Question

$A$ and $B$ are two square matrices, such that ${A^2}B = BA$ and ${\left( {AB} \right)^{10}} = {A^k} \cdot {B^{10}}$. Find the value of $k - 1020$.

Answer 1

Here, we will use the given information to find the values of ${\left( {AB} \right)^1}$, ${\left( {AB} \right)^2}$, and ${\left( {AB} \right)^3}$ in terms of $A$ and $B$. Then, rewriting the three equations, we will form a general formula for ${\left( {AB} \right)^n}$. Then, we will use the generalised formula and the given information to find the value of $k$. Finally, we will use the value of $k$ to simplify the expression $k - 1020$, and hence, obtain the required value.

Complete step-by-step answer:
First, we will find the value of ${\left( {AB} \right)^1}$.
Rewriting the expression, we get
$\Rightarrow {\left( {AB} \right)^1} = {A^1}{B^1}$
Rewriting 1 as $2 - 1$, we get
$\Rightarrow {\left( {AB} \right)^1} = {A^{2 - 1}}{B^1}$
We know that any number raised to power 1 is equal to itself.
Rewriting 2 as ${2^1}$, we get
$\Rightarrow {\left( {AB} \right)^1} = {A^{{2^1} - 1}}{B^1} \ldots \ldots \ldots \left( 1 \right)$
Now, we will find the value of ${\left( {AB} \right)^2}$.
Rewriting the expression, we get
$\Rightarrow {\left( {AB} \right)^2} = \left( {AB} \right)\left( {AB} \right)$
Removing the parentheses, we get
$\Rightarrow {\left( {AB} \right)^2} = ABAB$
Enclosing $BA$ in parentheses, we get
$\Rightarrow {\left( {AB} \right)^2} = A\left( {BA} \right)B$
It is given that ${A^2}B = BA$.
Substituting $BA = {A^2}B$ in the equation, we get
$\Rightarrow {\left( {AB} \right)^2} = A\left( {{A^2}B} \right)B$
Simplifying the expression, we get

\Rightarrow {\left( {AB} \right)^2} = {A^3}{B^2} \ldots \ldots \ldots \left( 2 \right) \\\\$$ Rewriting 3 as $$4 - 1$$, we get $$ \Rightarrow {\left( {AB} \right)^2} = {A^{4 - 1}}{B^2}$$ The number 4 is the square of 2. Rewriting 4 as $${2^2}$$, we get $$ \Rightarrow {\left( {AB} \right)^2} = {A^{{2^2} - 1}}{B^2} \ldots \ldots \ldots \left( 3 \right)$$ Next, we will find the value of $${\left( {AB} \right)^3}$$. Rewriting the expression, we get $$\Rightarrow {\left( {AB} \right)^3} = \left( {AB} \right)\left( {AB} \right)\left( {AB} \right) \\\ \Rightarrow {\left( {AB} \right)^3} = {\left( {AB} \right)^2}\left( {AB} \right) \\\\$$ Substituting $${\left( {AB} \right)^2} = {A^3}{B^2}$$ from equation $$\left( 2 \right)$$, we get $$ \Rightarrow {\left( {AB} \right)^3} = \left( {{A^3}{B^2}} \right)\left( {AB} \right)$$ Removing the parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}{B^2}AB$$ Rewriting $${B^2}$$ as $$BB$$, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}BBAB$$ Enclosing $$BA$$ in parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}B\left( {BA} \right)B$$ Substituting $$BA = {A^2}B$$ in the equation, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}B\left( {{A^2}B} \right)B$$ Removing the parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}B{A^2}BB$$ Rewriting $${A^2}$$ as $$AA$$, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}BAABB$$ Enclosing $$BA$$ in parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}\left( {BA} \right)ABB$$ Substituting $$BA = {A^2}B$$ in the equation, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}\left( {{A^2}B} \right)ABB$$ Removing the parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}{A^2}BABB$$ Enclosing $$BA$$ in parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}{A^2}\left( {BA} \right)BB$$ Substituting $$BA = {A^2}B$$ in the equation, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}{A^2}\left( {{A^2}B} \right)BB$$ Removing the parentheses, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^3}{A^2}{A^2}BBB$$ Simplifying the expression, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^7}{B^3}$$ Rewriting 7 as $$8 - 1$$, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^{8 - 1}}{B^3}$$ The number 8 is the cube of 2. Rewriting 8 as $${2^3}$$, we get $$ \Rightarrow {\left( {AB} \right)^3} = {A^{{2^3} - 1}}{B^3} \ldots \ldots \ldots \left( 4 \right)$$ Now, we will observe and generalise the equations formed. From equations $$\left( 1 \right)$$, $$\left( 3 \right)$$, and $$\left( 4 \right)$$, we have $${\left( {AB} \right)^1} = {A^{{2^1} - 1}}{B^1}$$ $${\left( {AB} \right)^2} = {A^{{2^2} - 1}}{B^2}$$ $${\left( {AB} \right)^3} = {A^{{2^3} - 1}}{B^3}$$ We can generalise the above to form a general formula for $${\left( {AB} \right)^n}$$. Thus, we get $${\left( {AB} \right)^n} = {A^{{2^n} - 1}}{B^n}$$ where $$n$$ is a natural number Now, we will find the value of $${\left( {AB} \right)^{10}}$$. Substituting $$n = 10$$ in the generalised formula $${\left( {AB} \right)^n} = {A^{{2^n} - 1}}{B^n}$$, we get $$ \Rightarrow {\left( {AB} \right)^{10}} = {A^{{2^{10}} - 1}}{B^{10}}$$ It is given that $${\left( {AB} \right)^{10}} = {A^k} \cdot {B^{10}}$$. Therefore, substituting $${\left( {AB} \right)^{10}} = {A^k} \cdot {B^{10}}$$ in the equation $${\left( {AB} \right)^{10}} = {A^{{2^{10}} - 1}}{B^{10}}$$, we get $$ \Rightarrow {A^k} \cdot {B^{10}} = {A^{{2^{10}} - 1}}{B^{10}}$$ Comparing the terms of the equations, we get $$\Rightarrow {A^k} = {A^{{2^{10}} - 1}} \\\ \Rightarrow k = {2^{10}} - 1 \\\\$$ Applying the exponent on the base, we get $$ \Rightarrow k = 1024 - 1 = 1023$$ $$\therefore $$ We get the value of $$k$$ as 1023. Finally, we will find the value of the expression $$k - 1020$$. Substituting $$k = 1023$$ in the expression, we get $$ \Rightarrow k - 1020 = 1023 - 1020$$ Thus, we get $$ \Rightarrow k - 1020 = 3$$ **$$\therefore $$ We get the value of the expression $$k - 1020$$ as 3.** **Note:** Here we are provided with matrices $$A$$ and $$B$$, and matrix multiplication is not commutative. This is why we cannot write $$AB$$ as $$BA$$ while simplifying the values of $${\left( {AB} \right)^2}$$ and $${\left( {AB} \right)^3}$$.