Solve the following equations by using Matrix inversion method

$3x+y-4z=0$

$-x+2y+z=4$

$2x-y+3z=8$

The given system of equation can be written as $AX=B$

Where $A=\left[\begin{array}{ccc}3& 1& -4\\ -1& 2& 1\\ 2& -1& 3\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{l}0\\ 4\\ 8\end{array}\right]$

$\left|A\right|=\left[\begin{array}{ccc}3& 1& -4\\ -1& 2& 1\\ 2& -1& 3\end{array}\right]=3(6+1)-1(-3-2)-4(1-4)$

$=21+5+12=38$

$\left|A\right|\ne 0\Rightarrow A^{-1}\text{ exists }$

cofactor of $3=(-1{)}^{1+1}\left|\begin{array}{cc}2& 1\\ -1& 3\end{array}\right|=6+1=7$

cofactor of $1=(-1{)}^{1+2}\left|\begin{array}{cc}-1& 1\\ 2& 3\end{array}\right|=-(-3-2)=5$

cofactor of $-4=(-1{)}^{1+3}\left|\begin{array}{cc}-1& 2\\ 2& -1\end{array}\right|=1-4=-3$

cofactor of $-1=(-1{)}^{2+1}\left|\begin{array}{cc}1& -4\\ -1& 3\end{array}\right|=-(3-4)=+1$

cofactor of $2=(-1{)}^{2+2}\left|\begin{array}{l}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right|=9+8=17$

cofactor of $1=(-1{)}^{2+3}\left|\begin{array}{cc}3& 1\\ 2& -1\end{array}\right|=-(-3-2)=5$

cofactor of $2=(-1{)}^{2+3}\left|\begin{array}{cc}3& 1\\ 2& -1\end{array}\right|=-(-3-2)=5$

cofactor of $-1=(-1{)}^{3+2}\left|\begin{array}{cc}3& -4\\ -1& 1\end{array}\right|=-(3-4)=1$

cofactor of $3=(-1{)}^{3+2}\left|\begin{array}{cc}3& 1\\ -1& 2\end{array}\right|=6+1=7$

$\mathrm{Adj}\mathrm{A}=(\text{ cofactors of }\mathrm{A}{)}^T={\left[\begin{array}{ccc}7& 5& -3\\ 1& 17& 5\\ 9& 1& 7\end{array}\right]}^T$

$=\left[\begin{array}{ccc}7& 1& 9\\ 5& 17& 1\\ -3& 5& 7\end{array}\right]$

$AX=B\Rightarrow X=A^{-1}B$

$X=\frac{\mathrm{Adj}A}{\det A}\cdot B=\frac{1}{38}\left[\begin{array}{ccc}7& 1& 9\\ 5& 17& 1\\ -3& 5& 7\end{array}\right]\left[\begin{array}{l}0\\ 4\\ 8\end{array}\right]$

$\frac{1}{38}\left[\begin{array}{c}0+4+72\\ 0+68+8\\ 0+20+56\end{array}\right]=\frac{1}{38}\left[\begin{array}{l}76\\ 76\\ 76\end{array}\right]=\left[\begin{array}{l}2\\ 2\\ 2\end{array}\right]$

$X=\left[\begin{array}{l}2\\ 2\\ 2\end{array}\right] \therefore \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}2\\ 2\\ 2\end{array}\right]$

$x=2,y=2,z=2$ is the solution for given system of equations