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Determine if the systems in Exercises 15 and 16 are consistent. Do not completely solve the systems.$$\begin{aligned} x_{1} &-2 x_{4}=-3 \\ 2 x_{2}+2 x_{3} &=0 \\ x_{3}+3 x_{4} &=1 \\-2 x_{1}+3 x_{2}+2 x_{3}+x_{4} &=5 \end{aligned}$$

Consistent.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 1

Systems of Linear Equations

Introduction to Matrices

Missouri State University

Baylor University

Lectures

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and this problem were given this system and were asked to determine if it's consistent, which by definition, means whether have city shits up as a first step ever in the system as this matrix right here now to the terminal that has a solution set, we're not gonna have to fully solve the Matrix. We're just gonna have to prove that it could be Britain introduced to root for. What that means is that these numbers over here going on diagonal are gonna correspond to our solution set and all the other numbers, with the exception of last column, of course are gonna be going to zero. So in this matrix, we can see that these three variables already equal to zero and where we're gonna try to do is ah, half this negative to this three. And this too, turned two zeros as well. So, for the negative, too, we can add two times were one 24 to get a new road for of zero three to negative three negative one. Now, for the three, we can add negative three over two times wrote to to row for to get a new roof or of 00 negative. One negative three. Negative one. Lastly, for the negative one, we can look at our row three right here and simply add a row three chiro for and we could see that that's gonna give us a new role for of simply all zeros. So now we have a new matrix of our original 1st 3 rows and then a new row, four of all zeros. And this right here is enough to prove that the system can be written in reduced reform. Because we have these zeroes over here in a triangular form and these numbers over here on the diagonal that are going to give us our solution set. Um, what having the zeroes in a triangle proves is that these numbers over here can also be turned into zero. And now this is gonna be our solution for X one. This is gonna be your social Rex, too x three and X four. So, again, the system is consistent

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