Solve each system of equations using matrices. Use Gaussian el...

Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

$\begin{cases}2x - y - z = 4 \x + y - 5z = -4 \x - 2y = 4\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations and we're asked to solve for it. So we recall from previous lessons that when we have a system of equations, we can create a matrix out of the equations, coefficients and constant terms. So the first equation there is going to give us our first row of our matrix. So the first row will be the coefficients three negative 11 in the constant term negative three. The second row will be the second equation one negative three negative 13. And the third row, the third equation seven negative two, negative two negative three. And now we can choose to use either Gaussian or Gauss Jordan elimination. I'm going to demonstrate Gauss Jordan elimination here and we are to solve for the system of equations. So what we want to do is do a series of transformations to get this matrix looking like one that has a diagonal of one's starting in the upper left hand corner surrounded by zeros for the first three columns. So the first column will be one 00, the second will be 010, and the third will be 001. We have our diagonal ones. And then the fourth column, that's where it's at because the fourth column will give us our in this case x, y and z solutions. So it's just really doing that series of transformations. So where do we start? First thing we want to do is get a one in the first column. So that's where this three is. How do we turn that three into a one. Well we're going to take the first row And we are going to multiply by the reciprocal of that target number. So we're going to multiply row one by one third in this case. So because we're replacing row one, we can copy down Rose two and three. That will give us one negative three negative and seven negative two negative two negative three. And now performing the transformation for real one multiplying everything by one third, three times one third is one negative one times one third is negative one third one times 1 3rd is 1 3rd and negative three times one third is negative one. And now we just want to get Zeros and the rest of column one. And we're going to use the one we just got in order to get Zeros and the other two spaces. Let's start with this one in row two. Using the one from row one, we're going to multiply row one in this case by a negative one. And add that to row two. Because we're replacing row two we can copy down Rose one and negative one third one third negative one and seven negative two negative two negative three. And now doing the work to replace road to negative one, times one is negative one plus one is the zero that we want negative one times negative one third is positive one third plus negative three gives us a negative eight thirds negative one times one third is negative one third plus negative one gives us negative four thirds negative one times negative one is positive one plus three gives us four. All right now we want a zero where this seven is to finish off column one and again we're going to use the row that has the one in the desired position which is row one. This time we're going to multiply row one times negative seven And then we will add that to row three. Because we're replacing row three we can copy down rows one and two. So we'll have one negative one third, one third negative one and zero negative eight thirds negative four thirds and four. And now doing the work to replace row three negative seven times one is negative seven plus seven is the zero that we want negative seven times negative one third gives us seven thirds plus negative two gives us one third negative seven times one third is negative seven thirds plus negative two is negative 3rd negative seven times negative one is positive seven plus negative three is four. The first column is done. Yay, All right now we want to get the one in place for the second column. That's where this negative eight thirds is. So we'll be replacing road to excuse me not road three And in order to do that this time we're going to multiply row two times the negative reciprocal of that number times negative 3/8. So that will become a positive one because we're replacing road to we can copy over rose one and three one negative one third, one third, negative one and 01 3rd negative 13 3rd and four. Now replacing road to zero times negative 3/8 is still zero negative eight thirds times negative 3/8 is the one that we want negative four thirds times negative 3/8 is one half And four times negative 3/8 is simplifies to negative three haps. All right now we've got that one in place in column two and we can use it to get zeros in the other spots in column two. So now let's work on row one. Oh, excuse me, I didn't put the transformation that we did up here. We took native 3/8 times row two. Okay, now what we're doing for row one, we are going to use the road that has the one in the desired position. That's row to this time. And in this case we'll multiply row two times one third And we'll add that to row one. Because we're replacing row one, we can copy down rows two and 3. And so we'll have half negative three halves and 01 3rd negative 13 3rd and four. Now doing the work to replace row one, multiplying row two by one third and adding that to row 10 times one third is zero plus one is still one one third times one is one third plus negative one third is the zero that we want. One third times one half is 1/6 plus one third is one half and one third times negative three halves is negative one half plus negative one is negative three halves. All right. One more zero to get in place for column two. That's where we have this one third. So we'll replace row three and again we'll work with row two to do it. This time we will multiply row two times negative one third And add that to row three. We can copy over rows one and half negative three halves and one half negative three halves. Those are very similar. And now replacing row 30 times negative one third is zero plus 001 times negative one third is negative one third plus one third is the zero that we want. One half times negative one third is negative 1/6 plus negative 3rd is negative nine halves and negative three halves times negative one third is one half plus four, gives us nine halves. Alright, that's looking nice. Now one more column to go. We need to get the one in place for column three. That's where this negative nine halves is So we'll be replacing row three and in order to get that to become a one. We're going to multiply by its negative reciprocal so that it would become a positive one. So we're taking negative 2/9 times row three. We can copy over rows one and half negative three halves and 011 half negative three halves. So similar. And now doing the work to replace row 30 times negative 2/9 is zero both times and negative nine halves times negative 2/ is a positive one but positive nine halves times negative two. Nineths is a negative one And row three is done. Now we can use this one to help us get zeros above it in column three. But look these two columns for rows one and two are identical. So we can actually transform those one halves into zeros at the same time. We can work on both row one & Row two with the same operation Because Road to Not Roll one. If we take row three the one that has the one in the desired position and we multiply it by negative one half and we add that two in the first case row one and then the second case row two. That will turn both of those one halves into zeros. So let's do those both at the same time. We can copy over row 3001, negative one because that one's not changing and now doing the work to replace rows one and two. Well negative one half times zero is zero plus zero is going to remain a zero for these two cases and negative one half times zero is zero plus one is going to remain a one for these two. That's the only difference for rows one and two now negative one half times one is negative one half plus one half is the zeros that we want here and negative one half times negative one is positive one half plus negative three halves. Well that's going to be negative one in both cases. And now look at this, we have our matrix in the desired form where we've got for the first three columns are diagonal ones. And then we recall that our fourth column gives us our x, y and Z solutions. So the solution set for this one is negative one negative one negative one. We look at our answer choices and that matches with answer choice. A well done. We'll catch you on the next one.

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$\begin{cases}2x - y - z = 4 \\x + y - 5z = -4 \\x - 2y = 4\end{cases}$

Key Concepts

Systems of Linear Equations

Matrix Representation of Systems

Gaussian Elimination and Gauss-Jordan Elimination

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