Answer to Question #144442 in Linear Algebra for Sourav Mondal

Question

Solve by Gaussian elimination method the following system of equations :

x+y+z+t=5

x-y+z+t=1

x + z + t = 3

Answer 1

make a matrix:

$\begin{pmatrix} 1& 1&1&1&&5 \\ 1&-1&1&1&&1\\ 1&0&1&1&&3 \end{pmatrix}$

make this matrix simple form:

line 3 minus line 2:

$\begin{pmatrix} 1& 1&1&1&&5 \\ 1&-1&1&1&&1\\ 0&1&0&0&&2 \end{pmatrix}$

line 2 minus line 1:

$\begin{pmatrix} 1& 1&1&1&&5 \\ 0&-2&0&0&&-4\\ 0&1&0&0&&2 \end{pmatrix}$

line 2 divided by -2:

$\begin{pmatrix} 1& 1&1&1&&5 \\ 0&1&0&0&&2\\ 0&1&0&0&&2 \end{pmatrix}$

lines 2 and 3 are equal, so we remove the third line:

$\begin{pmatrix} 1& 1&1&1&&5 \\ 0&1&0&0&&2\\ \end{pmatrix}$

this matrix has a simple form, back to system equations:

$x+y+z+t=5,$

$y=2;$

$x=5-2-z-t,$

$y=2;$

z and t are free variables

$z=h1,$

$t=h2;$

solution to the system of equations:

$x=3- h1-h2,$

$y=2,$

$z=h1,$

$t=h2;$

(where h1 and h2 are any real numbers)