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Find the value of the following determinant, where $i=\sqrt{-1}$  $\left| \begin{matrix} 2i & -3i \\ {{i}^{3}} & -2{{i}^{5}} \\\end{matrix} \right|$

Last updated date: 09th Aug 2024
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Hint: In order to solve this question, we have to simply expand the determinant in the usual way. After expanding the determinant, get the final expression. Calculate the values of higher powers of $i$ , by using the given value of $i$ . Put those calculated values into the final expression and get the answer.

Let the value of the determinant $\left| \begin{matrix} 2i & -3i \\ {{i}^{3}} & -2{{i}^{5}} \\ \end{matrix} \right|$ is equal to $I$ .
Then, we can also write it as $I=\left| \begin{matrix} 2i & -3i \\ {{i}^{3}} & -2{{i}^{5}} \\ \end{matrix} \right|$ .
Now we have to find the value of $I$ , for that we need to know how to expand $2\times 2$ determinant.
Let $D=\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|$ be a $2\times 2$ determinant.
Then, the value of determinant i.e. $D=ad-bc$ .
Similarly, using the same concept for $I=\left| \begin{matrix} 2i & -3i \\ {{i}^{3}} & -2{{i}^{5}} \\ \end{matrix} \right|$ , we get
$\Rightarrow I=2i\times (-2{{i}^{5}})-{{i}^{3}}\times (-3i)$
$\Rightarrow I=-4{{i}^{6}}+3{{i}^{4}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots .\text{ }\left( 1 \right)$
In order to get the value of $I$ , we have to find the fourth and sixth power of $i$ .
We know that $i=\sqrt{-1}$
Then ${{i}^{2}}=i\times i=\sqrt{-1}\times \sqrt{-1}={{(\sqrt{-1})}^{2}}=-1$
Similarly value of ${{i}^{4}}={{i}^{2}}\times {{i}^{2}}=-1\times -1=1\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text{ }\left( 2 \right)$
And finally value of ${{i}^{6}}={{i}^{4}}\times {{i}^{2}}=1\times -1=-1\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text{ }\left( 3 \right)$
Substituting the values of ${{i}^{4}}$ and ${{i}^{6}}$ in equation (1), we get
$\Rightarrow I=-4\times -1+3\times 1$
After simplifying above equation, we get
$\Rightarrow I=4+3=7$
Hence, the value of determinant $\left| \begin{matrix} 2i & -3i \\ {{i}^{3}} & -2{{i}^{5}} \\ \end{matrix} \right|$ is equal to 7.
So, the required value is 7.

Note: This question tests the understanding of both complex number as well as determinants. This is also a straightforward question. But one tricky part is that students often expand the determinant and then get the final expression and leave it assuming that expression as final answer. But as the value of $i$ is given so we have to calculate further the value of the final expression by using the value of $i$ and that is the answer.