Q.Solve the system $x+y+z=6,\ 2x+y-z=1,\ x-y+2z=5$ using the matrix method.

[!TLDR] Find $|A|=-7$, compute the full adjoint of the $3\times3$ coefficient matrix, then $ …

$A=\begin{pmatrix}1&1&1\2&1&-1\1&-1&2\end{pmatrix},\,B=\begin{pmatrix}6\1\5\end{pmatrix}$. Expanding along row 1: $|A|=1(2-1)-1(4+1)+1(-2-1)=1-5-3=-7 e0$. Computing all cofactors and transposing gives $\operatorname{adj}A=\begin{pmatrix}1&-3&-2\-5&1&3\-3&2&-1\end{pmatrix}$, so $A^{-1}=-\tfrac17\begin{pmatrix}1&-3&-2\-5&1&3\-3&2&-1\end{pmatrix}$. Then $X=A^{-1}B$: $x=-\tfrac17[1(6)-3(1)-2(5)]=-\tfrac17[6-3-10]=-\tfrac17(-7)=1$; $y=-\tfrac17[-5(6)+ …

Write $AX=B$, verify $|A|\ne0$, compute all nine cofactors and transpose to get $\operatorname{adj}A$, form $A^{-1}=\operatorname{adj} …

With nine cofactors to track, a single sign error in any one of them (especially the negative off-diagonal ones) will silently produce a wrong solution -- a …