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 trace of $A^3$ is $-18$, then the value of the determinant of $A$ is ___________

Answer 1

Explanation

Consider the matrix

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

Given that trace of $A$ is $3$

$\therefore a+d=3$

$\Rightarrow$ $d=3-a$

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

$$\begin{gathered} \Rightarrow A^3=\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right] \\ =\left[\begin{array}{cc}{a}^2+bc& 3b\\ 3c& cb+{\left(3-a\right)}^2\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 3-a\end{array}\right] \\ =\left[\begin{array}{cc}{a}^3+abc+3bc& a^{2}b+b^{2}c+9b-3ab\\ 3ac+c^{2}b+c{\left(3-a\right)}^2& 3bc+cb\left(3-a\right)+{\left(3-a\right)}^3\end{array}\right] \end{gathered}$$

Now the trace of $A^3$ is

$$\begin{gathered} Tr\left(A^3\right)=a^3+abc+3bc+3bc+cb\left(3-a\right)+{\left(3-a\right)}^3 \\ \Rightarrow -18=a^3+9bc+{\left(3-a\right)}^3 \\ \Rightarrow -18=a^3+9bc+27-27a+9a^2-a^3 \\ \Rightarrow -5=a^2-3a+bc.............................................\left(ii\right) \end{gathered}$$

Now, determinant of $A$ is

$$\begin{gathered} \left|A\right|=a\left(3-a\right)-bc \\ =-a^2+3a-bc \\ =-\left(-5\right) \\ =5 \end{gathered}$$

Hence the value of the determinant of $A$ is $5$