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September 27th, 2013, 04:56 AM   #1
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Linear systems help

Hello, I need to solve 2 tasks and I would like to get some help. I prefer Gaussian elimination.
In the first (shown second), I have got this result:
x1 - 2*x2 + x3 + x4 =a
-2x4 = b - a
and c=5a
but how can I continue?

In the second task (shown first), I got 2 equal lines:
a*x2 + (a+6)*x3 = 2
I'm not really experienced in matrices and linear equation systems yet (nor in English math ).
Attached Images
 mat1.png (4.8 KB, 142 views) mat2.png (4.4 KB, 142 views)

 September 27th, 2013, 08:36 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2228 In the task shown first, you shouldn't get equal lines (unless a = 2). There is no solution if a = -3. Can you post your attempts, so that we can identify any slips you have made?
 September 27th, 2013, 09:54 AM #3 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 Re: Linear systems help The problem is I don't know how to write matrices and other mathematical characters, but are you sure that you watched the mat1 attachment? Because the first picture is the second task.
 September 28th, 2013, 02:22 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2228 I referred to the task shown first, but named mat2.png. If you explain your work in words and using x1, x2, etc., I can improve the presentation for you and also identify any slips.
September 28th, 2013, 04:27 AM   #5
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Re: Linear systems help

I made a matrix (A1) from that equation system (mat2.png). The forum has some problems, so I made pictures

A1                   A2                   A3
1 1 -1 |1        1   1   -1 |1       1  1   -1 |1
2 3  a |3   ~   0   1 a+2 |1   ~   0 a a+6 |2
1 a  3 |2        0 a-1   4 |1        0 a a+6 |2

1st step (on A1 matrix)
3rd line - 1st line
2nd line - 2*(1st.line)

2nd step (on A2 matrix)
2nd line + 3rd line
3rd line + 2nd line

With these steps, I got the A3 matrix
and from that you can make equations:
1. x1  +  x2 -          x3 = 1
2.       a*x2 + (a+6)*x3 = 2
3.       a*x2 + (a+6)*x3 = 2

On the mat1.png, I made these steps: (starting with A1 matrix, what I made from the equations)

1 -2  1  1 |a
A1 = 1 -2  3 -4 |b
1 -2  1  5 |c

1 -2  1   1 |a
A2 = 2 -4   2  0 |b+a
-4  8 -4  0 |c-5a

1  -2  1  1 |a
A3 = 2 -4  2  0 |b+a
0   0  0  0 |c+2b-3a

1 -2  1   1  |a
A4 = 0  0  0 -2x4|b-a
0  0  0   0  |c+2b-3a

1st step (on A1)
2nd + 1st
3rd - 5*1st

2nd step (on A2)
3rd + 2*2nd

3rd step
2nd - 2*1st
and I got A4,
from that you can make these equations:
1. x1 - 2*x2 + x3 + x4 = a
2. -2*x4 = b-a
3. 0 = c + 2b - 3a
Attached Images
 A1 A2 A3.png (4.0 KB, 103 views) A1 A2 A3 A4.png (7.6 KB, 103 views)

 September 28th, 2013, 06:55 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2228 In your second step (for system in mat2.png), you make the second and third rows the same as you simultaneously do the same calculation to obtain the new values. You needed to produce another zero in the 3rd row. If you instead add (a-1)*2nd row to (-1)*3rd row, you get    A1                   A2                   A3 1 1 -1 |1        1  1   -1 |1       1  1      -1    |1 2 3  a |3   ~   0  1 a+2 |1   ~   0  1   a+2    |1 1 a  3 |2        0 a-1  4 |1        0  0 a²+a-6 |a-2 The third row now tells you that (a²+a-6)x3 = a-2, i.e., (a+3)(a-2)x3 = a-2. If a = -3, this equation has no solution. If a = 2, x3 can have any value, and the values of x1 and x2 can easily be found in terms of x3. For all other values of a, x3 = 1/(a + 3). Do you understand that? You should now be able to finish solving this system. In your working for the system shown in mat1.png, you have changed the second equation. Which version is correct?
 September 28th, 2013, 07:38 AM #7 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 Re: Linear systems help You mean the -2*x4? Sorry, it's simply -2 in the matrix, but the equation should be correct. The problem is I don't know what solution should I give? Because we have x1,x2,x3,x4 and a,b,c (parameters) as unknowns. How should the solution look like? we need to express the x's with the parameters? I will look after the mat2 task a bit later.
 September 28th, 2013, 02:32 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2228 I didn't mean the -2*x4. The original second equation has coefficients that do not match the second row of the matrix used in your posted attempt.
 September 29th, 2013, 04:02 AM #9 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 Re: Linear systems help Oh I see, the correct matrix is 1 -2 1  1 |a 1 -2 1 -1 |b 1 -2 1  5 |c 1step : 2nd-1st and 3rd-2nd 2.step : 3rd-3*2nd and I got 1 -2  1  1 |a 0  0  0 -2 |b-a 0  0  0  0 |c+2b-3a is this correct? I can express only x4 with b and a and I know that 0=c+2b-3a but what can I do with x1,x2,x3?
 September 29th, 2013, 06:34 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2228 In your second step, you should have put "+" instead of "-". Your final values are correct. You got 0 = c + 2b - 3a. If a, b and c do not satisfy that equation, the original equations are inconsistent and so have no solution. Assuming 0 = c + 2b - 3a, you can get x4 = (a - b)/2. Hence x1 - 2x2 + x3 = a - x4 = (a + b)/2. That means that x1, x2 and x3 can have any values that satisfy that equation. You can choose x1 and x2 arbitrarily, then use the equation to express x3 in terms of x1, x2, a and b.

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