# Thread: Non singular matrix proof

1. ## 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?

2. 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 $\displaystyle \det A=0\iff$ the characteristic polynomial of $\displaystyle A$ has free coefficient equal to zero $\displaystyle \iff$ zero is one of the eigenvalues of $\displaystyle A\iff\,A$ is singular.

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

Or we could even say $\displaystyle A$ is singular $\displaystyle \iff$ the rows (columns) of $\displaystyle A$ , when seen as vectors in $\displaystyle \mathbb{F}^n\,,\,\mathbb{F}=$ the definition field, are linearly dependent $\displaystyle \iff$ 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 $\displaystyle \iff \det A=0$ .

Choose yours...

Tonio