Question

3x+y+2z=3
2x-3y-z=-3
x+2y+z=4
solve using matrix method

Answer 1

Hello,
[tex]\left\{\begin{matrix} 3x+y+2z=3\\ 2x-3y-z=-3\\ x+2y+z=4 \end{matrix}\right.[/tex]


from the third you get:
x=4-2y-z

substituting in the first, we obtain:
3(4-2y-z)+y+2z=3;
12-6y-3z+y+2z=3;
-5y-z=3-12;
-z-5y=-9;
z+5y=9;
z=9-5y

substituting in the second one, we obtain:
2(4-2y-z)-3y-(9-5y)=-3;
8-4y-2z-3y-9+5y=-3;
8-2y-2(9-5y)-9=-3;
8-2y-18+10y-9=-3;
-19+8y=-3;
8y=19-3;
8y=16;
y=16:8=2;
y=2

therefore:
z=9-5y=9-5×2=9-10=-1
and
x=4-2y-z=4-(2×2)-(-1)=4-4+1=1

So you have:
x=1, y=2 , z=-1

bye :-)

Answer 2

Solving the system of simultaneous equations by using matrix method:

  3x + y + 2 z = 3 
  2x - 3 y - z = -3 
  x + 2 y + z = 4 

MATRIX METHOD:

$\left[\begin{array}{cccc}3&1&2&3\\2&-3&-1&-3\\1&2&1&4\end{array}\right] = \left[\begin{array}{cccc}0&-5&-1&-9\\0&-7&-3&-11\\1&2&1&4\end{array}\right] \\\\=\left[\begin{array}{cccc}0&-5&-1&-9\\0&8&0&16\\1&2&1&4\end{array}\right] = \left[\begin{array}{cccc}0&-5&-1&-9\\0&1&0&2\\1&2&1&4\end{array}\right] \\\\=\left[\begin{array}{cccc}0&-5&-1&-9\\0&1&0&2\\1&0&1&0\end{array}\right] =\left[\begin{array}{cccc}0&0&-1&1\\0&1&0&2\\1&0&1&0\end{array}\right]$

$=\left[\begin{array}{cccc}0&0&-1&1\\0&1&0&2\\1&0&1&0\end{array}\right] =\left[\begin{array}{cccc}0&0&-1&1\\0&1&0&2\\1&0&0&1\end{array}\right] \\\\So \: x=1, \: y=2 \:, z = -1$

x = 1, y =2, z = -1.