# Invertibility of a matrix.

Indicate why if a is a square matrix such that in each row and in each column one and only one element is non-zero then a is an invertible matrix.

I tried to encompass the problem by determinants, taking it to a form in which the main diagonal is zeros and the rest we all numbers are different from zero, but I can not find how to explain that the determinant of that matrix is ​​different from zero.

• What happens if you put it in row-echelon form? – saulspatz Feb 3 '19 at 1:57
• The columns are linearly independent (to wit, a reordering of the standard basis of $\Bbb R^n$). – Ivo Terek Feb 3 '19 at 1:58
• I can't see what happen when i put in the row-echelon form because i think there we already suppose some stuffs that we have to proof – user641010 Feb 3 '19 at 2:05
• Of all the terms in the determinant, exactly one is nonzero. – hardmath Feb 3 '19 at 2:06
• Reorder the rows (or columns if you prefer) so that your matrix is diagonal. – Robert Israel Feb 3 '19 at 2:16

We need to show that any column vector is not in the span of the others (i.e. they're linearly independent). Take one such column vector and suppose its nonzero element is in the $$i$$th row. Then in this row the other vectors have all $$0$$ entries by the condition, so their span contains of vectors with $$0$$ entries in the $$i$$th row too. Clearly, this column vector, having a nonzero element in that row, is not in the subspace.
Say your matrix looks like $$\begin{pmatrix} 0 & a & 0 \\ b & 0 & 0 \\ 0 & 0& c \end{pmatrix}$$. Then, you can simply rearrange the columns so that it looks like $$\begin{pmatrix} a& 0 & 0 \\ 0 & b & 0 \\ 0 & 0& c \end{pmatrix}$$. Once it is in this form, you should be able to tell that all of the columns are linearly independent. Thus, the matrix is invertible. (Note: It is important to assume that the matrix is square.)