Question

Find the value of the following determinant.
\[\left| {\begin{array}{*{20}{c}}
  {2i}&{ - 3i} \\
  {{i^3}}&{ - 2{i^3}}
\end{array}} \right|\] , where \[i = \sqrt { - 1} \]

Answer 1

Hint: We will solve for the determinant by using the general formula of determinant, comparing the values of both determinants we substitute the values to RHS and calculate the final value. In the end we substitute the values related to i.
* We use the formula to calculate the determinant of order $2 \times 2$
\[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc.\]

Complete step-by-step answer:
We are given the determinant \[\left| {\begin{array}{*{20}{c}}
  {2i}&{ - 3i} \\
  {{i^3}}&{ - 2{i^3}}
\end{array}} \right|\]
On comparing the values of the determinant with the general determinant \[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right|\]
We get \[a = 2i,b = - 3i,c = {i^3},d = - 2{i^3}\]
Substituting the values in \[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc.\]
We get \[\left| {\begin{array}{*{20}{c}}
  {2i}&{ - 3i} \\
  {{i^3}}&{ - 2{i^3}}
\end{array}} \right| = (2i)( - 2{i^3}) - ( - 3i)({i^3})\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {2i}&{ - 3i} \\
  {{i^3}}&{ - 2{i^3}}
\end{array}} \right| = - 4{i^4} + 3{i^4}\]
Taking common \[{i^4}\]from RHS
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {2i}&{ - 3i} \\
  {{i^3}}&{ - 2{i^3}}
\end{array}} \right| = {i^4}( - 4 + 3)\]

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {2i}&{ - 3i} \\
  {{i^3}}&{ - 2{i^3}}
\end{array}} \right| = - {i^4}\]
So the value of the determinant comes out to be \[ - {i^4}\].
We know \[i = \sqrt { - 1} \]
Squaring both sides of the equation we get
\[{i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1\] { as square cancels the under root )
Again squaring both sides of the equation we get
\[{i^4} = {\left( {{i^2}} \right)^2} = {\left( { - 1} \right)^2} = 1\].
Substituting the value of \[{i^4} = 1\] in the value of the determinant.
So the determinant value is \[ - (1) = - 1\].
Thus value of determinant is -1

Additional Information:
* Any matrix with a row having all elements as 0 will have its determinant is 0.
* The determinant of a diagonal matrix is the product of its diagonal entries.
* The determinant of a matrix is 0 if and only if its rows are linearly dependent, which means elements of a row can be written as a linear combination of elements of another row. If the rows are linearly independent, then the determinant is non-zero.

Note: Students are likely make mistake of thinking that value of determinant can be calculated from both ways i.e. (ad-bc) and (bc-ad), which is wrong, we can only calculate the value of determinant from one side i.e. (ad-bc).