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Question 01Solve the given equation by Gauss elimination method X1 + 2xz 3x3 =-1-3x1 + xz - 9x3 = -75 x1 + 3x24x3 = 2...

Question

Question 01Solve the given equation by Gauss elimination method X1 + 2xz 3x3 =-1-3x1 + xz - 9x3 = -75 x1 + 3x24x3 = 2

Question 01 Solve the given equation by Gauss elimination method X1 + 2xz 3x3 =-1 -3x1 + xz - 9x3 = -7 5 x1 + 3x2 4x3 = 2



Answers

Solve using Gauss-Jordan elimination. $$ \begin{aligned} 3 x_{1}+5 x_{2}-x_{3} &=-7 \\ x_{1}+x_{2}+x_{3} &=-1 \\ 2 x_{1} &+11 x_{3}=& 7 \end{aligned} $$

What's all this system of equations using the gallus Jordan elimination method? So let's get to this into an augmented matrix first with coefficients negative 2131 negative 4 to 1 negative 31 with the constant negative 701 And we want to reduce this matrix. So let's start by taking uh negative one half of row one so that we can get a one in our first element. So our new row one will be one negative one half, negative three halves, positive seven halves. Those two and 3 are unchanged. I actually want zeros below the one in column wand. So we can take road to my s row one. Put this in road to near row one and three are unchanged in our new road to. It's going to be zero, negative seven half seven has negative seven hands. Now let's get a zero in row three by just doing row three minus row one and putting that in row three. So rose one and two are unchanged in row three becomes zero, negative five halves, five halves, negative five halves. Okay, now we want to one in road to so we can do negative 2/7 times road to so we'll one unchanged And row three unchanged. And our new road to would be zero 1 -1 1. Now we want zeros above and below the one. And so what we could do is grow one plus half of grow too And put this in row one. So our new row one Will be 10 -2, 4 Rose two and 3 unchanged. Mhm. Okay, Now we want to zero below the one in road to actually row three And we can achieve this by taking row three And adding five half of the world too. In putting this in row three, So one and two unchanged in our new row three will be all zeroes. And so we are now in reduced form and we can interpret this x up one minus two times, except three equals four, except to minus except three equals one, and then this is zero equals zero. So this is a dependent solution. So we can define parameter since both of these equations involved except three, let's say except three is S an ex of one minus two. S. Is equal to foreign, except to minus S. Is equal to one, or accept one equals To s plus four And accept two equals s plus one. And so our solution would be to S plus four comma, S plus one. Comma S. Yeah.

Solve this system using the gastro in elimination method. So let's get this into an augmented matrix first with the coefficient to negative 13 to 1, negative one, augmented with a constant 07 negative one. And let's work on reducing this. So let's take capital grow one and we get one negative one half zero. Then let's work on getting zeros below it so we can take road to and subtract three times row one and put this in row one. So um excuse me wrote to. So our new road to will become 07 halves seven. Yeah, yeah. Now let's get a zero in row three, column one by just subtracting row three minus row one in putting this in row three. Some rose one mm hmm two unchanged in our new row three will be zero negative one half negative one. Now let's get a one in, wrote to column two by taking to seventh of road to. So rose one and three are unchanged and our new road to is going to be 01 into. Now let's work on getting zeros above and below. So let's take row one and add it to half of road to and put that in row one. So our new row one will be 101 throw two and three unchanged. And now as good as zero in row three column too, so it's below that one, and we can achieve this by taking a row three and adding to it half of the road to and putting this in row three. So our rows one and two unchanged, row three then just becomes all zeros and so were solved except one is one and except two is 212 is a solution. No.

We want to solve the system of equations. Even the Gauss Jordan elimination methods. So let's get to an augmented matrix first with coefficients two negative 13 to 1, negative one with the constant 07 negative two. We want to reduce this matrix. So let's get the element in row one, column one to be a one by taking half of world one. That would be one negative one, half zero. And then we'll work on getting zeros below the one. So let's get this 3 to 0 by taking road to and subtracting three times row one, I'm putting this in row two, So rose one and 3 unchanged. And our new road two is 07 half seven. And then we can get a zero in row three by just subtracting rope, row three minus row one And putting this in row three. And our new uh wow, There was one into unchanged And our new row three Will be 0 -1/2 negative too. Now that we've achieved this, let's get a one in. Um our second row, we can do this by taking 2/7 of road to come on was commissioned. Mhm. 2/7 of road to would be 01. Um To Yeah. Okay, now let's work on getting zeros above and below that one. So let's say in row one We can take no one and add half of row two And rose two and 3 unchanged in our new row one is 101. Now let's get a zero below the one by taking road to and um Well let's do it this way, Let's say we'll take row three and add to it, Half of row one have to grow to. I'm sorry. We're gonna put this in row three. So rose one and two unchanged. Our new road three will be 00 -1. What? And I will want to take um Yeah the opposite of row ones of it are row three. Yeah So that it's a positive one and now we not need to get zeros above the one. So let's say row 1 -83 and put this in row one. Yeah So those two and three unchanging a new row one is 100. And when you have one more you need to get rid of that too. And so we can take road to And subtract two times row three and put that in row two. Thank you. So Rose one and three unchanged in our new road to Louis 010. We're in reduced form. We interpret this X. Of one is zero X. Of 20 and zero equals one. Which means this is inconsistent. There is no solution to this system. Okay.

He's using the journal elimination method. We need to get this into an argument from matrix with coefficients to negative five, negative three, negative 4, 10 to 6 negative 15, negative one with 76 negative 19. And we're gonna get this in reduced row form. So let's take one half of row one and put that into our new row one. Which will be uh one negative five halves, negative three halves, seven halves. Yeah, Rose two and 3 unchanged. Then let's get Zeros below the one. So Let's put let's take road to and add four times. Row one. Uh huh. And we're gonna put this into row two by the way. And so row two will become 00 negative for 20. Rose one In three are unchanged. Now, I want to get a zero in row three. So let's take row three and subtract six times, grow one and put that in row three. So our new row three Will be 008, negative 40 Rose one into unchanged. No, we want to get a one in road to. So let's let's end we want it to be positive. So let's scale take negative 1/4 of road to. So we'll have a new road two of 001, negative five rose three and one unchanged. And now we want to Get a zero above and below the one. So let's take row one and add 3/2 of road to and put that in row one. So we'll have a new road one of one, negative five have zero negative four rows two and three are unchanged. Um How you one or 0 below the one. And so let's take row three and subtract eight times row two and put that in row three. So our new row three will be all zeros. The road to in one are unchanged. So now we're in a reduced form which would mean X one minus five halves. X two is equal to negative four. X sub three Is equal to -5. So let's let except to be T. Then we can write Our solution as except one equals positive Vibe has T -4, or we can write it like five halves, t minus four comma t comma negative five. Yeah. Mhm.


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