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Suppose $A$ is a $3 \times 3$ matrix and $y$ is a vector in $\mathbb{R}^{3}$ such that the equation $A \mathbf{x}=\mathbf{y}$ does not have a solution. Does there exist a vector $\mathbf{z}$ in $\mathbb{R}^{3}$ such that the equation $A \mathbf{x}=\mathbf{z}$ has a unique solution? Discuss.

No

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 5

Solution Sets of Linear Systems

Introduction to Matrices

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Lectures

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problem. We're given the Matrix three by three matrix and the vector y and such data a eight times acts. It was why has no solution. And meanwhile, we want to We want to figure out that whether there exists a vector Z in our three such that the equation the X people see has a unique social. All right, so first, let's consider the each of the reduce station of form of eight. So let's say a first. It has to be three by three matrix. Now consider reduced agent for reduce station form should be one the erosion of you needed by one and with some stuff over here and zero over here, so Oh, I would just leave this part blank. All right, so this is our eight, and we're considering system a axe. It was why. So this is Tom. This system here must have a solution right here. Because no matter which, why we're taking here, we have already writing a new page. So consider Hey, X, it was why No. Put a here, all zeros and some stuff and, uh, yeah, no bearable. Here. Victor ags X Y x two x three And why want y two by three. So I should be white too. And the white three. Now, remember that, uh, this system must have a have a solution, because we we have right here in the meadow, no matter which, which, which is why we're taking We can always take X three equals flight three and next to can be determined by X three and X one can be determined by x two and next three. So we're done. So we have to make he ordered to make this system none consistent. We have to make it, at least warned. Roll all zeros. So let's say in this case, we have the last rule, all zeros, and this should be one. This is zero, and we have some stuff over here. Uh, say I'll just draw like this. All right, so right now we consider a x equal. See, So we need to make this system consistent, but you see from here that we can it's impossible to have a unique access here because we have who are taking. We're taking a Z. That is, that makes this system consistent. We have to make the last entry zero right now. We have again. We have to, uh, three unknowns, which is X one plus some cooperation. That's a a 13 times x three because Z y and next two plus a 23 times eight x three will see to. Now we have here x one, the person 1st 1st on noon on X to the second unknown Onda X three, the third unknown. And we only have two you wishes here. So that means this this system has infinite many solutions. So it it's just impossible to have a unique, unique solution in this case. So that means that means we cannot. That means we cannot have.

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