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Linear Algebra (finding values of a constant k to give no/unique/infinite solutions) 
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#1
Jan2709, 05:21 AM

P: 49

First time poster. Have attempted the problem, but keep hitting dead ends and have no idea how to proceed.
1. The problem statement, all variables and given/known data Determine the values of k for which the system of linear equations has (i) no solution vector, (ii) a unique solution vector, (iii) more than one solution vector (x,y,z): kx + y + z = 1 x + ky + z = 1 x + y + kz = 1 2. The attempt at a solution The only approach I can think of is reducing it to row echelon form, but given the k's I'm finding it impossible to do so. Putting it in augmented matrix form and reducing it as best I can, I get: [k 1 1  1] [0 (k1) (1k)  0] [1 1 k  1] The above is getting me nowhere. I've tried taking out factors of k, but I'm not sure if that's the way to go because the end result is overly complicated. Any help? 


#2
Jan2709, 03:50 PM

P: 1,294

You need to reduce it before you can do anything and yes that can be messy.
Once, you have reasonably reduced refer to when a matrix has no solution, unique solution, or infinite. 


#3
Feb109, 05:29 AM

P: 49

I've been trying that but I just can't seem to reduce to anything doable. I've got it down to row echelon form, but I'm left with the bottom line of [0 0 k^2+2k+3k3}
I don't know if I've done something wrong, but with that last line I just can't solve it. Can someone help me out here or at least give me a push in the right direction? This question is doing my head in... 


#4
Feb109, 03:41 PM

P: 126

Linear Algebra (finding values of a constant k to give no/unique/infinite solutions)
Do the determinant of the coefficient matrix, and you get a polynomial. Case by case, replace in the original matrix with the roots of the polynomial and solve. You'll get 0 or infinite solutions depending on the root.



#5
Feb209, 06:08 AM

P: 49

Thanks mate, worked great.
If you don't mind me asking, what's the theory behind the method? Also, would there be a way to extend it to find a unique solution (in an example which has one), or would row reduction be possible in such a situation? 


#6
Feb209, 09:42 AM

P: 126

Let's call A the coefficient matrix of a system of linear equations.
If det(A) = 0, that means at least one of the rows is a linear combination of the other rows. That's why we look for the roots in the determinant. Now one row is linear combination of two others, for example, row 3 = row 1 + row 2 (for the coefficients). Now if the constants in row 1 and 2 add to the constant in row 3, some variables are free (infinite solutions). If the constants in row 1 and 2 don't add to the constant in row 3, there's a contradiction and there's no solution. Now if AX = C (where X, C are vectors ; X is the unknown vector and C is the constant vector) and det(A) is not 0...  you can solve the system with Gauss triangularization  you can find A^{1} (since det(A) is not 0) and X = A^{1}C  you can use Cramer's rule to do it only with determinants Unless you're solving a system where many coefficients are symbols, the best is Gauss. Cramer's rule would be useful if you had to solve something like: ax + 2by + 3az = 1 2ax + ay + 5cz = d 2x + 3cy + 5bz = 2d Because you can do it with Gauss, but... ahem... x = (5ab  15 c^{2}  6a^{2}d  10b^{2}d + 9acd + 20bc d)/(6 aLet's call A the coefficient matrix of a system of linear equations. If det(A) = 0, that means at least one of the rows is a linear combination of the other rows. That's why we look for the roots in the determinant. Now one row is linear combination of two others, for example, row 3 = row 1 + row 2 (for the coefficients). Now if the constants in row 1 and 2 add to the constant in row 3, some variables are free (infinite solutions). If the constants in row 1 and 2 don't add to the constant in row 3, there's a contradiction and there's no solution. Now if AX = C (where X, C are vectors ; X is the unknown vector and C is the constant vector) and det(A) is not 0...  you can solve the system with Gauss triangularization  you can find A^{1} (since det(A) is not 0) and X = A^{1}C  you can use Cramer's rule to do it only with determinants Unless you're solving a system where many coefficients are symbols, the best is Gauss. Cramer's rule would be useful if you had to solve something like: ax + 2by + 3az = 1 2ax + ay + 5cz = d 2x + 3cy + 5bz = 2d Because you can do it with Gauss, but if det(A) is not 0... ahem... [tex]x = \frac{5 a b  15 c^2  6 a^2 d  10 b^2 d + 9 a c d + 20 b c d}{6 a^2 + 5 a^2 b  20 a b^2 + 18 a^2 c + 20 b c  15 a c^2}[/tex] [tex]y = \frac{10 a b + 10 c  6 a d + 12 a^2 d + 5 a b d  10 a c d}{6 a^2 + 5 a^2 b  20 a b^2 + 18 a^2 c + 20 b c  15 a c^2}[/tex] [tex]z = \frac{2 a + 6 a c + 2 a^2 d + 4 b d  8 a b d  3 a c d}{6 a^2 + 5 a^2 b  20 a b^2 + 18 a^2 c + 20 b c  15 a c^2}[/tex] So it would be kind of easy to get lost in the process. PS: I had this semester's linear algebra exam today... it went damn well :) 


#7
Feb209, 09:07 PM

P: 49

Thanks heaps for that springo.



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