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In Exercises $7-12$ , describe all solutions of $A \mathbf{x}=0$ in parametric vector form, where $A$ is row equivalent to the given matrix. $$\left[\begin{array}{cccc}{1} & {3} & {-3} & {7} \\ {0} & {1} & {-4} & {5}\end{array}\right]$$

$\mathbf{x}=\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}}\end{array}\right]=x_{3}\left[\begin{array}{c}{-9} \\ {4} \\ {1} \\ {0}\end{array}\right]+x_{4}\left[\begin{array}{r}{8} \\ {-5} \\ {0} \\ {1}\end{array}\right]$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 5

Solution Sets of Linear Systems

Introduction to Matrices

Missouri State University

McMaster University

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In mathematics, the absolu…

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in this example, we have a matrix equation, A X equals zero and we have some work that has already been done on the coefficient matrix. Say it's Rory equivalent to the following matrix we see displayed here using this information, provided we're going to completely solve this system. Well, that means we also have to write to the Matrix A and Roll reduced echelon form. So let's look at our pivots for this pivot column. We have a zero below, and so we're all set. But in this pivot column, we have one row operation to make, which is going to be to eliminate that three you see here. So let's start off by reading the following The Matrix A, which is Roy Clint to the one displayed, will also be Royko Flint to the following. I'm going to copy Row 201 negative 45 and if we multiply? Wrote to buy negative three Added to Row one. We obtain one than zero nine and a negative eight. Let's say what this tells us so far. Recall we're taking a X equals zero vector. We never augmented this matrix here. It's the coefficient matrix, but still we can imagine that the zero vector would be there and no one then tells us that x one plus nine times x three minus eight x four equals zero. Row two tells us that x two minus four x three plus five x four Looks like I forgot my variable X here on the x four. And so we have this equation set to zero as well. Now notice that the variables X three and X four are both free variables. I don't describe this solution set. First, we're going to say that the solution set is a vector X, which is equal to x one x two x three x four and we can do the following next. If we go to the equations that we have displayed here, will be able to solve for X one by subtracting nine x three and adding eight x four. Then do the same for X two for X to weaken solve in this equation for x two. By doing the following at four x three to both sides of that equation and subtract five x four from both sides. Now x three and x four are free and that what that means is that X three is just equaled. X three saw a copy of here. Likewise, X four equals x four. I don't copy it here for the sake of organization. Now we want to write this in Parametric Vector form and this turns out to be a technique that is very valuable towards other ideas in linear algebra. So we're going to work this out step by step. First we put down the scaler x three and X for these air the unknowns. But they're considered parameters now and these parameters multiply. Plus, there particular called vectors. How we find the column vectors is once again a similar theme in linear algebra. We just take coefficients. So start here for the extra variable that coefficients are negative. 941 and zero Because we have it missing down here. Then for the x four variable, go and pull these coefficients, which are eight negative five than a zero and a one. Then finally, this is the Parametric vector form for this equation. We should say one extra thing, however, just to remind herself that x three and X four parameters will say x three x four are in our So as we vary those two parameters over all real numbers, then this is the prayer metric vector form for the complete solution set.

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