In Exercises 21–38, solve each system of equations using matrices...

Question

In Exercises 21–38, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

System of equations for exercise 29 in college algebra, chapter on matrices.

Accepted Answer

Welcome back. I am so glad you're here. We are given the system of equations and we're asked to solve for it. Well, typically we have an X value, a Y value of Z value and that's equal to some constant term. But if you notice the first two equations here, the equal sign is floating in the middle and not right before the constant term. So we need to do a little bit of work on these first two equations to get them so that they have the X, Y and Z on the same side on the left hand side. So for the first equation we can subtract two Z from both sides and then this will read two, X plus y minus two, Z equals negative 12. And then the second term we can subtract two Y from both sides and we can add Z to both sides. And then that will read five x minus two, Y plus Z equals negative 15. And the third one's fine. Two, x minus y plus eight, Z equals 24. And now what we can do to solve for this. So we recall from previous lessons that we can create a matrix using the coefficients and constant terms from this system of equations. So we can take the first equation and that becomes the first row of our matrix using the coefficients negative two in the constant term negative 12. And then the second equation becomes the second row, five negative 21 negative 15. And the third equation the third row two negative 18, 24. And then we are told we can use either Gaussian elimination or Gauss Jordan elimination, I'm going to demonstrate Gauss Jordan elimination. And the way that works is we're going to do a series of transformations and we're going to get this matrix transformed so that the first three columns have a diagonal of ones starting in the upper left hand corner and are surrounded by zeros. So the first column will be 100, the second will be 010, and the third will be 001. And then the fourth column, that's where the money is. In this case, it will be our X, y and Z solutions. So what we need to do for our series of transformations is start with column one. And we're going to start by getting this one in position. So we need to turn this two into a one. How do we do that? We're going to take row one and we are going to multiply it by the reciprocal of that target number. So we're going to multiply it by one half in this case. And because we're replacing row one, we can copy down rose two and three. So that will be -2, 1 -15, two negative 18 24. And now doing the work to replace row 12 times one half is 11 times one half is one half negative two times one half is negative one and negative 12 times one half is negative six. Okay, we've got that one in position and now we use that one to get zeros in the rest of column one. So we want this five in row two to become a zero. And we use the one with the one, the row with the one in the desired position. So that's row one to get the zero there in this case will multiply row one times negative five and then add that to row two. Because we're replacing road too, we can copy down rose one and 311 half negative one, negative six and two negative 24. And now doing the work to replace road to we're going to take row one, times negative five and add it to row two negative five times one is negative five plus five is the zero that we want. Ye negative five times one half is negative five halves plus negative two is negative nine halves negative five times negative one is positive five plus one is a positive six and negative five times negative six is positive, 30 plus negative 15 is positive 15. Okay, one more position in column one we need a zero in row three. Where this too is and again we're going to use the row that has the one in the desired position. That's row one and in this case we'll multiply it by negative two And add that to row three because we're replacing row three. We can copy down rows one and half negative one negative six and zero negative nine halves 15. Now replacing row three we're going to take negative two times row one. So negative two times one is negative two plus two is the zero that we want negative two times one half is negative one plus negative one is negative two negative two times negative one is positive two plus eight is 10 and negative two times negative six is a positive 12 plus 24 is 36. Okay, column one is done. Yay. Now we want to get the one in the desired position for column two and that's where this not negative nine halves is in road to in order to get that negative nine halves to become a one. We're going to multiply by its negative reciprocal by negative 2/9. Because we're replacing road to we can copy over rose one and 311 half negative one negative 60 negative 2 36. Now replacing road to negative 2/9 times zero is zero negative nine halves, times negative 2/9 is the one that we want. Ye six times negative 2/9 is negative four thirds And 15 times negative 2/9 is negative 10/3. Okay now we need zeros in the rest of the positions of column two. So this one half in row one is our next target. And we're going to use the road that has the one in the desired position. That's row two and in this case we'll multiply row two times negative one half And add that to row one. Because we're replacing row one, we can copy rows two and three. So that will be 01, negative four thirds and negative 13th and zero negative 2, 10 36. Now replacing row one negative one half times zero is zero plus one is still one negative one half times one is negative one half plus one half is the zero that we want ye negative one half times negative four thirds is two thirds plus negative one is negative one third and negative one half times negative 13th is five thirds plus negative six is negative 13 3rd. Okay, one more zero to go in column two. We want to replace row three where that negative two is we're going to use row to to help us get a zero there. In this case we'll multiply row two times two and then add that to row three because we're replacing row three, we can copy over rows one and 210 negative one third negative 13 3rd and 01 negative four thirds and negative 13th. And now replacing Row three, zero times two is zero plus zero is 01 times two is two plus negative two is the zero that we want negative four thirds times two is negative eight thirds plus 10 is 3rd and negative 13th times two is negative, 23rd plus 36 is 88 3rd. All right. One column to go. We've got this now. We want a one in the desired position of column three. That's where this 22 3rd is. So we'll be replacing row three again and we'll be multiplying by the reciprocal of that targeted number. So 3/ Times Row three. We can copy over rows one and 210 negative one third negative 13 3rd and 01 negative four thirds negative 13th. And now replacing row 30 times 3. 20 seconds is zero. Both times 22 3rd times 3, 20 seconds is one year and 88 3rd times 3. 20 seconds is four. What a nice number. And now we need to just get zeros in the rest of the positions for column three. Let's work on row one. We're going to take the road that has the one in the desired position. That's row three. And in this case we'll multiply row three times one third And add that to row one. We can copy down rose two and three. So that will be 01, negative four thirds negative 13th and 0014. Now doing the work to replace row 11, 3rd time 00 plus one is 11, 3rd time 00 plus zero is 01 3rd, times one is one third plus negative one third is the zero that we want and one third times four is four thirds plus negative 13 3rd is negative three. All right. One more. We've got this last spot in row to where we want zero. We'll use roe three to get it will multiply row three times four thirds And add that to row two. So replacing row two, we can copy over Rose one and 3100, negative three and 0014. Now replacing row 24 3rd times 00 plus 004 3rd times zero is zero plus one is 14 3rd times one is four thirds plus negative. Four thirds is the zero that we want. Ye and four thirds times four is 16 3rd plus negative. 13th is to look at this. We have got in the first three columns are diagonal of ones. And then that fourth column corresponds with in this case are x, y and Z solutions. So our solutions set is native 3, 2 and four scroll up just a little bit and we look at our answer choices and this matches with answer choice. D. Well done. We'll catch you on the next one

In Exercises 21–38, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Key Concepts

Systems of Equations

Matrices

Gaussian Elimination

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