11 = 3, B12 = 1, B13 = 2 and B21 = 2, B22 = 1, B23 = 2 and B31 = 6, B22 = 2, B33 = 5
$\therefore adj.B = \left[ {\begin{array}{*{20}{c}} 3&1&2 \\ 2&1&2 \\ 6&2&5 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&2&6 \\ 1&1&2 \\ 2&2&5 \end{array}} \right]$
$\therefore {B^{ - 1}} = \frac{1}{{\left| B \right|}}\left( {adj.B} \right) = \frac{1}{1} $ $= \left[ {\begin{array}{*{20}{c}} 3&2&6 \\ 1&1&2 \\ 2&2&5 \end{array}} \right]$
From eq. (i), ${\left( {AB} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 3&2&6 \\ 1&1&2 \\ 2&2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right]$
$\Rightarrow {\left( {AB} \right)^{ - 1}} $ $= \left[ {\begin{array}{*{20}{c}} {9 - 30 + 30}&{ - 3 + 12 - 12}&{3 - 10 + 12} \\ {3 - 15 + 10}&{ - 1 + 6 - 4}&{1 - 5 + 4} \\ {6 - 30 + 25}&{ - 2 + 12 - 10}&{2 - 10 + 10} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 9&{ - 3}&5 \\ { - 2}&1&0 \\ 1&0&2 \end{array}} \right]$
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