Find the adjoint of the matrix A = left[begin | vedclass

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(D) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}ight]$.The adjoint of a matrix is the transpose of the cofactor

Vedclass · Accepted Answer

$\left[\begin{array}{rrr}3 & 1 & -11 \ -12 & 5 & -1 \ 6 & 2 & 5\end{array}ight]$

Vedclass · Answer

(D) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}ight]$.The adjoint of a matrix is the transpose of the cofactor matrix. Let $C_{ij}$ be the cofactor of the element $a_{ij}$.$C_{11} = +\left|\begin{array}{cc}3 & 5 \ 0 & 1\end{array}ight| = 3 - 0 = 3$$C_{12} = -\left|\begin{array}{cc}2 & 5 \ -2 & 1\end{array}ight| = -(2 + 10) = -12$$C_{13} = +\left|\begin{array}{cc}2 & 3 \ -2 & 0\end{array}ight| = 0 - (-6) = 6$$C_{21} = -\left|\begin{array}{cc}-1 & 2 \ 0 & 1\end{array}ight| = -(-1 - 0) = 1$$C_{22} = +\left|\begin{array}{cc}1 & 2 \ -2 & 1\end{array}ight| = 1 - (-4) = 5$$C_{23} = -\left|\begin{array}{cc}1 & -1 \ -2 & 0\end{array}ight| = -(0 - 2) = 2$$C_{31} = +\left|\begin{array}{cc}-1 & 2 \ 3 & 5\end{array}ight| = -5 - 6 = -11$$C_{32} = -\left|\begin{array}{cc}1 & 2 \ 2 & 5\end{array}ight| = -(5 - 4) = -1$$C_{33} = +\left|\begin{array}{cc}1 & -1 \ 2 & 3\end{array}ight| = 3 - (-2) = 5$The cofactor matrix is $\left[\begin{array}{ccc}3 & -12 & 6 \ 1 & 5 & -1 \ -11 & -1 & 5\end{array}ight]$.Taking the transpose,we get $	ext{adj}(A) = \left[\begin{array}{ccc}3 & 1 & -11 \ -12 & 5 & -1 \ 6 & 2 & 5\end{array}ight]$.

Find the adjoint of the matrix $A = \left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$.

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