Question

solve by matrix method x + y + z = 7 \\ 2x - y + z = 5\\ 3x + y - z = 8

Answer 1

Hello,
$\left\{\begin{matrix} x + y + z = 7\\ 2x - y + z = 5\\ 3x + y - z = 8 \end{matrix}\right.$

from the first you get:
x=7-y-z

substituting in the second one, we obtain:
2(7-y-z)-y+z=5;
14-2y-2z-y+z=5;
-3y-z=5-14;
-3y-z=-9;
3y+z=9;
z=9-3y

substituting in the third, we obtain:
3(7-y-z)+y-(9-3y)=8;
21-3y-3z+y-9+3y=8;
21-3y-3(9-3y)+y-9+3y=8;
21-3y-27+9y+y-9+3y=8;
10y-15=8;
10y=8+15;
10y=23;
y=23/10

therefore:
z=9-3y=9-3×23/10=9-69/10=90-69/10=21/10
and
x=7-23/10-21/10=70-23-21/10=26/10=13/5

So you have:
x=13/5, y=23/10 , z=21/10

bye :-)

Answer 2

Solving the system of simultaneous equations by using matrix method:

    x + y +  z = 7
  2x -  y + z = 5
  3x + y - z = 8

MATRIX METHOD:

$\left[\begin{array}{cccc}1&1&1&7\\2&-1&1&5\\3&1&-1&8\end{array}\right] =\left[\begin{array}{cccc}1&1&1&7\\2&-1&1&5\\5&0&0&13\end{array}\right] \\\\=\left[\begin{array}{cccc}1&1&1&7\\3&0&2&12\\5&0&0&13\end{array}\right] =\left[\begin{array}{cccc}1&1&1&7\\0&-3&-1&-9\\1&0&0&13/5\end{array}\right] \\\\=\left[\begin{array}{cccc}0&1&1&22/5\\0&-3&-1&-9\\1&0&0&13/5\end{array}\right] =\left[\begin{array}{cccc}0&-2&0&-23/5\\0&-3&-1&-9\\1&0&0&13/5\end{array}\right]$

[tex]=\left[\begin{array}{cccc}0&1&0&23/10\\0&3&1&9\\1&0&0&13/5\end{array}\right]\\\\=\left[\begin{array}{cccc}0&1&0&23/10\\0&0&1&21/10\\1&0&0&13/5\end{array}\right] [/tex]

So answer is :  x = 13/5;    y = 23/10 ,   z = 21/10.

It is preferable to give some numbers which are easy to calculate.