Find matrix determinant

Question

How do I reduce this matrix to row echelon form and hence find the determinant, or is there a way that I am unaware of that finds the determinant of this matrix without having to reduce it row echelon form given this is all I know and there exists no additional information.

$\left[ \begin{array}{ccc} 1+x & 2 & 3 & 4 \\ 1 & 2+x & 3 & 4 \\ 1 & 2 & 3+x & 4 \\ 1 & 2 & 3 & 4+x \\ \end{array} \right]$

Answer 1

Let $A$ be the matrix in your question, then $x$ is an eigenvalue since $A-xI$ is clearly singular and the dimension of its corresponding eigenspace is $3$, hence $\lambda_{1} = \lambda_{2} = \lambda_{3} = x$. The trace is equal to the sum of its eigenvalues, so $\lambda_{4} = x+10$. Finally, the determinant is equal to the product of the eigenvalues so $\operatorname{det}(A) = x^{4}+10x^{3}$.

Answer 2

Subtract $2\times$ first column from the second column, $3\times$ first column from the third and $4\times$ first column from the fourth column.

You get

$$\left|\begin{matrix}1+x & -2x & -3x & -4x\\ 1 & x & 0 & 0\\ 1 & 0 & x & 0\\ 1 & 0 & 0 & x\end{matrix}\right|$$

Answer 3

Assume $f(x)=\Delta$. It's a fourth degree polynomial.

$C_1\to C_1+C_2+C_3+C_4\implies x+10 $ is a factor.

$f(0)=0\implies 0$ is a root.// Repeated Rows

$f'(0)=0+0+0+0\implies 0$ is again repeated.// Repeated Rows

$f''(0)=0\implies 0$ is again repeated.// Repeated Rows for last time.

This all $\implies f(x)=x^3(x+10)$. Its obvious that coefficient of $x^4$ must be $1$.

For differentiation, differentiate one row and treat others as constant(like product rule). Add them all. It's pretty easy to see that they have same rows(without even writing them).

Answer 4

If you know how value of determinant is influenced by elementary row/column operations (see ProofWiki) then you could start by adding all other columns to the last one (which does not change the determinant) and the rest is relatively easy:

$$ \begin{vmatrix} 1+x & 2 & 3 & 4 \\ 1 & 2+x & 3 & 4 \\ 1 & 2 & 3+x & 4 \\ 1 & 2 & 3 & 4+x \end{vmatrix}= \begin{vmatrix} 1+x & 2 & 3 & 10+x \\ 1 & 2+x & 3 & 10+x \\ 1 & 2 & 3+x & 10+x \\ 1 & 2 & 3 & 10+x \end{vmatrix}= (x+10) \begin{vmatrix} 1+x & 2 & 3 & 1 \\ 1 & 2+x & 3 & 1 \\ 1 & 2 & 3+x & 1 \\ 1 & 2 & 3 & 1 \end{vmatrix}= (x+10) \begin{vmatrix} x & 0 & 0 & 1 \\ 0 & x & 0 & 1 \\ 0 & 0 & x & 1 \\ 0 & 0 & 0 & 1 \end{vmatrix}=(x+10)x^3 $$