Determinant of matrix that contains a column of same numbers

Question

Given $4\times 4$ matrix:

$$\begin{vmatrix} x & x+a& 0& 1\\ x+a& x& a& 1\\ 0& a& x& 1\\ 1& -1& 1& 1 \end{vmatrix}=x+a$$

Is it safe to assume that the determinant equals $0$ because there is a column than contains the same numbers(most right) and using Gaussian elimination technique I can turn it into zeroes and then the determinant is $0$. Is it correct?

Answer 1

No, it is not correct, unless that “same number” is $0$. For instance$$\begin{vmatrix}1&0&0&\ldots&0&a\\0&1&0&\ldots&0&a\\0&0&1&\ldots&0&a\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&a\\0&0&0&\cdots&0&a\end{vmatrix}=a.$$

Answer 2

Fix $a$ and let $p$ be the value of the determinant which is a polynomial in $x$. Note that $p(0) = p(-a) = 0$ and so $x(x+a)$ must divide $p$. Also note that $p$ has degree 2.

A little more grind shows that $p(a)=-2(1+a)^2$ from which we conclude that $p(x) = -2(1+a)x(x+a)$.

Alternatively, let WA do the work.

Answer 3

No. Try to think about what happens if you were to carry out the Gaussian elimination you propose. You can turn all but one of the $1$s into a $0$, but your proposed last step wouldn't work -- you can't subtract a row from itself when doing Gaussian elimination.