make a matrix:
( 1 1 1 1 5 1 − 1 1 1 1 1 0 1 1 3 ) \begin{pmatrix}
1& 1&1&1&&5 \\
1&-1&1&1&&1\\
1&0&1&1&&3
\end{pmatrix} ⎝ ⎛ 1 1 1 1 − 1 0 1 1 1 1 1 1 5 1 3 ⎠ ⎞
make this matrix simple form:
line 3 minus line 2:
( 1 1 1 1 5 1 − 1 1 1 1 0 1 0 0 2 ) \begin{pmatrix}
1& 1&1&1&&5 \\
1&-1&1&1&&1\\
0&1&0&0&&2
\end{pmatrix} ⎝ ⎛ 1 1 0 1 − 1 1 1 1 0 1 1 0 5 1 2 ⎠ ⎞
line 2 minus line 1:
( 1 1 1 1 5 0 − 2 0 0 − 4 0 1 0 0 2 ) \begin{pmatrix}
1& 1&1&1&&5 \\
0&-2&0&0&&-4\\
0&1&0&0&&2
\end{pmatrix} ⎝ ⎛ 1 0 0 1 − 2 1 1 0 0 1 0 0 5 − 4 2 ⎠ ⎞
line 2 divided by -2:
( 1 1 1 1 5 0 1 0 0 2 0 1 0 0 2 ) \begin{pmatrix}
1& 1&1&1&&5 \\
0&1&0&0&&2\\
0&1&0&0&&2
\end{pmatrix} ⎝ ⎛ 1 0 0 1 1 1 1 0 0 1 0 0 5 2 2 ⎠ ⎞
lines 2 and 3 are equal, so we remove the third line:
( 1 1 1 1 5 0 1 0 0 2 ) \begin{pmatrix}
1& 1&1&1&&5 \\
0&1&0&0&&2\\
\end{pmatrix} ( 1 0 1 1 1 0 1 0 5 2 )
this matrix has a simple form, back to system equations:
x + y + z + t = 5 , x+y+z+t=5, x + y + z + t = 5 ,
y = 2 ; y=2; y = 2 ;
x = 5 − 2 − z − t , x=5-2-z-t, x = 5 − 2 − z − t ,
y = 2 ; y=2; y = 2 ;
z and t are free variables
z = h 1 , z=h1, z = h 1 ,
t = h 2 ; t=h2; t = h 2 ;
solution to the system of equations:
x = 3 − h 1 − h 2 , x=3- h1-h2, x = 3 − h 1 − h 2 ,
y = 2 , y=2, y = 2 ,
z = h 1 , z=h1, z = h 1 ,
t = h 2 ; t=h2; t = h 2 ;
(where h1 and h2 are any real numbers)
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