Solving variables in a matrix for a specific determinant

Question

The matrix is as follows:

$$ A = \begin{pmatrix} 0 & x & 1 & 2 \\ x & 1 & 1 & x \\ 1 & x & x & 1 \\ 1 & x & 1 & x \end{pmatrix} $$

What I want to do is to find all the solutions for the equation: $$\det(A) = 0$$

At first I attempted to simplify it into a polynomial, but ending up with a 4th degree term makes me wonder if there's any easier way of solving this? You can easily see that the rows/columns would be linearly independent if $x$ is equal to $1$. But I'm having a hard time realizing any other solutions this way.

Have you guys got any idea? Any help would be much appreciated!

Answer 1

$$A = \begin{vmatrix} 0 & x & 1 & 2 \\ x & 1 & 1 & x \\ 1 & x & x & 1 \\ 1 & x & 1 & x \end{vmatrix}\stackrel{R_2-xR_3\,,R_4-R_3}=\begin{vmatrix} 0 & x & 1 & 2 \\ 0 & 1-x^2 & 1-x^2 & 0 \\ 1 & x & x & 1 \\ 0 & 0 & 1-x & x-1 \end{vmatrix}= $$

$$\begin{vmatrix} x & 1 & 2 \\ 1-x^2 & 1-x^2 & 0 \\ 0 & 1-x & x-1 \end{vmatrix}=(1-x^2)(1-x)\begin{vmatrix} x & 1 & 2 \\ 1 & 1 & 0 \\ 0 & 1 & \!\!-1 \end{vmatrix}=(1-x)^2(1+x)\left[-x+2+1\right]=$$

$$=-(x-1)^2(x+1)(x-3)$$

Answer 2

By the row operations $l_2\leftarrow l_2-xl_3$ and $l_4\leftarrow l_4-l_3$ and developing along the first column we get

$$\Delta=\det\begin{bmatrix}x&1&2\\1-x^2 &1-x^2&0\\0&1-x&x-1\end{bmatrix}$$ Now $c_2\leftarrow c_2+c_3$ and we develop along the third column we get $$\Delta=(x-1)(x(1-x^2)-3(1-x^2)=-(x-1)^2(x-3)(x+1)$$

Answer 3

The most straightforward idea would be to write down your polynomial, and divide it by $x-1$ to obtain a third-degree polynomial with roots wchich are easy to guess.

Another idea would be to perform some vector/column operation on the matrix to simplify the expression for determinant from the very beginning:

1. subtract the third line from the fourth one

2. subtract the first column from the fourth one

3. subtract the second column from the third one

4. subtract the fourth line from the first one

Now you should obtain a rather sparce matrix, and its determinant should be easily found a factorised into a product of monimials.

If you still need help, ask in comments.