First we'll form an augmented matrix for the system:
$\begin{bmatrix} 0 & 2 & 4 & 2 & | & -5 \\ 0 & 5 & 10 & 5 & | & -15 \\ 1 & 2 & 2 & -3 & | & 3 \end{bmatrix}$
Swap the first and third rows to move the 1 to the upper left corner:
$\begin{bmatrix} 1 & 2 & 2 & -3 & | & 3 \\ 0 & 5 & 10 & 5 & | & -15 \\ 0 & 2 & 4 & 2 & | & -5 \end{bmatrix}$
We'll use the first row to eliminate the 2 in the second row by subtracting twice the second row from the third row, and then scale down the second row by 1/5:
$\begin{bmatrix} 1 & 2 & 2 & -3 & | & 3 \\ 0 & 1 & 2 & 1 & | & -3 \\ 0 & 0 & 0 & 0 & | & 1 \end{bmatrix}$
Since we have a row with 0's on the variable side and a non-zero constant on the other side, and thus an inconsistency, the system has no solution.
Solve the following system of linear equations:
2x2+4x3+2x4 = −5
5x2+10x3+5x4 = −15
x1+2x2+2x3−3x4 =3
If the system has no solution, demonstrate this by giving a row-echelon form of the augmented matrix for the system.
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