Q.Find $A^{-1}$ by adjoint method and by elementary transformations if $A = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$.

Step 1 (adjoint method): $A=\begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$. Cofactors: $A_{11}=\begin{vmatrix}1&2\\2&4\end{vmatrix}=0$; $A_{12}=-\begin{vmatrix}-1&2\\1&4\end{vmatrix}=-(-4-2)=6$; $A_{13}=\begin{vmatrix}-1&1\\1&2\end{vmatrix}=-2-1=-3$; $A_{21}=-\begin{vmatrix}2&3\\2&4\end{vmatrix}=-(8-6)=-2$; $A_{22}=\begin{vmatrix}1&3\\1&4\end{vmatrix}=4-3=1$; $A_{23}=-\begin{vmatrix}1&2\\1&2\end{vmatrix}=0$; $A_{31}=\begin{vmatrix}2&3\\1&2\end{vmatrix}=4-3=1$; $A_{32}=-\begin{vmatrix}1&3\\-1&2\end{vmatrix}=-(2+3)=-5$; $A_{33}=\begin{vmatrix}1&2\\-1&1\end{vmatrix}=1+2=3$.

Step 2: Cofactor matrix $\begin{bmatrix}0&6&-3\\-2&1&0\\1&-5&3\end{bmatrix}$, so $\text{adj}\,A=\begin{bmatrix}0&-2&1\\6&1&-5\\-3&0&3\end{bmatrix}$.

Step 3: $|A|$, expanding along row 1: $|A|=1(0)+2(6)+3(-3)=0+12-9=3$. So $A^{-1}=\dfrac13\begin{bmatrix}0&-2&1\\6&1&-5\\-3&0&3\end{bmatrix}=\begin{bmatrix}0&-\tfrac23&\tfrac13\\2&\tfrac13&-\tfrac53\\-1&0&1\end{bmatrix}$. …

Find $A^{-1}$ by the adjoint formula, and separately by a chain of elementary row transformations; b …

• Getting two DIFFERENT final answers from the two methods and not realising this signals an arithmetic error somewhere (the inverse is always unique) …