Determinant of an Invertible MatrixDate: 08/04/99 at 03:37:14 From: Mike Smith Subject: Determinants I've been having problems understanding what determinants have to do with matrix invertibility. Could you give me an explanation as well as a proof of why a matrix is invertible if and only if the determinant is non-zero? I would really like to know. Mike Smith. Date: 08/04/99 at 07:54:11 From: Doctor Floor Subject: Re: Determinants Dear Mike, Thanks for your question: Let us consider the following general nxn matrix: [a11 a12 ... a1n] M = [a21 a22 ... a2n] [ : : : ] [an1 an2 ... ann] We define the cofactor of a11, call it c11, by the determinant of the matrix found when we leave row 1 and column 1 out of M, so: |a22 ... a2n| c11 = | : : | |an2 ... ann| In almost the same way, the cofactor cpq of apq is found as the determinant of the matrix when we leave row p and column q out of M. Now we have to use the rule of signs: If p+q is even then cpq = determinant, if p+q is odd then cpq = -determinant. Doing this we find that: ap1*cp1 + ap2*cp2 + ap3*cp3 + ... + apn*cpn = det(M) .....[1] for any 1 <= p <= n. Also, we find that: aq1*cp1 + aq2*cp2 + aq3*cp3 + ... + aqn*cpn = 0 ..........[2] when 1 <= p <= n and 1 <= q <= n, while p and q are unequal. Expression [2] is the determinant of the matrix you get when in matrix M you replace row p by row q. The matrix then has two equal rows, so clearly its determinant must be 0. We find the inverse of M as: [c11/det(M) ... cn1/det(M)] M^(-1) = [c12/det(M) ... cn2/det(M)] [ : : ] [c1n/det(M) ... cnn/det(M)] because now M*M^(-1) = I (the nxn identity matrix) (check this). This computation of the inverse of M clearly can't work when det(M) = 0. The fact that a matrix with determinant 0 really can't have an inverse can be seen from the following: We know that, when det(M) = 0, then row n is a linear combination of rows 1, 2, 3, ..., n-1. Now suppose that matrix N is the inverse of matrix M, so M*N = I (the identity matrix). Note that the nth row of M*N is now a linear combination of the rows 1, 2, 3, ..., n-1 of M*N too. The elements in the nth column of rows 1, 2, 3, ..., n-1 of I are all zero, so that means that the nth element of the nth row of M*N should be zero, too, being a linear combination of zeroes. But the nth element of the nth row of I is one, and we have a contradiction. I hope this cleared things up. If you have more questions, don't hesitate to send them to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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