The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 × 2 matrix…

Let $A=\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right] \Rightarrow \left|A\right|=3a-a^2-bc$

$A^3=\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+bc& ab+3b-ab\\ ac+3c-ac& cb+(3-a{)}^2\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+bc& 3b\\ 3c& cb+(3-a{)}^2\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]$

$tr\left(A^3\right)=a^3+abc+3bc+3bc+3bc-abc+(3-a{)}^3=-18$

$a^3+9bc+{\left(3-a\right)}^3=-18$

$9bc+27-9a(3-a)=-18$

$bc+3-3a+a^2=-2$

$3a-a^2-bc=5 \Rightarrow \left|A\right|=5$