Practicing Success
The adjoint of the matrix $A=\left[\begin{array}{cc}1 & 2 \\ 3 & -5\end{array}\right]$ is |
$\left[\begin{array}{cc}1 & -2 \\ -3 & -5\end{array}\right]$ $\left[\begin{array}{cc}-5 & -2 \\ -3 & 1\end{array}\right]$ $\left[\begin{array}{cc}2 & 1 \\ -3 & 5\end{array}\right]$ $\left[\begin{array}{cc}-1 & 3 \\ 2 & 5\end{array}\right]$ |
$\left[\begin{array}{cc}-5 & -2 \\ -3 & 1\end{array}\right]$ |
$A=\left[\begin{array}{cc} 1 & 2 \\ 3 & -5 \end{array}\right]$ The co-factors are given by, $A_{11}=-5 ; A_{12}=-3 ; A_{21}=-3 ; A_{22}=1$ Let, $B ⥂~=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{cc}-5 & -3 \\ -2 & 1\end{array}\right]$ adj. (A) = B' = $\left[\begin{array}{cc}-5 & -2 \\ -3 & 1\end{array}\right]$ Hence (2) is the correct answer. |