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Question: Find the value of the following determinant, where \( i=\sqrt{-1} \) \( \left| \begin{matrix} ...

Find the value of the following determinant, where i=1i=\sqrt{-1}
2i3i i32i5 \left| \begin{matrix} 2i & -3i \\\ {{i}^{3}} & -2{{i}^{5}} \\\ \end{matrix} \right|

Explanation

Solution

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 ii , by using the given value of ii . Put those calculated values into the final expression and get the answer.

Complete step-by-step answer :
Let the value of the determinant 2i3i i32i5 \left| \begin{matrix} 2i & -3i \\\ {{i}^{3}} & -2{{i}^{5}} \\\ \end{matrix} \right| is equal to II .
Then, we can also write it as I=2i3i i32i5 I=\left| \begin{matrix} 2i & -3i \\\ {{i}^{3}} & -2{{i}^{5}} \\\ \end{matrix} \right| .
Now we have to find the value of II , for that we need to know how to expand 2×22\times 2 determinant.
Let D=ab cd D=\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right| be a 2×22\times 2 determinant.
Then, the value of determinant i.e. D=adbcD=ad-bc .
Similarly, using the same concept for I=2i3i i32i5 I=\left| \begin{matrix} 2i & -3i \\\ {{i}^{3}} & -2{{i}^{5}} \\\ \end{matrix} \right| , we get
I=2i×(2i5)i3×(3i)\Rightarrow I=2i\times (-2{{i}^{5}})-{{i}^{3}}\times (-3i)
I=4i6+3i4. (1)\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 II , we have to find the fourth and sixth power of ii .
We know that i=1i=\sqrt{-1}
Then i2=i×i=1×1=(1)2=1{{i}^{2}}=i\times i=\sqrt{-1}\times \sqrt{-1}={{(\sqrt{-1})}^{2}}=-1
Similarly value of i4=i2×i2=1×1=1 (2){{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 i6=i4×i2=1×1=1 (3){{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 i4{{i}^{4}} and i6{{i}^{6}} in equation (1), we get
I=4×1+3×1\Rightarrow I=-4\times -1+3\times 1
After simplifying above equation, we get
I=4+3=7\Rightarrow I=4+3=7
Hence, the value of determinant 2i3i i32i5 \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 ii is given so we have to calculate further the value of the final expression by using the value of ii and that is the answer.