Question

Solve the following equations
$5x-6y+4z=15$
$7x+4y-3z=19$
$2x+y+6z=46$

Answer 1

The above equations can be written in matrix form as
$\left[\begin{array}{ccc}5& ;-6& ;4\\ 7& ;6& ;-3\\ 2& ;1& ;6\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ = $\left[\begin{array}{c}15\\ 19\\ 46\end{array}\right]$ or AX = B
|A| = D = $\left|\begin{array}{ccc}5& ;-6& ;4\\ 7& ;4& ;-3\\ 2& ;1& ;6\end{array}\right|$
Apply $C_1-2C_2,C_3-6C_2$ to make two zeros in $R_3$
$D=\left|\begin{array}{ccc}17& ;-6& ;40\\ -1& ;4& ;-27\\ 0& ;1& ;0\end{array}\right|=-\left|\begin{array}{cc}17& ;40\\ -1& ;-27\end{array}\right|=-[-459+40]=419\ne 0$.
$\therefore$ The matrix A is non-singular or rank of matrix A is
3. We will have a unique solution and the equations are consistent.
By Crammer's rule
$\frac{x}{D_1}=\frac{y}{D_2}=\frac{z}{D_3}=\frac{1}{D}$
Where $D_1$ is obtained from D by replacing the first column of D by b's i.e. $15,19,46$
$D_1=\left|\begin{array}{ccc}15& ;-6& ;4\\ 19& ;4& ;-3\\ 46& ;1& ;6\end{array}\right|=15\left(27\right)-19(-40)+46\left(2\right)=1257$
$\therefore \frac{x}{1257}=\frac{y}{1676}=\frac{z}{2514}=\frac{1}{419}$
$\therefore x=3,y=4,z=6$.