Question

Given the system of equations:

$\begin{array}{cc}& 5x-6y+4z=15\\ & 7x+4y-3z=19\\ & 2x+y+6z=46\end{array}$
Find $x,y,z$.

Answer 1

From the equation-

$\left[\begin{array}{ccc}5& -6& 4\\ 7& 4& -3\\ 2& 1& 6\end{array}\right]\times \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}15\\ 19\\ 46\end{array}\right]$

or, $AX=Y$

Where $A=\left[\begin{array}{ccc}5& -6& 4\\ 7& 4& -3\\ 2& 1& 6\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],Y=\left[\begin{array}{c}15\\ 19\\ 46\end{array}\right]A^{-1}=\frac{adjA}{\left|A\right|}adj\left(A\right)={\left[\begin{array}{ccc}27& -48& -1\\ -(-40)& 22& -17\\ 2& 43& 62\end{array}\right]}^T=\left[\begin{array}{ccc}27& 40& 2\\ -48& 22& 43\\ -1& -17& 62\end{array}\right]\left|A\right|=5(24+3)+6(42+6)+4(7-8)=135+288-4=419$

or $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{419}\times \left[\begin{array}{ccc}27& 40& 2\\ -48& 22& 43\\ -1& -17& 62\end{array}\right]\times \left[\begin{array}{c}15\\ 19\\ 46\end{array}\right]$

or $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{419}\times \left[\begin{array}{c}1257\\ 1676\\ 2514\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 6\end{array}\right]\therefore X=3,Y=4,Z=6$