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Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot in each column. Explain why the system has a unique solution.

See Explanation.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

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for this question. We want to find the values of a for which this system equations has no solution. Exactly one solution and infinitely many solutions, So we'll start with no solution. So first things first. We want to write out our Matrix from the given equation, so we have one. Next one plus two x two plus x three is equal to two to x one minus two x two plus three x three is equal to one and we have one x one plus two x two minus a squared minus three x three is equal to a So before we go have been started. Let's remind ourselves of what the Matrix looks like. That has no solution. So the requirement is that the reduce special inform has a one in the right most column. So let's do so matrix simplification. Here, our first step is gonna be extinct, are three and said that equal to our three minus are one. So if we do that first column stays the same. Second column. Cesaire Sorry, First road stays the same. Second row stays the same, and the third row become 00 And then we have negative a squared minus three minus one. And then over here we have a minus two. So what we want to dio is we want to figure out a value such that this is equal to zero, that this is not able to see her because that will give us and inconsistent system of equation. So let's first simplify this turn over here. So we have negative a square minus three minus one equal to negative. A squared plus three on this one is equal to a negative. A squared plus two right. And we want it to be equal to zero. So let's go ahead and do that now and solve for the A that will make this condition true. So a squared is gonna be able to to because we moved the a square to the other side. Then we take the square root but both sides and we get that a has to equal the square root of two. OK, so that's our first requirement. Now let's go ahead to our second requirement, which states we want a minus two to not be equal zero. So if a minus two can't be able to zero, we want any thio not equal to two. These two requirements agree with each other, since two is not equal to the square to. So we know that for no solution to exist, we want a to be equal to the square of two. But for the next one, for exactly one solution, we know that system equations will have exactly one solution if the reduced excellent form of The Matrix has no free variables. So we'll go ahead and we'll take this simplified matrix here and we'll just go ahead and put it on the other page so we don't have to repeat our row operations. So we have 1212 two Negative, too. 31 00 negative. A squared plus two and then a minus two. Okay, so let's continue so reduction, so we can get a closer to produce a short form. So to do that, uh, let's first take our second row and said it equal to the first row plus the second row. And this will help us knock out this term right here. So when we do that, we get this matrix. First row stays the same second row since we're out of it, becomes 30 for three and then the bottom row stays the same. Okay, so now let's go ahead and do another row operation. So this row operation will help us knock out this turn right here to make it a zero. So to do that, we want to take our one and said it equal to three are one plus. Negative or two not will. Give us this Major X right here. We'll get zero six Negative. 13 this rural states. Same. And the last room will also save them. Okay, I know it will Dio is we'll just go ahead and swap these two roads so it's closer to reduce social reform. So 30 for three 06 negative. 13 00 negative. A squared plus two and a minus two. Okay, so we want no free variable. So what that means is we want to make sure that this entry is not equal to zero. Okay, So in order to that, we set a squared plus two and not the equal to zero We saw for a squared. When we get that a and not be equal to the spirit of two, and if that condition holds, that'll make sure that there is exactly one solution and this is sort of operation. Okay, So onto the last word, I don't want to find value for a for which there are infinitely many solutions. And this means that the reduce national inform has at least one free variable so greater than or equal to 13 variables. So by the previous implication that we've already done a few times, I'll just right are reflex of broad matrix. So 30 for three 06 negative. 1300 negative. A squared was two and a minus two. And we just got that matrix from right here. That's where we got it from. Okay, So if we want at least one free variable, that means that we want toe Look at this road right here. And we want to make sure that this is equal to zero and that this is equal to zero. So in order to do that, we sent negative a squared close to you. Well, zero. And we saw for that. And we get that, um well, so yeah, we want to sell for that. And then next up, we won't solve Bernie minus two is equal to zero. So we saw this really quickly. We know that we want eight equal to and so we're gonna plug this, eh Into this equation will see we have negative a squared plus two equals zero That gives this negative two squared was too equal to zero. And then this gives us negative four plus two has to be equal to zero. And then we send finest and see that negative too has toe equal to zero. But we know that that's not true. So we know that this part must be wrong and that part's not true, which means that there are no values for an infinite solution.

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