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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 $3 \times 2$ matrix with two pivot positions.

a.) Nob.) No

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 5

Solution Sets of Linear Systems

Introduction to Matrices

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in this example, We're dealing with the Matrix A. We're told it's of size three by two and contains to pivot positions. Let's start out this problem by indicating what the format of a would look like using just generic elements. So I have three rows. I'm going to use a solid bullet to represent an arbitrary real number. We have two columns salt Stop here and here's our three by two matrix. We have to pivot positions, and that means, at best will have a pivot here and a pivot here for two rows, which are Pivot rose. Next, let's consider the system or matrix equation. A X equals the zero vector. So this is, ah, homogeneous matrix equation, and we know X equals does. Your vector itself is a solution, So this is consistent because it has the trivial solution here. The question then becomes is there were Let's Fraser of slightly more dramatically are there nontrivial solutions to determine if there are nontrivial solutions? We go to the columns. We know that column one is a pivot column, as is calling to Let's write down here, both columns of a our pivot cones. This is a very important observation in the sense that now that we know that both columns are pivot columns, we can conclude immediately that there are no free variables. So if there are no free variables, we know that the system A X equals zero will have a unique solution. But here's one solution. Already the trivial solution. This then tells us there are no nontrivial solutions that's phrasing it with a double negative. So if that makes you uncomfortable, you could say there is only the trivial solution. Let's consider a different system. Next with the same matrix a size three by two with two pivot positions, consider the system eight times X equals some arbitrary vector B B is unspecified. And so for this particular system, we want to know, Is this always consistent? So is this system always consistent, always being a very important word here, well, to determine if a system is consistent or always consistent in this context. Now we turn to the rose as opposed to the columns when we're considering the nontrivial solution. So going row wise, no one is a pivot row. Row two is also a pivot row, but Row three is not so Row three has no pivot, so we can come down here and say, Not all Rose of a Our Pivot Rose. This gives us a set of logically equivalent statements. Now we know now that the columns of a cannot span all of our three or three comes from this dimension we know no. All vectors be in. Our three can be written as a linear combination of a but these air, just different ways to rephrase what we're really trying to get at. Since not all rows are pivot rose, we can now conclude. Therefore, a X equals B is inconsistent for some vector be in our three. It's not saying it's always inconsistent. For example, of B is a zero vector. We certainly have a solution that trivial one. And for non trip for non zero vectors be there's likely solutions. What it is saying is we can absolutely find a particular vector B to make the system inconsistent

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