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

Hint: 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
\[\Rightarrow {\left( {AB} \right)^2} = A{A^2}BB \\
   \Rightarrow {\left( {AB} \right)^2} = {A^3}{B^2} \ldots \ldots \ldots \left( 2 \right) \\\]
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 \[{\left( {AB} \right)^3}\].