In Exercises 3–5, solve each system of equations using matrices. ...

Question

In Exercises 3–5, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Accepted Answer

Welcome back. I'm so glad you're here. We've got a system of equations and we get to choose either Gaussian elimination or Gauss Jordan elimination to provide a solution to solve for x, y and Z. And so I'm going to show you Gaussian elimination with this example and that will look like in the end we're working towards getting a diagonal of one's starting in the upper left hand and then zeros below them and then just other numbers here, above the ones. And what this will tell us is our Z value on the third row. And we'll use back substitution to go back and solve for Y and then for X. So step one here, now that we know where we're going, we're going to rewrite our system of equations as a coefficient matrix. So the first row, two for X, one for y, one for Z and six for that constant term three for X. One for Y. And then very careful note, there's a zero for that Z value in the second row and then that equals three. The constant term X on the third row is two, Y is negative three and again Z is zero and the constant term is two. Typically you'll see a matrix written with these brackets and this line and I'm not going to write the brackets in line for everyone here just to save a little bit of space. So how do we go about doing this? Well, we want to start off by getting a one in the first column in the correct place. So that's up here where this too is we want to get a one there and then we're going to continue working on the column, getting the zeros in their places. And we use the road that has a one in it in order to get those zeros there. So first thing we're going to do step one for our Gaussian elimination, we're going to replace row one and we'll do so by Multiplying this first row by 1/2. So 1/2 times row one. Because we're replacing row one, we can just copy over rows two and three. So we've got 31032 negative 302. And now we'll start with that multiplication for row 11 half times two is one. That's the one we're looking for one half times one is one half, one half times one is one half still and one half times six is three. Okay, we've got the one in the desired place and now we're going to continue with that first column and get the zeros into place. So let's work on this three. That means first step to here, we're replacing row two And as I mentioned before, we're going to use the row that has the one in the desired position. So that's the first row here to get that three into a zero. So we're going to multiply row one by negative three -3 times row one and then add that to row two because we're replacing road to, we can copy over rose one and three. So we've got 11 half one half three and two negative 302. And now multiplying and adding negative three times one is negative three plus three equals zero. That we want negative three times one half is negative three halves plus one gives us negative one half, negative three times one half is again negative three halves. But this time times plus zero gives us negative three halves negative three times three is negative nine plus three, gives us negative six. And we've done it. We've got our zero in that desired spot. Now let's continue with the first column and get a zero to replace this too. So for step three we're replacing row three and we do that by working with the row that has the one in the desired spot. Roll one. We're going to multiply that times negative two. So negative two times row one and add that to row three. Because we're replacing row three we can copy over rows one and two. So we've got 11 half one half three zero negative one half, negative three halves and negative six. And now let's do the multiplication. In addition negative two times one is negative two plus zero. Is the zero that we want negative two times one half is negative two halves or negative one plus negative three gives us negative four negative two times one half again is negative one plus zero gives us negative one negative two times three is negative six plus two, gives us negative four. And we've got the zero in the spot that we need. So now we're going to look to the second column and we want to get the one in the desired spot. So that's in the second row. First step four, we're replacing row two and we'll do that by multiplying Row two by -2. So -2 times row two. Because we are replacing road too, we can copy over rose one and 311 half, one half three separate this a little bit And -4 -1 -4. And now do that, multiplication negative two times zero is zero negative two times negative one half is one negative two times negative three halves is three negative two times negative six is 12. We've got the one in the spot that we need it. And now we continue with that column and what do we need? We need a zero below the one and we achieve that zero by working with the row that has the one in the spot. We need it. So that's the second row and we're going to multiply Row two x 4. So for step five we're replacing row three and we do so by multiplying row two times four and then adding that to row three because we're replacing row three. We can copy over rows one and two. So we've got 11 half, one half 3013, 12. Now we'll do the multiplication. In addition four times zero is zero plus zero is 04 times one is four plus negative four is the zero that we want. Four times three is 12 plus negative one, gives us 11 and four times 12 is 48 plus negative four, gives us 44. We're close guys. This is almost in the desired diagonal of ones. Now, we've just got to get this 11 into a one. We do that by working with this row and we're going to multiply row three times 1/11. So for step six, we're working with row three, replacing row three by multiplying row three times 1/11. Because we're replacing row three, we can copy over rows one and 211 half. One half 3013 and 12. And now 1/11 time 00 1/11 times zero is still 01, 11 times 11 is one and 1/11 times 44 is four. And that is the desired Gaussian elimination and now we can use substitution to solve for X, Y and Z. What we've got here is a system of equations of coefficients. So we can rewrite this as one, X. So I'm just going to write X plus one half Y plus one half, Z equals three. And I'm just stealing this from this final matrix, we had the next row. Well there's zero excess. We've got one. Y. So Y plus three, Z equals 12. And for the third row we've got Z equals four. We've got our Z value already Z equals four. And now we can solve for why in row two by substituting Of four and for the Z. So rewriting wrote to, we've got Y plus three times four equals 12. Or in other words why? Plus 12 equals 12, solving for Y. We've got Y equals zero. Got two of them. Now let's solve for X. Back in that first row, we've got X plus one half of zero plus one half of four equals three. Once this first row in our system of equations. And so we've got X plus will one half times zero is zero and one half of four is two. So X plus two equals three. We want to get X by itself. So we've got X equals three minus two, which is one, so X equals one, Y equals zero and Z equals four. We scroll back a little to look at all of our answer choices and this matches with answer choice. C Well done, we'll catch you on the next one

In Exercises 3–5, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Key Concepts

Matrices

Gaussian Elimination

Back-Substitution

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