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Repeat Exercise 5 for a nonzero $3 \times 2$ matrix.

Thus, the possible echelon forms of a non-zero $3 \times 2$ matrix are:$\left[\begin{array}{cc}{\mathbf{}} & {*} \\ {0} & {0} \\ {0} & {0}\end{array}\right],\left[\begin{array}{ll}{0} & {\mathbf{}} \\ {0} & {0} \\ {0} & {0}\end{array}\right]$ and $\left[\begin{array}{ll}{\mathbf{}} & {*} \\ {0} & {\mathbf{}} \\ {0} & {0}\end{array}\right]$

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 were asked to solve the giving when your system, by using gassy and elimination So Gaussian elimination asks us to put the Matrix and Groeschel on form before we begin to solver are variable. So the first step is to write the Matrix from the equations that a given to our first information reads to X one was two x two was too extreme is equal to zero. The 2nd 1 reads negative to x one plus five. Next to was to x three is equal to one. The last one reads a That's one, but sex to a spore at score or sorry for X three is equal to negative lunch. So now we want to simplify this nature. So our first step is going to be to take our two. It's an unequal to our one plus are too, and that'll make this term equal to zero, which will help us get it and the desired for So this matrix becomes the following. Their next up is going to be to try to get this to be zero. So in order to do that, we need to take our three. That unequal to four are one minus R three. When we do that, we get this resulting major now to some funds even further. We sent our three set it equal to our three miners are too. And this gives us this major. Now we're lost up is we want to make sure that the leading entries are all one. So to do this, we need to take our one that is equal to 1/2 are one and said or two and equal that unequal toe 17 are too. Once we do that, we get this nature. Okay, so now we're ready to start solving for variable so we'll go ahead and copy dot matrix over here. And the first thing we want to note is, if we have any free variables, We didn't see that x one Russell leading entry X two as a leading entry, but x three doesn't. So we know that our through our free variable is x three. So we want to solve for X one and x two in terms of x three. So let's go ahead and start doing back. So, for that's two, our second row reads next to us or seven x three is equal to 17 since we want to solve for X two in terms of X three, you get X two is equal to 17 minus 47 x three. Now I want to solve for X one. So to do that, we'll go ahead and take a look at our first road here. So we have X one plus x two plus X three is equal to zero. We want to solve for X alone in terms of X Therese. Oh, but we're gonna want to do is you substitution to get rid of this x two term here. So we're gonna substitute this value for X stew for this value back suit. So that gives us X one +17 minus +47 x three plus X three is equal to zero. And this sense of simplifying to X one is equal to negative one over seven minus 11 7 x three and now we're ready to go ahead and right. Our final solution. So x one and X two is equal to our constant. So x one has ah, 1/7 term. X two has a what's in the term plus X three shows are free variable and ex three's coefficient for X one is negative, 11 7th and for ex to its negative forces, and that is our final solution.

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