Question

If $X = \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]$ , then what is the value of ${X^n}$ ?
A. $\left[ {\begin{array}{*{20}{c}}
  {3n}&{ - 4n} \\
  n&{ - n}
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
  {2 + n}&{5 - n} \\
  n&{ - n}
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
  {{3^n}}&{{{\left( { - 4} \right)}^n}} \\
  {{1^n}}&{{{\left( { - 1} \right)}^n}}
\end{array}} \right]$
D. None of these

Answer 1

Hint: To evaluate any power of the matrix, matrix multiplication is performed over that matrix again and again, as per the power. For example, ${A^2} = AA$ and ${A^3} = AAA$ , where $A$ is a square matrix.

Complete step by step Solution:
Given is matrix $X$ such that:
$X = \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]$
The question is asking us to evaluate the value of ${X^n}$.
Let us evaluate it for $n = 2$, that is, let’s evaluate ${X^2}$ and compare our result with each option.
${X^2} = XX$
Substituting the value of $X$,
${X^2} = \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 1}
\end{array}} \right]$
Performing Matrix Multiplication,
\[{X^2} = \left[ {\begin{array}{*{20}{c}}
  {3 \times 3 + \left( { - 4} \right) \times 1}&{3 \times \left( { - 4} \right) + \left( { - 4} \right) \times \left( { - 1} \right)} \\
  {1 \times 3 + \left( { - 1} \right) \times 1}&{1 \times \left( { - 4} \right) + \left( { - 1} \right) \times \left( { - 1} \right)}
\end{array}} \right]\]
On simplifying further, we get:
\[{X^2} = \left[ {\begin{array}{*{20}{c}}
  5&{ - 8} \\
  2&{ - 3}
\end{array}} \right]\] … (1)
Now, let us evaluate the value of each option for $n = 2$ and compare it with the result obtained in (1),
For option A:
$\left[ {\begin{array}{*{20}{c}}
  {3n}&{ - 4n} \\
  n&{ - n}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {3\left( 2 \right)}&{ - 4\left( 2 \right)} \\
  {\left( 2 \right)}&{ - \left( 2 \right)}
\end{array}} \right]$
Simplifying further, we get:
$\left[ {\begin{array}{*{20}{c}}
  {3n}&{ - 4n} \\
  n&{ - n}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  6&{ - 8} \\
  2&{ - 2}
\end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}}
  5&{ - 8} \\
  2&{ - 3}
\end{array}} \right] = {X^2}$
For option B:
$\left[ {\begin{array}{*{20}{c}}
  {2 + n}&{5 - n} \\
  n&{ - n}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {2 + 2}&{5 - 2} \\
  2&{ - 2}
\end{array}} \right]$
Simplifying further, we get:
$\left[ {\begin{array}{*{20}{c}}
  {2 + n}&{5 - n} \\
  n&{ - n}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  4&3 \\
  2&{ - 2}
\end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}}
  5&{ - 8} \\
  2&{ - 3}
\end{array}} \right] = {X^2}$
For option C:
$\left[ {\begin{array}{*{20}{c}}
  {{3^n}}&{{{\left( { - 4} \right)}^n}} \\
  {{1^n}}&{{{\left( { - 1} \right)}^n}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {{3^2}}&{{{\left( { - 4} \right)}^2}} \\
  {{1^2}}&{{{\left( { - 1} \right)}^2}}
\end{array}} \right]$
Simplifying further, we get:
$\left[ {\begin{array}{*{20}{c}}
  {{3^n}}&{{{\left( { - 4} \right)}^n}} \\
  {{1^n}}&{{{\left( { - 1} \right)}^n}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  9&{16} \\
  1&1
\end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}}
  5&{ - 8} \\
  2&{ - 3}
\end{array}} \right] = {X^2}$
This means that none of the options will provide the value of ${X^n}$, for any value of $n$.

Therefore, the correct option is (D).

Note: To evaluate the value of a matrix, raised to a power, matrix multiplication is performed on the matrix over and over again till the power is obtained, that is, for a square matrix, $A$, ${A^n} = A \times A \times A \ldots n{\text{ times}}$.