Question

If $A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]$ then ${A^k} = \left[ {\begin{array}{*{20}{c}}
  {1 + 2k}&{ - 4k} \\
  k&{1 - 2k}
\end{array}} \right]$
Where k is any positive integer.

Answer 1

Hint-In this question, we use the concept of Arithmetic Progression. We know the nth term of an A.P. is ${a_n} = a + \left( {n - 1} \right)d$ . In this question we also use the concept of multiplication of two matrices.

Complete step-by-step answer:
Given, $A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]$
Now, we have to find the values ${A^2}$, ${A^3}$ and ${A^4}$ to identify the type of series followed by each element of the matrix.
${A^2} = \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]$
We use the concept of multiplication of two matrices.
$
  {A^2} = \left[ {\begin{array}{*{20}{c}}
  {3 \times 3 + \left( { - 4} \right) \times 1}&{3 \times \left( { - 4} \right) + \left( { - 4} \right) \times \left( { - 1} \right)} \\
  {1 \times 3 + \left( { - 1} \right) \times 1}&{1 \times \left( { - 4} \right) + \left( { - 1} \right) \times \left( { - 1} \right)}
\end{array}} \right] \\
  {A^2} = \left[ {\begin{array}{*{20}{c}}
  {9 - 4}&{ - 12 + 4} \\
  {3 - 1}&{ - 4 + 1}
\end{array}} \right] \\
  {A^2} = \left[ {\begin{array}{*{20}{c}}
  5&{ - 8} \\
  2&{ - 3}
\end{array}} \right] \\
$
Now, we find the value of ${A^3}$
$
  {A^3} = \left[ {\begin{array}{*{20}{c}}
  5&{ - 8} \\
  2&{ - 3}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right] \\
  {A^3} = \left[ {\begin{array}{*{20}{c}}
  {15 - 8}&{ - 20 + 8} \\
  {6 - 3}&{ - 8 + 3}
\end{array}} \right] \\
  {A^3} = \left[ {\begin{array}{*{20}{c}}
  7&{ - 12} \\
  3&{ - 5}
\end{array}} \right] \\
 $
 Now, we find the value of ${A^4}$
$
  {A^4} = \left[ {\begin{array}{*{20}{c}}
  7&{ - 12} \\
  3&{ - 5}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right] \\
  {A^4} = \left[ {\begin{array}{*{20}{c}}
  {21 - 12}&{ - 28 + 12} \\
  {9 - 5}&{ - 12 + 5}
\end{array}} \right] \\
  {A^4} = \left[ {\begin{array}{*{20}{c}}
  9&{ - 16} \\
  4&{ - 7}
\end{array}} \right] \\
 $
Now, we observe which type of series is followed by each element of the matrix.
First row and first column of A, ${A^2}$, ${A^3}$ and ${A^4}$
3, 5, 7, 9…………….
Above series is in A.P with common difference, d=2 and first term, a=3
Now, nth term of an A.P. is ${a_n} = a + \left( {n - 1} \right)d$
$
   \Rightarrow {a_k} = 3 + \left( {k - 1} \right) \times 2 \\
   \Rightarrow {a_k} = 3 + 2k - 2 \\
  {a_k} = 1 + 2k,{\text{where }}k{\text{ is any positive integer}} \\
$
First row and second column of A, ${A^2}$, ${A^3}$ and ${A^4}$
-4, -8, -12, -16…………….
Above series is in A.P with common difference, d=-4 and first term, a=-4
Now, nth term of an A.P. is ${a_n} = a + \left( {n - 1} \right)d$
$
   \Rightarrow {a_k} = - 4 + \left( {k - 1} \right) \times \left( { - 4} \right) \\
   \Rightarrow {a_k} = - 4 - 4k + 4 \\
  {a_k} = - 4k,{\text{where }}k{\text{ is any positive integer}} \\
 $
Second row and first column of A, ${A^2}$, ${A^3}$ and ${A^4}$
1, 2, 3, 4…………….
Above series is in A.P with common difference, d=1 and first term, a=1
Now, nth term of an A.P. is ${a_n} = a + \left( {n - 1} \right)d$
$
   \Rightarrow {a_k} = 1 + \left( {k - 1} \right) \times 1 \\
   \Rightarrow {a_k} = 1 + k - 1 \\
  {a_k} = k,{\text{where }}k{\text{ is any positive integer}} \\
 $
Second row and Second column of A, ${A^2}$, ${A^3}$ and ${A^4}$
-1, -3, -5, -7…………….
Above series is in A.P with common difference, d=-2 and first term, a=-1
Now, nth term of an A.P. is ${a_n} = a + \left( {n - 1} \right)d$
$
   \Rightarrow {a_k} = - 1 + \left( {k - 1} \right) \times \left( { - 2} \right) \\
   \Rightarrow {a_k} = - 1 - 2k + 2 \\
  {a_k} = 1 - 2k,{\text{where }}k{\text{ is any positive integer}} \\
$
So, it’s proved ${A^k} = \left[ {\begin{array}{*{20}{c}}
  {1 + 2k}&{ - 4k} \\
  k&{1 - 2k}
\end{array}} \right],{\text{where }}k{\text{ is any positive integer}}$
Note-In such types of problems we use the concept of series first we have to find the value ${A^2}$, ${A^3}$ and ${A^4}$ and then observe which type of series followed by each element of matrices. If it is arithmetic progression then we use the nth term of an A.P. So, we will get the required answer.