# Prove that the determinant of a matrix is zero

Hi I need some help with this question:

Let $$A$$ be an $$n \times n$$ matrix, let $$i, j, k$$ be pairwise distinct indices, $$1 \leq i, j, k \leq n$$, and let $$\lambda,\mu \in \mathbb R$$ be arbitrary real numbers. Suppose that $$a_k$$, the $$k-$$th row vector of $$A$$, is equal to $$\lambda a_i + \mu a_j$$, where $$a_i, a_j ∈ \mathbb R^n$$ denote the $$i-$$th and the $$j-$$th row vectors of $$A$$ respectively. Prove that $$\det(A) = 0$$.

I think I need to split the matrix up into two separate ones then use the fact that one of these matrices has either a row of zeros or a row is a multiple of another then use $$\det(AB)=\det(A)\det(B)$$ to show one of these matrices has a determinant of zero so the whole thing has a determinant of zero. So I was wondering is there a way to split these matrices up so it suits my method?

Hint: The demerminant of a non-invertible matrix remains invariant under elementary row operations. Performing such row operations can you create a row with all components equal to zero? What would it mean for the determinant of a matrix, if there is a row only with zeros?

• Determinant does not remain invariant under row operations. – Artem Nov 14 '16 at 15:18
• Yh only when you swap rows but dw I've done it thanks. – Thomas Nov 14 '16 at 15:28
• It is Determinant – El borito Jan 6 at 6:18