Question

Express the following equation and solve them by finding the inverse of coefficient matrix using adjoining method.
$x+y+z=6$
$2x+y+2z=10$
$3x+3y+4z=21$

Answer 1

$\begin{array}{c}x+y+z=6,2x+y+2z=10,3x+3y+4z=21\\ A=\left(\begin{array}{ccc}1& 1& 1\\ 2& 1& 2\\ 3& 3& 4\end{array}\right)\times =\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}6\\ 10\\ 21\end{array}\right]\\ \Rightarrow \left|A\right|=1\left(-2\right)-1\left(2\right)+1\left(3\right)\\ \Rightarrow \left|A\right|=-2-2+3=-1\ne 0\\ \to Aisvertible\\ \Rightarrow co-factorof\left|A\right|are\\ \Rightarrow A_1=-2,A_2=-2,A_3=3\\ \Rightarrow A_{21}=-1,A_{22}=1,A_{23}=0\\ \Rightarrow A_{31}=1,A_{32}=0,A_{33}=-1\\ adjA=\left(\begin{array}{ccc}-2& -1& 1\\ -2& 1& 0\\ 3& 0& -1\end{array}\right)\\ \Rightarrow x=A^{-1}B\\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-1}\left(\begin{array}{ccc}-2& -1& 1\\ -2& 1& 0\\ 3& 0& -1\end{array}\right)\left[\begin{array}{c}6\\ 10\\ 21\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-1\left[\begin{array}{c}-1\\ -2\\ -3\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\\ \therefore x=1,y=2,andz=3\end{array}$