# Linear Algebra (finding values of a constant k to give no/unique/infinite solutions)

by rafehi
Tags: algebra, constant, linear, solutions, values
 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-1C - 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 c2 - 6a2d - 10b2d + 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-1C - 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... $$x = \frac{5 a b - 15 c^2 - 6 a^2 d - 10 b^2 d + 9 a c d + 20 b c d}{-6 a^2 + 5 a^2 b - 20 a b^2 + 18 a^2 c + 20 b c - 15 a c^2}$$ $$y = \frac{-10 a b + 10 c - 6 a d + 12 a^2 d + 5 a b d - 10 a c d}{-6 a^2 + 5 a^2 b - 20 a b^2 + 18 a^2 c + 20 b c - 15 a c^2}$$ $$z = \frac{-2 a + 6 a c + 2 a^2 d + 4 b d - 8 a b d - 3 a c d}{-6 a^2 + 5 a^2 b - 20 a b^2 + 18 a^2 c + 20 b c - 15 a c^2}$$ 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 :)