Answer to Question #106055 in Linear Algebra for Rutuja

Question

Check whether following system of equations has a solution.
3x+2y+6z+4w=4
x+2y+2z+w=5
x+z+3w=3

Answer 1

First we construct and transform a matrix for this system:

$\begin{pmatrix} 3 & 2 & 6 & 4 &|& 4 \\ 1 & 2 & 2 & 1 &|& 5 \\ 1 & 0 & 1 & 3 &|& 3 \end{pmatrix} \overset{R_1 - 3R_2, R_2 - R_3}{=} \begin{pmatrix} 0 & -4 & 0 & 1 &|& -11 \\ 0 & 2 & 1 & -2 &|& 2 \\ 1 & 0 & 1 & 3 &|& 3 \end{pmatrix} \overset{R_3 \to R_1, R_1 \to R_3}{=} \\ = \begin{pmatrix} 1 & 0 & 1 & 3 &|& 3 \\ 0 & 2 & 1 & -2 &|& 2 \\ 0 & -4 & 0 & 1 &|& -11 \end{pmatrix} \overset{R_3 + 2R_2}{=} \begin{pmatrix} 1 & 0 & 1 & 3 &|& 3 \\ 0 & 2 & 1 & -2 &|& 2 \\ 0 & 0 & 2 & -3 &|& -7 \end{pmatrix}\\$

$z = \frac{3w - 7}{2}\\ y = \frac{-z + 2w + 2}{2} = -\frac{z}{2} + w + 1 = \frac{7 - 3w}{4} + w + 1 = \frac{w + 11}{4}\\ x = -z - 3w + 3 = \frac{7 - 3w}{2} -3w + 3 = \frac{13 - 9w}{2}$

So a solution of the system exists.