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

Here we are asked to find the value of a given matrix raised to the power eight. We will first find the relation between the given two matrices, then we will raise the power step by step, that is first we will raise the power to two, then after simplification, we will again raise it by two and repeat till we reach the required power value.

Complete step by step answer:
It is given that 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]
We aim to find the value of ${A^8}$. As we can see that the options are all related to the matrix $B$ first we need to find the relation between the matrices $A$ and $B$.
Consider the matrix $A$, \left[ {\begin{array}{*{20}{c}} i&{ - i} \\\ { - i}&i; \end{array}} \right]
we can see that the imaginary number $i$ is common in all the elements in this matrix so let’s take it out commonly.
A = i\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]
Now we can see that the matrix $A$ is now a matrix $B$ multiplied by an imaginary number $i$.
That is $A = iB$. Thus, we have found the relationship between the matrices. Using that relation we will find the value of ${A^8}$.
Consider the relation that we have found $A = iB$ now raising the power to two on both sides of this relation we get
${A^2} = {\left( {iB} \right)^2}$
$\Rightarrow {A^2} = - {B^2}$
Now let us find the matrix - {B^2} = - \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right].
- {B^2} = - {\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]^2}
- {B^2} = - \left[ {\left( {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right)} \right]
- {B^2} = - \left[ {\left( {\begin{array}{*{20}{c}} {1 + 1}&{ - 1 - 1} \\\ { - 1 - 1}&{1 + 1} \end{array}} \right)} \right]
On simplifying this we get
- {B^2} = - \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right] = - 2B
Therefore, we get ${A^2} = - 2B$. Now let us raise the power to two on both sides we will get ${\left( {{A^2}} \right)^2} = {\left( { - 2B} \right)^2}$.
On simplifying this we get
${A^4} = 4{B^2}$
Since $- {B^2} = - 2B \Rightarrow {B^2} = 2B$ we get
${A^4} = 8B$
Now again raising the power to two we get
${\left( {{A^4}} \right)^2} = {\left( {8B} \right)^2}$
On simplifying this we get ${A^8} = 64{B^2}$. We already know that $- {B^2} = - 2B \Rightarrow {B^2} = 2B$ using this we get
${A^8} = 128B$. Thus, we have found the required value.
Let’s see the options. option (a) $4B$ is an incorrect option as we got ${A^8}$ is $128$ times the matrix $B$.
Option (b) $128B$ as we got the same value in the above calculation.
Option (c) $- 128B$ is an incorrect option as we got ${A^8}$ is $128$ times the matrix $B$, not $- 128$ times.
Option (d) $- 64B$ is an incorrect option as we got ${A^8}$ is $128$ times the matrix $B$.

Hence, option (b) $128B$ is the correct answer.

Note:
In complex numbers, there is a special number called an imaginary number that is denoted by $i$ and it is also called iota. The imaginary number was formed by taking the square root to minus one that is $\sqrt { - 1} = i$. Therefore, when we square the imaginary number, we get minus one that is ${i^2} = \sqrt { - 1} \times \sqrt { - 1} = - 1$.