Matrices-back substitution

TyErd · Oct 31, 2011

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?

Ray Vickson · Oct 31, 2011

TyErd said: ↑

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?

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

TyErd · Oct 31, 2011

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

TyErd · Oct 31, 2011

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??

Mark44 · Oct 31, 2011

TyErd said: ↑

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

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

TyErd said: ↑

if the third is also an equation that means there must be an answer right?

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.

TyErd said: ↑

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??

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Matrices-back substitution

TyErd

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Ray Vickson

TyErd

TyErd

Mark44

Staff: Mentor

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