Question

The solution set of the system of equations $x+y-z=6;3x-2y+z=-5;x+3y-2z=14$ is $(x,y,z)$ then $x+y+z$ is equal to

• A. 2
• B. 2.00
• C. 2.0

Answer 1

Answer: (A), (B), (C)

Using Cramer's rule,
$D=\left|\begin{array}{ccc}1& 1& -1\\ 3& -2& 1\\ 1& 3& -2\end{array}\right|=-3$

$D_x=\left|\begin{array}{ccc}6& 1& -1\\ -5& -2& 1\\ 14& 3& -2\end{array}\right|$
$R_1\to R_1+R_2$
$R_3\to R_3+2R_2$
$\Rightarrow D_x=\left|\begin{array}{ccc}1& -1& 0\\ -5& -2& 1\\ 4& -1& 0\end{array}\right|=-3$


$D_y=\left|\begin{array}{ccc}1& 6& -1\\ 3& -5& 1\\ 1& 14& -2\end{array}\right|$
$R_1\to R_1+R_2$
$R_3\to R_3+2R_2$
$\Rightarrow D_y=\left|\begin{array}{ccc}4& 1& 0\\ 3& -5& 1\\ 7& 4& 0\end{array}\right|=-9$


$D_z=\left|\begin{array}{ccc}1& 1& 6\\ 3& -2& -5\\ 1& 3& 14\end{array}\right|=6$

Then, $x=\frac{D_x}{D}=\frac{-3}{-3}=1$

$y=\frac{D_y}{D}=\frac{-9}{-3}=3$

$z=\frac{D_z}{D}=\frac{6}{-3}=-2$
$\therefore x+y+z=2$