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

Hint: Denote $A = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)$ and then do matrix multiplication. Use the formula ${(x - y)^3} = {x^3} - 3{x^2}y + 3x{y^2} - {y^3}$. Determinant of a $2 \times 2$ matrix denoted by $\left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)$ is ${a_{11}}{a_{22}} - {a_{21}}{a_{12}}$.

Formula Used:
${(x - y)^3} = {x^3} - 3{x^2}y + 3x{y^2} - {y^3}$
Determinant of $\left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)$ is ${a_{11}}{a_{22}} - {a_{21}}{a_{12}}$.

Complete step by step Solution:
Let $A = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)$. Since trace = 3, $a + d = 3$
${A^2} = \left( {\begin{array}{*{20}{c}}
  {{a^2} + bc}&{b(a + d)} \\
  {c(a + d)}&{bc + {d^2}}
\end{array}} \right)$
${A^3} = \left( {\begin{array}{*{20}{c}}
  {{a^3} + 2abc + bcd}&{{a^2}b + {b^2}c + abd + b{d^2}} \\
  {{a^2}c + acd + b{c^2} + c{d^2}}&{abc + 2bcd + {d^3}}
\end{array}} \right)$
The trace of ${A^3}$ is -18. Therefore,
${a^3} + 2abc + bcd + abc + 2bcd + {d^3} = - 18$
Since $d = 3 - a$,
${a^3} + 2abc + bc(3 - a) + abc + 2bc(3 - a) + {(3 - a)^3} = - 18$
${a^3} + 2abc + 3bc - abc + abc + 6bc - 2abc + 27 - 27a + 9{a^2} - {a^3} = - 18$
$9{a^2} - 27a + 9bc + 27 = - 18$
Solving further,
${a^2} - 3a + bc = - 5$
$|A| = ad - bc = a(3 - a) - bc = 3a - {a^2} - bc$
$|A| = - ({a^2} - 3a + bc)$
$|A| = 5$
The determinant of A is 5.

Note: Instead of substituting $d = 3 - a$ after finding ${A^3}$ we can substitute it before finding ${A^2}$.
We will then get ${A^3} = \left( {\begin{array}{*{20}{c}}
  {{a^3} + abc + 3bc}&{{a^2}b + {b^2}c + 9b - 3ab} \\
  {3ac + b{c^2} + c{{(3 - a)}^2}}&{3bc + bc(3 - a) + {{(3 - a)}^3}}
\end{array}} \right)$. From this we can equate ${a^3} + abc + 3bc + 3bc + bc(3 - a) + {(3 - a)^3}$ to the trace.