How to prove surjectivity of determinant

Question

It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?

Answer 1

Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.

Answer 2

Hint:

Given $r\in \Bbb R $, take the diagonal matrix $\begin{bmatrix}r&0&\cdots & 0\\ 0&1&\cdots & 0\\ \vdots&\vdots&\ddots&\vdots\\0 & 0 & \cdots &1\end{bmatrix} $.

Answer 3

Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)

So, given $r\in\Bbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $\frac rd$, to get a matrix with determinant $r$.