Answer to Question #86856 in Linear Algebra for Yara Sayed Ahmad

Question

The following is the reduced row echelon form of the augmented matrix for a system of linear equations.

1 0 -3 0 0 | 2
0 1 -1 0 2 | 6
0 0 0 1 1 | 3
0 0 0 0 0 | 0

a) How many variables were in the original system of equations, and what tells you this?

b) How many equations were in the original system of equations, and what tells you this?

c) State the solution (unique or general as appropriate), if it exists, for the system using x1,x2,x3,… for variables. If no solution exists, explain why no solution exists.

Answer 1

a) There are 5 variables in the original system of equations, because of 6 columns in the augmented matrix. As the last column means constant terms, rest columns correspond to variables.

b) There were 4 equations in the original system because of 4 rows in the augmented matrix.

c) Rank of the matrix is 3. Hence, the general solution has (5-3) free parameters:

$$x3= \frac{x1}{3}-\frac{2}{3},x4= -\frac{x1}{3}+x2-\frac{7}{3},x5= \frac{x1}{3}-x2+\frac{16}{3}$$

Where $x1, \, x2$ free parameters