4. An algebra problem by Sompong Chuisurichy

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

Then we get,

$A^2 + 7A + 31I =\begin{pmatrix} a^2+7a+bc+31 & ab+bd+7b \\ ca+cd+7c & d^2+bc+7d+31 \end{pmatrix} =0$

so we have 4 equations,

$a^2+7a+bc+31=0 \cdots(1)$

$ab+bd+7b=0 \cdots(2)$

$ca+cd+7c=0 \cdots(3)$

$d^2+bc+7d+31=0 \cdots(4)$

Dividing eq (2) or eq (3) by $c$ we get,

$a+b+7=0 \cdots(5)$

Subtracting eq (4) from eq (1) we get,

$a^2-d^2+7a-7d=(a-d)(a+d+7)=0$

$\Rightarrow a-d=0$

$\Rightarrow a=d$

Substitute $a=d$ in eq (5) we get,

$a=d= - \frac{7}{2}$

Substitute value of $a$ in eq (1) we get,

$bc= - \frac{75}{4}$

Now,

$det(A+I)= (a+1)(d+1)-bc= \fbox{25}$

Characteristic polynomial of A = $x^2 + 7x + 31$ , to get the characteristic polynomial of $A+I$ we transform the roots by putting $y= x + 1$ so $x = y-1$ now the characteristic polynomial of $A+I$ as $(y-1)^2 + 7(y-1) + 31 = y^2 +5y + 25$ so $det(A + I ) = 25$

trick here is to use Cayley Hamilton theorem