Question

The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2x2 matrix such that the trace of A is 3 and the trace of A³ is −18, then the value of the determinant of A is

Answer 1

Let us have a matrix $A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right].$

We have to find $|A|.$

Trace of a matrix is the sum of main diagonal entries. Here it is equal to 3.

$\longrightarrow a+d=3$

$\longrightarrow (a+d)^3=3^3$

$\longrightarrow a^3+d^3+3ad(a+d)=27$

$\longrightarrow a^3+d^3+9ad=27\quad\quad\dots(1)$

On finding $A^3,$

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

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

I'm just writing main diagonal entries of $A^3$ instead of writing it completely.

• $(A^3)_{11}=a^3+2abc+bcd$

• $(A^3)_{22}=abc+2bcd+d^3$

The trace of $A^3$ is -18.

$\longrightarrow a^3+2abc+bcd+abc+2bcd+d^3=-18$

$\longrightarrow a^3+d^3+3bc(a+d)=-18$

$\longrightarrow a^3+d^3+9bc=-18\quad\quad\dots(2)$

Subtracting (2) from (1),

$\longrightarrow [a^3+d^3+9ad]-[a^3+d^3+9bc]=27-(-18)$

$\longrightarrow 9(ad-bc)=45$

$\longrightarrow\underline{\underline{|A|=5}}$

Hence 5 is the answer.