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Suppose $A$ is an $n \times n$ matrix with the property that the equation $A \mathbf{x}=\mathbf{0}$ has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation $A \mathbf{x}=\mathbf{b}$ must have a solution for each $\mathbf{b}$ in $\mathbb{R}^{n}$ .

Hence, the equation $A \mathbf{x}=\mathbf{b}$ must have a solution for each b in $\mathbb{R}^{n}$ .

Algebra

Chapter 2

Matrix Algebra

Section 3

Characterizations of Invertible Matrices

Introduction to Matrices

Missouri State University

University of Michigan - Ann Arbor

Lectures

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problem. Um, we want to show that if a X equals zero has only the trivial solution, she will solution. Then way. Want to show Dad? Hey, X, Because be must I have this ocean for every Be okay. Now, The way to do it without the convertible matrix serum is to observe the structure off A. So we consider let's consider the reduced row ish in form off, eh? Say a prime. Okay, so scuse me. So that means, um, a prime times X equals zero. This system has has no pre variable because this system has has only has only intriguing solution that it means the system has no free bearable. Because we we know from our previous material we know that, uh, the system has nontrivial solution. It has at least a work repairable. So he's just a negation of the statement. So has I'll just write down here just for your information, A prime axe times. They're a prime axe equal zero has has on Lee Trivial solution. It's equivalent to say that this system has no free bearable. No, for you bearable. Okay, So for the Moore Demi's matrix A has, um, has an people in positions. Of course. We're assuming Matrix a had Matrix A's On By and Square matrix. Yeah, this is our assumption. Um, people positions. Excuse me, people. Positions. And this is our conclusion here. Now, from here, we can say, um, since there are n people, people in positions and the matrix structure off, eh, Um, or we consider a prime and B and be seen as some some some values on the diagonal and zero for the lower triangular. An appetite for dinner. We have some numbers. I don't. I just, uh, using darts to represent them. So, hey, in terms off a a prime time zags, we can see that of all the terrible here we are using a, uh, using this form of a prime then our nest our last variable se X end must be determined by the value we're taking for B. So that means So we're saying be one. So let's suppose this is our vector B until fiend is equal to a prime times axe. And this is some sort of early on the diagonal times the vector X. So that means we were supposed defector ads his ex Juan x two txn Okay, So that means the last the last term Other day that I'll be there diagonal If we multiply, multiply this This value by the nest. Last entry off. This factor acts. We have some. Aye, aye aye. Which is the entry for the last in the center of the diagonal of ape. Right? So? So there's a II times the last the last available off the vector acts, which is X end is exactly the end. So we can solve for accent here. So that is bien divided by a II and so on so forth. So on so forth we can solve for everything Go bearable x one Until we were done, we have done for X one because, um, there's no free bearable. So that's why we can do this. So that means xn. That means acts the vector must exist. So what that means for the more damages the system a axe? It was B has a solution or every single b we're taking the knob should be from are in

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