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Using Cramer's rule to solve linear equations with complex coefficients?

  1. Dec 24, 2006 #1

    I got down Cramer's rule down fine, now I need to extend it to include equations that have complex coefficients. Do I let each matrix entry be something like, "5 + 2i" or is there something more than that?

    For example,

    say we have

    (2+3i)x + (5+3i)y + (9-6i)z = 10 + i
    (4+3i)x + (5-3i)y + (9-6i)z = 5 + i
    (6+2i)x + (4+3i)y + (5+6i)z = 10 + 2i

    Perhaps I let each entry be just the coefficient infront of x, y and z?
  2. jcsd
  3. Dec 25, 2006 #2


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    You got it: it's that simple. Just put complex numbers as elements of the determiniants you calculate for Cramer's rule.
  4. Dec 27, 2006 #3


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    here is cramers rule: if A is a square matrix with entries in any commutative ring, not necesarily a field, and if adj(A) is its "adjoint matrix" (whose (j,i) entry is (-1)^(i+j) times the determinant of the matrix obtained from A by deleting the ith row and jth column of A),
    then adj(A).A = A.adj(A) = d.I, where d is he determinant of A, and I is the identity matrix.

    as a corollary, if d is invertible, then a solution of the vector equation Av=w, is v = d^(-1)adj.A.w.

    for example the entries can be complex numbers. or polynomials with entries in a field. or integers.
  5. Dec 28, 2006 #4


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    as a corollary, although not every equation can be solved, you can always solve, if not for a v that gives you w, at least for a v that gives you dw, where d = detA.

    e.g. if the 2x2 matrix has rows [ 2 3], [3, 1], then d = 2.1 - 3.3 = -7.

    thus adjA is the matrix with rows [ 1 -3], [ -3 2]. then A.adjA = [-7 0], [0 -7]. so given AX = w, if we set X = adjA.w, at least we get AX = dw.

    i.e. if w = [1 1], and we set X =[-2 -1], then A.w = [-7 -7], instead of [1 1].

    so at least you can solv e for some vector on the line joining w to the origin, although some times this only gives [0 0].
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