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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
If $\text{A}^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\text{and B}=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}, $find $(AB)^{-1}$
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Answer

Given: $\text{A}^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix} \text{and B}=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix} $
Since, $(AB)^{-1}= B^{-1}A^{-1}$ [Reversal law] ......(1)
Now $\left|\text{B}\right|=\begin{vmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{vmatrix}=1(3 - 0) -2(- 1 -0) + (-2)(2 - 0)=3+2-4=1\neq0$
Therefore,$ B^{-1}​​​​​​​$​​​​​​​exists.
$\therefore$
$B_{11} = 3, B_{12} = 1, B_{13} = 2$ and $B_{21} = 2, B_{22} = 1, B_{23}= 2$ and $B_{31} = 6, B_{32} = 2, B_{33} = 5$
$\therefore\text{adj. B}=\begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}$
$\therefore\text{B}^{-1}=\frac{1}{\text{|B|}}(\text{adj. B})=\frac{1}{1}\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}$
From eq. $(1), (AB)^{-1} =$ $\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
$\Rightarrow $ $(\text{AB})^{-1}=\begin{bmatrix}9-30+30&-3+12-12&3-10+12\\3-15+10&-1+6-4&1-5+4\\6-30+25&-2+12-10&2-10+10\end{bmatrix}=\begin{bmatrix}9&-3&5\\-2&3&1\\1&0&2\end{bmatrix}$

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