Question

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2\times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is $-18,$ then the value of the determinant of $A$ is

Answer 1

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\cdots \left(1\right)$
Given $Tr\left(A\right)=3$
$\Rightarrow a+d=3$
$\Rightarrow d=3-a$
Putting in equation $\left(1\right),$
$A=\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]$

Also, $Tr{\left(A\right)}^3=-18$
$A^3=\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]$
$\Rightarrow A^3=\left[\begin{array}{cc}{a}^2+bc& 3b\\ 3c& cb+{(3-a)}^2\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]$
$\Rightarrow A^3=\left[\begin{array}{cc}{a}^3+abc+3bc& a^{2}b+b^{2}c+9b-3ab\\ 3ac+c^{2}b+c{(3-a)}^2& 3bc+cb(3-a)+{(3-a)}^3\end{array}\right]$
$Tr\left(A^3\right)=a^3+abc+3bc+3bc+3bc-abc+{(3-a)}^3=-18$
$\Rightarrow a^3+9bc+{(3-a)}^3=-18$
$\Rightarrow a^3+9bc+27-a^3-3\cdot 3a(3-a)=-18$
$\Rightarrow a^2-3a+bc=-5\cdots \left(2\right)$

Now, $\left|A\right|=a(3-a)-bc$
$=3a-a^2-bc=5$ $[$From equation $\left(2\right)]$