# Non singular matrix proof

• Apr 27th 2010, 09:27 PM
newtomath821
Non singular matrix proof
a matrix is non singular only if its det does not equal zero. Calculate its inverse.

How do I go about proving this? I can only think of a counter example where matrix is singular given identical rows or columns or multiples of each other, which will generate a det of 0.

What do you think?
• Apr 28th 2010, 02:11 AM
tonio
Quote:

Originally Posted by newtomath821
a matrix is non singular only if its det does not equal zero. Calculate its inverse.

How do I go about proving this? I can only think of a counter example where matrix is singular given identical rows or columns or multiples of each other, which will generate a det of 0.

What do you think?

It all depends on what you already know: we can say that $\det A=0\iff$ the characteristic polynomial of $A$ has free coefficient equal to zero $\iff$ zero is one of the eigenvalues of $A\iff\,A$ is singular.

Or we can say that $A$ is singular $\iff \ker A\neq \{0\}\iff \exists 0\neq v\,\,\,s.t.\,\,\,Av=0\iff$ zero is one of the eigenvalues of $A\iff A$ is singular.

Or we could even say $A$ is singular $\iff$ the rows (columns) of $A$ , when seen as vectors in $\mathbb{F}^n\,,\,\mathbb{F}=$ the definition field, are linearly dependent $\iff$ when bringing the matrix to echelon form (and thus at most multiplying the matrix's determinant by a non-zero factor) we get a row of zeros $\iff \det A=0$ .

Choose yours...(Nod)

Tonio