Quadratic Equation with Matrices

All four combinations of signs for $\pm$ : $\left( \begin{array}{cc} \frac{1}{2} \left(-\sqrt{21}\pm 5\right) & 0 \\ 0 & \frac{1}{2} \left(-\sqrt{21}\pm 5\right) \\ \end{array} \right)$

The solution of: $x^2+5 x+1=0$ is $\frac{1}{2}(-\sqrt{21}\pm 5)$ .

With $m=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$ , solutions of $m.m+5 m+\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)=\left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)$ are:

$\left( \begin{array}{cc} a & b \\ \frac{-a^2-5 a-1}{b} & -a-5 \\ \end{array} \right)$

$\left( \begin{array}{cc} \frac{1}{2} \left(-\sqrt{21}\mp 5\right) & 0 \\ c & \frac{1}{2} \left(\sqrt{21}\pm 5\right) \\ \end{array} \right)$

and the previous four solutions. The solution immediately above is two solutions: chose either the top or bottom sign.

Let $M = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$ . Then $M^2 = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} = \begin{bmatrix} a^2 & 0 \\ 0 & d^2 \end{bmatrix}$ , and

$M^2 + 5M + I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

becomes

$\begin{bmatrix} a^2 & 0 \\ 0 & d^2 \end{bmatrix} + 5\begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

which leads to equations $a^2 + 5a + 1 = 0$ and $d^2 + 5d + 1 = 0$ , which solve to $a = \frac{-5 \pm \sqrt{21}}{2}$ and $d = \frac{-5 \pm \sqrt{21}}{2}$ .

Since there are $2$ options for both $a$ and $d$ , there are $2 \times 2 = \boxed{4}$ possible matrices that satisfy the given conditions:

$\begin{bmatrix} \frac{-5 + \sqrt{21}}{2} & 0 \\ 0 & \frac{-5 + \sqrt{21}}{2} \end{bmatrix}$ , $\begin{bmatrix} \frac{-5 + \sqrt{21}}{2} & 0 \\ 0 & \frac{-5 - \sqrt{21}}{2} \end{bmatrix}$ , $\begin{bmatrix} \frac{-5 - \sqrt{21}}{2} & 0 \\ 0 & \frac{-5 + \sqrt{21}}{2} \end{bmatrix}$ , and $\begin{bmatrix} \frac{-5 - \sqrt{21}}{2} & 0 \\ 0 & \frac{-5 - \sqrt{21}}{2} \end{bmatrix}$ .

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For the bonus, let $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ . Then $M^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & b(a + d) \\ c(a + d) & bc + d^2 \end{bmatrix}$ , and

$M^2 + 5M + I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

becomes

$\begin{bmatrix} a^2 + bc & b(a + d) \\ c(a + d) & bc + d^2 \end{bmatrix} + 5\begin{bmatrix} a & b \\ c & d \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

which leads to equations $a^2 + bc + 5a + 1 = 0$ , $b(a + d) + 5b = 0$ , $c(a + d) + 5c = 0$ , and $bc + d^2 + 5d + 1 = 0$ , which lead to multiple solutions for all non-zero entries as long as $d = -a - 5$ and $bc = -d^2 - 5d - 1$ , one solution being $M = \begin{bmatrix} -\frac{5}{2} & \frac{\sqrt{21}}{2} \\ \frac{\sqrt{21}}{2} & -\frac{5}{2} \end{bmatrix}$ .