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. $\begin{cases}3x + 4y + 2z = 3 \4x - 2y - 8z = -4 \x + y - z = 3\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given a system of equations and asked to solve it using Gaussian elimination. So we recall from previous lessons that when we have a system of equations, we can create a matrix for that. And the matrix is going to be the coefficients and constant terms of our system of equations. So the first row of our matrix will be the first equations, coefficients of four, negative two, -2, and then the constant zero. And then our second row will be our second equations coefficients in constant terms. So 363 and nine. And our third row will be 9 12 6, 24. So now we've got a matrix set up and for Gaussian elimination we want to have the matrix looking something like this. We want diagonal ones starting in the upper left hand corner that will have zeros below it? In that column? The second column will have something and then there's that one and a zero below it. The third column will have something and something and then R one and then the fourth column will be something, something. And then our Z solution and once we know what our Z solution is, we can go back and use substitution to solve for R. Y. And then our X. So let's get started on looking having this matrix look like the diagonal ones. We'll start with this one in the upper left hand corner. So we're going to replace row one. How are we going to do that? How do we get this to be a one? We will multiply everything in row one times 1/4. So we'll take 1/4 times row one. Because we're replacing row one. We can copy down Rose two and three. So let's do that 36399, 12, 6, 24. And now multiplying row one times 1/4 4 times 1/ is the one that we want -2 times 1 4th is negative 1/2 negative two times 1/4 is still negative one half and zero times 1/4 is zero. Okay, we've got the first one done now we want a zero below it in column one. So we need to turn this three into a zero. We're going to replace row to how do we get that? Three into a zero. We're going to do some work on the road that has the one in the desired position for our column. So that's row one and we're going to take row one in this case and multiplied by negative three. So negative three times, row one and then add that to row two. Because we're replacing road to, we can copy down Rose one and three. So we've got one negative one half negative one, half zero and row 39 6 24. Now doing the work for row two negative three times one is negative three plus three is the zero that we want negative three times negative one half is a positive three halves positive three halves plus six is 15 halves negative three times negative one half is positive three halves positive three halves plus three is positive nine halves negative three times zero is zero plus nine is nine. Alright, we've got 10 in place and now we want to finish off the column and get a zero where that nine is. So we're replacing row three. We do that by doing some work on the road that has the one in the desired spot. So that's rho one in this case we'll multiply row one times negative nine And then add that to row three. Because we're replacing row three, we can copy down rows one and 2. So that will be one negative one half negative one half, zero and 0, 15 halves, nine halves and nine. So many fractions. Okay, now replacing row three negative nine times one is negative nine plus nine is the zero that we want, negative nine times negative one half is positive. Nine halves plus 12 is halves negative nine times negative. One half is positive. Nine halves plus six is 21 halves and negative nine times 00 plus 24 is 24. Okay, that first column is done. Now we want a one in this second row, second column position. So we're replacing row two in order to get that 15 halves to be one. We're going to take row two and multiply it by its reciprocal. So multiply everything in their times to 15th. So, so to 15th times, row two. Because we're replacing row two, we can copy over rose one and three. So one negative one half negative one half zero and 0. 33 halves 21 halves and 24. Alright. Doing the work for row 20 times to 15 0, 15 halves times to 15th is the one that we want. A nine halves times to 15th. The choose cancel out. So that's 9/15 and that's going to simplify the 3/ and nine times to 15th. That's 18, 15th, which simplifies to 6/5. Alright, we've got the one, therefore, column two. Now let's get this zero in the second column, third row position. We're going to replace row three. We're going to do that by doing some work on the road that has the one in the desired spot. That's row two, we'll take row two And we'll multiply that by -33/2 and we'll take that negative 33 halves times, row two and then add that to row three scooch ng down a bit because we're replacing row three, we can copy down rows one and two. So we've got one, negative one half, negative one, half zero and row 201, 3/5 6, 5th. And now doing the work for row three negative 33 halves, times zero is zero plus zero is zero, negative. 33 halves times one is negative, 33 halves plus positive 33 33 halves is the zero that we want there negative 33 halves times 3/5 is negative 99 10th plus 21 halves is 3/ and taking negative 33 halves times 6/5 is negative 99 5th plus 24 gives us 21 fifths. Okay, now we are almost there. We just need to get a one in this third row, third column position. So to replace row three, we want to one there. We're going to multiply everything by that. Reciprocal. So multiplying everything by five thirds five thirds times row three. Because we're replacing row three, we can copy down rows one and two. So one negative one half negative one half, 001, 3/5 6 5th. And now four row 30 times five thirds is zero and again 0 3/5 times five thirds is the one that we want and 21 5th times five thirds. Well, the fives cancel out, that's 21 3rd not simplifies to seven. All right now we know that this is a matrix with the corresponding coefficients and constant terms for our system of equations. So we can rewrite this as one, x minus one half, y minus one half Z equals zero for our first equation and our second equation is the second row. There are zero access. So we can just start with y plus 3/ Z equals 6/5. And our third equation the third row, there are no excess and no wise. So it's just Z and there's just one Z. So Z equals seven. And there we go. We got our Z. So we have solved for a Z. I'll write that up here so I can copy that back up in a few minutes when we solved for Y and X. So now in our second equation where we have a Z, we can substitute a seven. So rewriting equation two, we've got Y plus 3/5 not times Z. But now times seven equals 6/5. And doing the multiplication 3/5 times seven is Y plus 21 5th equals 6/5. And now we can subtract 21 5th from both sides. And we get why equals 6/ minus 21 5th is negative three. So we've solved for Y. And the and now we can use substitution into equation one, we know Ry and rz and we can solve for X. So our first equation we have x minus one half not times Y but now times negative three minus one half, not times Z. But times seven I wanted to switch colors there seven and that equals zero. Doing the multiplication, we've got negative one half times negative three is going to be x plus three halves and then negative one half, times seven is going to be minus seven halves equals zero. Well positive three halves minus seven halves is going to give us a minus four halves equals zero. Well negative four halves, that's negative two. So we've got x minus two equals zero and we can add two to both sides and we get X equals two. So we now have our solutions and we can drag these back up to the top so we can see which one this matches with will be a mess for a second. I do I get back up and putting this where we've got space. Okay, now looking at our answer choices for X equals to Y equals negative three and Z equals seven. We match with answer choice A 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. $\begin{cases}3x + 4y + 2z = 3 \\4x - 2y - 8z = -4 \\x + y - z = 3\end{cases}$

Key Concepts

Systems of Linear Equations

Gaussian Elimination

Row Operations and Consistency

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