Question

Solve system of linear equations, using matrix method.
$2x+y+z=-1$ ,$x-2y-z=\frac{3}{2}$ ,$3y-5z=9$

Answer 1

$2x+y+z=-1$

$x-2y-z=\frac{3}{2}$

$3y-5z=9$

$\left[\begin{array}{ccc}2& 1& 1\\ 1& -2& -1\\ 0& 3& -5\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 3/2\\ 9\end{array}\right]$

$A$ $B$ $C$

$AX=B$

$X=A^{-1}B$

$A^{-1}=\frac{Adj\left(A\right)}{|A|}$

$|A|=2[10+3]-1[-5]+1\left[3\right]$

$=26+5+3=34$

$Adj\left(A\right)={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^T$

$A_{11}=\left|\begin{array}{cc}-2& -1\\ 3& -5\end{array}\right|(-1{)}^{1+1}=13$

$A_{12}=\left|\begin{array}{cc}1& -1\\ 0& -5\end{array}\right|(-1{)}^{1+2}=-5$

$A_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}1& -2\\ 0& 3\end{array}\right|=3$

$A_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}1& 1\\ 3& -5\end{array}\right|=8$

$A_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& -1\\ 0& -5\end{array}\right|=-5$

$A_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}2& 1\\ 0& 3\end{array}\right|=-6$

$A_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}1& 1\\ -2& -1\end{array}\right|=1$

$A_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|=3$

$A_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}2& 1\\ 1& -2\end{array}\right|=-5$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ -5& -5& 3\\ 3& -6& -5\end{array}\right]\left[\begin{array}{c}-1\\ 3/2\\ 9\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{ccc}-13& +12& +9\\ 5& -\frac{15}{2}& +27\\ -3& -9& -45\end{array}\right]$

$x=\frac{8}{34},y=\frac{24.5}{34},z=\frac{-57}{34}$