In Exercises 1 - 24, use Gaussian Elimination to find the comp...

Question

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. $\begin{cases}5x + 8y - 6z = 14 \3x + 4y - 2z = 8 \x + 2y - 2z = 3\end{cases}$

Accepted Answer

welcome back. I am so glad you're here. We are given the system of equations and we are asked to solve for it using Gaussian elimination. So we recall from previous lessons that when we have a system of equations, we can rewrite it as a matrix. So we can take the coefficients in the constant terms from each equation and put those as the rows of our matrix. So the first equation becomes our first row. So we will have 10, 16 negative 12, 28. And then the second equation becomes our second row of our matrix. So that will be nine, 12 -6, 12. And the third row will be 24 negative 46. And there we have our matrix and now we can remember what we've learned about Gaussian elimination in previous lessons and what we want is to get this matrix having a diagonal row of ones. So the first column will be one and then zeros below it. The second column will be something and then a one and a zero below it. The third column will be something, something and then a one, C, r diagonal row of ones. And then the fourth column will be something something. And this third one will be our Z value. And once we've solved for Z we can use substitution to go back in and solve for Y. And then solve for X. Alright, so getting our matrix to have that diagonal row of ones. We want to start in the upper left hand corner. We want to turn this tin into a one. So we'll be replacing row one. And how do we do that? There are multiple ways but I am going to take row one and subtract row two. And that will replace our row one. Because we are replacing row one. We can copy down Rose two and three. So that will be 9 12 negative 6, 12 to 4, negative 46. And I'm not going to copy all the brackets because that will save a little bit of space and time now for row one minus road too. So that will be 10 minus nine. Which is the one that we want. 16 minus 12 is four negative 12 minus negative six is negative 12 plus six, which is negative six. And then 28 minus 12 is 16. Alright, We have our replace row one. And now we want to work on the column. We want to get a zero below that one. We'll be replacing row two. In order to do that in order to get a zero there. We're going to work with the road that has the one in the desired spot which is row one. We're going to take row one multiply by negative nine. And then we will add that to row two because we're replacing row two. We can copy over rose one and three. So 14 negative six negative 6 16 and negative 46. And then row two will be negative nine times one is negative nine plus nine is zero, negative nine times four is negative 36 plus 12 is going to be a negative 24 negative nine times negative six is a positive 54 minus six is a positive and negative nine times 16 is negative 144 plus 12 gives us a negative 132. Okay, we've got our zero in that position and now we want to get a zero in the last position of column one. We will be replacing row three and in order to do that in order to get the zero there, we're going to work with row one that has the one in the spot And we are going to take row one, multiplied by -2. And we are going to add that to row three because we're replacing row three, we can copy down rows one and two. So that will be 14, negative 6, 16 0, negative 24 48 negative 132. And now taking row one times negative two and adding that to row three, negative two times one is negative two plus two is the zero that we want negative two times four is negative eight plus four is negative four, negative two times negative six is positive 12 minus four is a positive eight and negative two times 16 is negative, 32 plus six is negative 26. Okay, now our first column is complete and we want to get a one in our desired position in the second column, which is where this negative 24 is. So we're going to replace row two. And in order to do that, we are going to multiply everything in row two times the reciprocal of that number and it needs to be positive. So it'll be the negative reciprocal. So negative one 24th times row two. And because we're replacing road to, we can copy down rose one in 314 negative 6, 16 0, negative 48, negative 26. And multiplying road two times negative one, 24th, zero times negative one, 24th is zero, negative 24 times negative one, 24th is the one that we want, 48 times negative one. 24th is going to be a negative two And -132 times negative one, is going to be 11/2. Okay, now we need a zero in the second column, third row, where that negative four is, we're going to replace row three And in order to do that in order to get a zero, there we're going to work with the road that has the one in the desired position which is road to and will multiply row two times four And then add that to row three. Because we are replacing row three, we can copy down rows one and negative 6, 16 01 negative 2, 11 halves. Now, doing the work to replace row 34 times zero is zero plus zero is 04 times one is four plus negative four is the zero that we want, four times negative two is negative eight plus eight. Well that is 04 times halves is 22 plus A negative 26 is going to be a negative four. Well, interesting. Look at this last row we recall from previous lessons that this corresponds with an equation in our system of equations. This is zero, X is plus zero wise plus zero. Zs equals negative four. And there's just no situation in which that is possibly true because this is impossible. Then our answer is going to be that there is no solution which matches with answer choice D. Well done. We'll catch you on the next one.

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. $\begin{cases}5x + 8y - 6z = 14 \\3x + 4y - 2z = 8 \\x + 2y - 2z = 3\end{cases}$

Key Concepts

Systems of Linear Equations

Gaussian Elimination

Row Operations and Row-Echelon Form

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