1.

If A=\(\begin{bmatrix}-8&2\\6&-3\end{bmatrix}\) and B=\(\begin{bmatrix}2&1\\1&7\end{bmatrix}\). Find (AB)^-1.(a) –\(\frac{1}{432}\) \(\begin{bmatrix}-27&6\\9&14\end{bmatrix}\)(b) \(\frac{1}{432}\) \(\begin{bmatrix}27&6\\9&14\end{bmatrix}\)(c) \(\frac{1}{432}\) \(\begin{bmatrix}-27&6\\9&14\end{bmatrix}\)(d) \(\frac{-1}{432}\) \(\begin{bmatrix}27&6\\9&14\end{bmatrix}\)This question was addressed to me in an online interview.The question is from Determinants in section Determinants of Mathematics – Class 12

Answer»

Correct OPTION is (c) \(\FRAC{1}{432}\) \(\BEGIN{bmatrix}-27&6\\9&14\end{bmatrix}\)

The explanation: Given that, A=\(\begin{bmatrix}-8&2\\6&-3\end{bmatrix}\) and B=\(\begin{bmatrix}2&1\\1&7\end{bmatrix}\)

∴AB=\(\begin{bmatrix}-8×2+2×1&-8×1+2×7\\6×2+(-3)×1&6×1+(-3)×7\end{bmatrix}\)=\(\begin{bmatrix}-14&6\\9&27\end{bmatrix}\)

adj(AB)=\(\begin{bmatrix}27&-6\\-9&-14\end{bmatrix}\)

|AB|=27×(-14)-(-9)×(-6)=-378-54=-432

(AB)^-1=\(\frac{1}{|AB|}\) adj AB=\(\frac{1}{-432} \begin{bmatrix}27&-6\\-9&-14\end{bmatrix}\)=\(\frac{1}{432} \begin{bmatrix}-27&6\\9&14\end{bmatrix}\).



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