Matrix and Determinant
78803
If \(A=\left[\begin{array}{ccc}4 & 3 & 2 \\ -1 & 2 & 0\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 2 \\ -1 & 0 \\ 1 & -2\end{array}\right]\) then \((\mathbf{A B})^{-1}=\)
1 \(\frac{1}{6}\left[\begin{array}{cc}-2 & -4 \\ 3 & 3\end{array}\right]\)
2 \(\frac{1}{6}\left[\begin{array}{ll}2 & 4 \\ 3 & 3\end{array}\right]\)
3 \(\frac{1}{6}\left[\begin{array}{cc}-2 & 4 \\ 3 & -3\end{array}\right]\)
4 \(\frac{1}{6}\left[\begin{array}{cc}2 & -4 \\ -3 & 3\end{array}\right]\)
Explanation:
(A) : Given that,
\(A=\left[\begin{array}{ccc}4 & 3 & 2 \\ -1 & 2 & 0\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 2 \\ -1 & 0 \\ 1 & -2\end{array}\right]\)
\(\mathrm{AB}=\left[\begin{array}{ccc}4 & 3 & 2 \\ -1 & 2 & 0\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -1 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}4+(-3)+2 & 8+0+(-4) \\ (-1)+(-2)+0 & -2+0+0\end{array}\right]\)
\(\therefore \mathrm{AB}=\left[\begin{array}{cc}3 & 4 \\ -3 & -2\end{array}\right]\)
Now, \(|\mathrm{AB}|=-6-(-12)=6 \neq 0\)
\(\operatorname{adj}(A B)=\left[\begin{array}{cc}-2 & -4 \\ 3 & 3\end{array}\right]\)
\(\therefore(\mathrm{AB})^{-1}=\frac{1}{|\mathrm{AB}|} \operatorname{adj}(\mathrm{AB})=\frac{1}{6}\left[\begin{array}{cc}-2 & -4 \\ 3 & 3\end{array}\right]\)