invertible matrices?

Question

how do i prove that a square matrix with two identical rows is not invertible?

Answer 1

We first deal with 2x2 matrices. Let the matrix be written as

|a, b|
|a, b|,

where a, b ∈ R, the real numbers (or C, the complex numbers, if we're dealing with that field). An invertible matrix has non-zero determinant. But the determinant of the matrix given above is

ab - ab = 0.

Hence it cannot be invertible.

Now we introduce an induction hypothesis, whereby all n x n matrices with two identical rows are not invertible. Let M be an (n + 1) x (n + 1) matrix and let the k-th row be one of the rows that is not identical to another (if one cannot pick one, then the case is trivial); observe that its determinant is calculated as

Σ [(-1)^(i + k + 2)] * m(1, i) * det(M(k, i))

the sum being taken from i = 1, ..., (n + 1), where m(k, i) is the i-th element of the k-th row, M(k, i) is the n x n matrix obtained by eliminating the k-th row and the i-th column from M, and det() denotes the determinant of a matrix. However, since each M(k, i) has two identical rows, we know det(M(k, i)) = 0 for i = 1, ..., (n + 1). Thus the determinant of M is zero, and so it's not invertible.

Hence, by induction, the result follows.

Answer 2

right here is an hassle-free thank you to coach that no longer each 2 invertible matrices upload as much as a diverse. Take any invertible matrix A. Now, the destructive of A is yet another invertible matrix. The sum A + (-A) is the empty matrix, that's singular. QED.

Answer 3

If a square matrix, A, has 2 identical rows, its
determinant is 0.
But A^-1 = A_adj/ det(A),
so A^-1 doesn't exist.

Answer 4

a square (n x n) matrix with 2 identical rows is
equivalent to an (n-1 x n) matrix...does this give ya a hint?
:)

Answer 5

det = 0 so not invertible

suppose its n*n .. you have at most n-1 basis vectors locked into the matrix, and there is another whole dimension that is projected onto the n-1 dimensional plane ... if you tried to invert it ... you are projecting from n-1 dimensions to n dimensions and you haven't got enough information for a one-one mapping.