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Mathematics Question on Determinants

Find adjoint of each of the matrices [1−12\235−201]\begin{bmatrix}1&-1&2\\\2&3&5\\\\-2&0&1\end{bmatrix}

Answer

Let A=[1−12\235−201]\begin{bmatrix}1&-1&2\\\2&3&5\\\\-2&0&1\end{bmatrix}
we have

A11=∣35\01∣\begin{vmatrix}3&5\\\0&1\end{vmatrix}=3-0=3

A12=-∣25−21∣\begin{vmatrix}2&5\\\\-2&1\end{vmatrix}=-12

A13=∣23−20∣\begin{vmatrix}2&3\\\\-2&0\end{vmatrix}=0+6=6

A21=-∣−12\01∣\begin{vmatrix}-1&2\\\0&1\end{vmatrix}=-(-1-0)=1

A22=∣12−21∣\begin{vmatrix}1&2\\\\-2&1\end{vmatrix}=1+4=5

A23=-∣1−1−20∣\begin{vmatrix}1&-1\\\\-2&0\end{vmatrix}=-(0-2)=2

A31=∣−12\35∣\begin{vmatrix}-1&2\\\3&5\end{vmatrix}=-5-6=-11

A32=-∣12\25∣\begin{vmatrix}1&2\\\2&5\end{vmatrix}=-(5-4)=-1

A33=∣1−1\23∣\begin{vmatrix}1&-1\\\2&3\end{vmatrix}=3+2=5

Hence adj A=[A11A21A31\A12A22A23\A13A23A33]\begin{bmatrix}A_{11}&A_{21}&A_{31}\\\A_{12}&A_{22}&A_{23}\\\A_{13}&A_{23}&A_{33}\end{bmatrix}=[31−11 −125−1 625]\begin{bmatrix}3&1&-11\\\ -12&5&-1\\\ 6&2&5\end{bmatrix}