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Determinants — Applied Maths STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsDeterminants5 Marks
Question
Given $A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right], B^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right],$ compute $(A B)^{-1}.$
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Answer

$|A|=5(-1)+4(1)=-1$
$\begin{array}{lll}C_{11}=-1 & C_{21}=8 & C_{31}=-12 \\ C_{12}=0 & C_{22}=1 & C_{32}=-2 \\ C_{13}=1 & C_{23}=-10 & C_{33}=15\end{array}$
$\begin{aligned} A^{-1} & =\left[\begin{array}{ccc}1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15\end{array}\right] \\ (A B)^{-1} & =B^{-1} A^{-1}\end{aligned}$
$\begin{array}{l}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{ccc}1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15\end{array}\right] \\ =\left[\begin{array}{lll}-2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42\end{array}\right]\end{array}$

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