Find matrix determinant 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.
$\left[
  \begin{array}{ccc}
   1+x  & 2 & 3 & 4 \\
    1 & 2+x & 3 & 4 \\
    1 & 2 & 3+x & 4 \\
    1 & 2 & 3 & 4+x \\
  \end{array}
\right]$
 A: 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 $\operatorname{det}(A) = x^{4}+10x^{3}$.
A: 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
$$\left|\begin{matrix}1+x & -2x & -3x & -4x\\
1 & x & 0 & 0\\
1 & 0 & x & 0\\
1 & 0 & 0 & x\end{matrix}\right|$$
A: 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).
A: 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:
$$
\begin{vmatrix}
   1+x  & 2 & 3 & 4 \\
    1 & 2+x & 3 & 4 \\
    1 & 2 & 3+x & 4 \\
    1 & 2 & 3 & 4+x
\end{vmatrix}=
\begin{vmatrix}
   1+x  & 2 & 3 & 10+x \\
    1 & 2+x & 3 & 10+x \\
    1 & 2 & 3+x & 10+x \\
    1 & 2 & 3 & 10+x
\end{vmatrix}=
(x+10)
\begin{vmatrix}
   1+x  & 2 & 3 & 1 \\
    1 & 2+x & 3 & 1 \\
    1 & 2 & 3+x & 1 \\
    1 & 2 & 3 & 1
\end{vmatrix}=
(x+10)
\begin{vmatrix}
    x & 0 & 0 & 1 \\
    0 & x & 0 & 1 \\
    0 & 0 & x & 1 \\
    0 & 0 & 0 & 1
\end{vmatrix}=(x+10)x^3
$$
