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In Exercises $29-32,$ (a) does the equation $A \mathbf{x}=0$ have a nontrivial solution and (b) does the equation $A \mathbf{x}=\mathbf{b}$ have at least onesolution for every possible $\mathbf{b} ?$$A$ is a $3 \times 3$ matrix with two pivot positions.

a. A non-trivial solutionb. No.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 5

Solution Sets of Linear Systems

Introduction to Matrices

Oregon State University

McMaster University

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Lectures

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Okay, So in this problem, work even a is a three bite the remain trees with two people, two positions. Okay, there are two questions we need to figure out here. First question is, that's a X equals zero have a nontrivial solution. So we first consider hey, inform off. Reduced a penchant for So that should be 11 one. And with some stuff here because it is a reducing nation form. So all all the entries, um, here will be 00 and zero. Okay, So now we only allow two people positions here. And the matrix I'm writing down here has has three people in positions, and we have to delete at least one those things, um, we have We have more people to positions here, so a se So this one does not work. So let's say they do the new me shakes. They say the new Matrix is so we need to delete at least a warrant road. That means at least one row has to be always ears. So let's say the last row is all yours. So right now we have a 11 and all zeros and with some stuff over here, All right. So you can see from in this matrix that that there exists at least one people, one pre bearable. Not people. Just there exist at least one at these one. Pre venerable. Because we have X one times the coefficient here and with some stuff, either. And here should be zero with some cooperation. I say we have some coefficient. Um, you brought off x three say a 33 as the coefficient of X three. He caused zero. And the second row keeps us x to us A. Sorry, uh, let's say, um, here the coefficient here that the coalition be one a 13 So that's a 13 And this 8 to 3 times x three equal zero So x one x two can be expressed in terms off x three. So ex Tereza free bearable. That means this is stone. We call our effect from over textbook, that is, except a axe equals zero has a nontrivial solution. You can only if there exists of free variable. And here repairable, it's extreme. So the system has country view as a non trivial solution. So the first question is down. Now, The second question second question is, Does the equation X equals? B have at least one solution for anybody, so let's check. So if we have X one plus a 13 as before, that's three course. Some, let's say some counseling. Be one and x two plus a 33 Sorry, 8 to 3 a 23 x three. Because at some B two now we still have a heaven entry. I have 1/3 entry for Vector B. No says since we're we're considering any sector B. So the third entry can be anything but this case. Seems we since the last row is all zero. So that means the last the next entry be has to be so that the three has to be zero. But we've But right now we're considering all considering all the bees, that means the three can be anything. So that means we cannot have. That means that the equation here, the system we have here we're having here doesn't have at least one solution for anybody because we take like we take a B one b 21 So the system doesn't work because we have all zeros in the last row, over a and no matter which, no matter which act we're taking the NASA rover people wear considering the boat matrix multiplication Here, the last roll we'll always be zero. This is the last rule is always zero When we're taking the last entry here as one. And this stuff, they're not equal. So that means this system doesn't have at least one solution for any.

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