# Matrices-back substitution

1. Oct 31, 2011

### TyErd

1. The problem statement, all variables and given/known data

I have attached the question

2. Relevant equations

3. The attempt at a solution

I think the answer is no solution because there is 5 variables but only 3 equations. Is that correct?

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2. Oct 31, 2011

### Ray Vickson

No, that is not the reason. Look at the final row; it is shorthand for an equation involving x1, x2, x3, x4, x5. What is the equation?

RGV

3. Oct 31, 2011

### TyErd

do you mean 0x1 + 0x2 + 0x3 + 0x4 + 0x5 = 4?

4. Oct 31, 2011

### TyErd

if the third is also an equation that means there must be an answer right? So because we have to assign variables and we have 5unknowns and 3 equations that must mean two values will be variables right?? so the answer has to be B yeah??

5. Oct 31, 2011

### Staff: Mentor

For what values of the variables x1, x1, x2, x3, and x4 will this be a true statement?

Not necessarily. There are three possibilities for a system of equations (which are here represented by an augmented matrix):
1) a unique solution
2) multiple solutions
3) no solution.