Answer to Question #125832 in Linear Algebra for hitendra

Question

 Check whether the following system of equations has a solution. (6)

3x+2y+6z+4w =4

x+2y +2z +w =5

x+z+ 3w =3

Answer 1

Let's consider the matrix of the system:

$\begin{pmatrix} 3 & 2 & 6 & 4 & |4 \\ 1 & 2 & 2 & 1 & |5 \\ 1 & 0 & 1 & 3 & |3 \end{pmatrix}$

Using Gauss method (= making transformations that do not change the system), we can get this matrix (it still coresponds to the given system):

$\begin{pmatrix} 2 & 0 & 0 & 9 & |13 \\ 0 & 4 & 0 & -1 & |11 \\ 0 & 0 & -2 & 3 & |7\,\,\, \end{pmatrix}$

The rank of the matrix of coefficient is equal to the rank of extended matrix. (An extended matrix is ​​a matrix obtained from a matrix of coefficients to which a column of free terms is attached.) This means that, according to the Rouché–Capelli theorem, the system has a solution.

From this particular matrix we can see that system has the infinite number of solutions. We can leave w as a parameter, and express x, y, z through it. So we get:

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