I was reviewing this test and came over this question:
How many distinct solutions $x$ are there for the equation det($A-xI$) $=0$ where det stands for determinant, $I$ is the identity matrix, and $A$ is the matrix
$$ \begin{bmatrix} 1 & 0 & -3 & 0 \\ 0 & 2 & 0 & 1 \\ 1 & 0 & 3 & 0 \\ 1 & -2 & 0 & 1 \\ \end{bmatrix} $$ It looks really computational (4 x 4 matrix determinant), so I am asking you guys if there is an elegant way to solve this. Thanks in advance.
To evaluate $\left|\begin{array}{cccc} 1-x & 0 & -3 & 0\\ 0 & 2-x & 0 & 1\\ 1 & 0 & 3-x & 0\\ 1 & -2 & 0 & 1-x \end{array}\right|$, identify a row or column with as few nonzero entries as possible, e.g. the rightmost column. The determinant can be written as $\left|\begin{array}{ccc} 1-x & 0 & -3\\ 1 & 0 & 3-x\\ 1 & -2 & 0 \end{array}\right|+\left(1-x\right)\left|\begin{array}{ccc} 1-x & 0 & -3\\ 0 & 2-x & 0\\ 1 & 0 & 3-x \end{array}\right|$. Now let's repeat: $$\left|\begin{array}{cc} 0 & -3\\ 0 & 3-x \end{array}\right|+2\left|\begin{array}{cc} 1-x & -3\\ 1 & 3-x \end{array}\right|+\left(1-x\right)\left(\left|\begin{array}{cc} 0 & -3\\ 2-x & 0 \end{array}\right|+\left(3-x\right)\left|\begin{array}{cc} 1-x & 0\\ 0 & 2-x \end{array}\right|\right)=2((1-x)(3-x)+3)+(1-x)(2-x)(3+(3-x)(1-x)).$$You'll notice one quadratic obviously divides that; the whole quartic simplifies to $(4-3x+x^2)(6-4x+x^2)$. You can now find all eigenvalues from the quadratic formula; there are four roots in $\Bbb C$< none of them real because $$a=1,\,b=-4,\,c=6\implies b^2-4ac<0;\,a=1,\,b=-3,\,c=4\implies b^2-4ac<0.$$