Question

If A = \left( {\begin{array}{*{20}{c}} i&{ - i} \\\ { - i}&i \end{array}} \right) and B = \left( {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right) then ${A^8}$ equals
(a) $4B$
(b) $128B$
(c)$- 128B$
(d)$- 64B$

Answer 1

As we know that the above question is of matrices. A matrix is a set of numbers arranged in rows and columns so they form a rectangular array. The numbers are called the entries or elements of the matrix. The term $i$ is an imaginary number and it is called iota and it has the value of $\sqrt {- 1}$. These numbers are not really imaginary. In this question we will first square both the given numbers and then use the basic exponential formula to solve this.

Complete step by step solution:
As per the given we have been givenA = \left( {\begin{array}{*{20}{c}} i&{ - i} \\\ { - i}&i \end{array}} \right) and B = \left( {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right) so from this we have to find the value of ${A^8}$. We know that both the matrices are idempotent matrices, which means that when these matrices are multiplied by itself, it gives the same value.
Here we have $A = iB$, we will now square both the sides, so we have: ${A^2} = {(iB)^2}$ i.e. ${A^2} = {i^2}{B^2}$
And we know the value of $i = \sqrt { - 1}$, so when it is squared it gives the value $- 1$, so we can say $- {B^2}$ It gives \left( {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right). So we can write it as $- 2B$.
We will square it again and we get: ${A^4} = {( - 2B)^2}$ i.e. ${A^4} = 4{B^2}$, again we can get the power in front by the basic rule of exponential logarithm. So it gives ${A^4} = 4(2B) = 8B$.
We will square it again and this time we have:${({A^4})^2} = {(8B)^2} \Rightarrow {A^8} = 64{B^2}$, we will get the power again in the front and it gives :${A^8} = 2(64B) = 128B$.
Hence the correct option is (b)$128B$.

Note: We should be careful while solving this kind of questions and we need to be aware of the matrices, their properties and the exponential formulas. We should note that the value of ${i^2}$ is mistakenly taken as $1$, which is a completely wrong value and it may lead to the wrong answers. We should be careful with complex numbers.