Math Is Fun Forum

EPhillips1989

Member

Registered: 2007-11-03

Posts: 29

matrices

let A be a nxn matrix with a row of zeros can anyone prove that A is not invertible

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: matrices

do you have any theorems at your disposal? There should be a quicker way to prove it.

Anyway, here is one way to do it. Suppose There is some matrix B such that AB = I, by definition of matrix multiplication, we have that

[i]ij = [AB]ij = ∑ [A]ik[b]kj from k = 1 to n. Now suppose this matrix A has a row of zeros in the m'th collum, the identity matrix would have a 1 at the m'th row and the m'th collumn, so we have that

1 = [i]mm = [AB]mm = ∑ [A]mk[b]km  from k = 1 to n

thus the summation has a total value of 1. But note for all k, we have that [A]mk = 0 (because this exists in the mth row and thus has a value of zero), so we have this summation is the same as
∑ 0*[b]km from k = 1 to n, = ∑ 0 from k = 1 to n = 0. Thus this entry cannot possible be the (m,m)th entry of the identity matrix, therefore the matrix A cannot have an inverse.

Last edited by mikau (2007-11-23 10:30:41)

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: matrices

The determinant of the matrix is 0 and so it is not invertible.
(Assuming you're allowed to use that implication)

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