Question

If $A = \left[ {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&2&0 \\
  0&0&2
\end{array}} \right]$ , then ${A^5} = $
A. $5A$
B. $10A$
C. $16A$
D. $32A$

Answer 1

Hint: We can simply determine any power of a matrix using the nth power of a matrix expression. Matrix powers frequently follow a pattern. As a result, we can calculate any power of a matrix without performing all the multiplications if we can identify the order in which the powers of the matrix follow. This implies that we don't need to calculate all of the powers of a matrix in order to obtain a formula that gives us the nth power.

Complete step by step Solution:
We have been provided with the matrix $A = \left[ {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&2&0 \\
  0&0&2
\end{array}} \right]$
To find its 5th power i.e., ${A^5}$, we have to multiply the given matrix to itself and repeat the procedure again till we reach the 5th power of the matrix to get a relation between matrix $A$ and ${A^5}$.
So, starting with ${A^2} = A \cdot A = \left[ {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&2&0 \\
  0&0&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&2&0 \\
  0&0&2
\end{array}} \right]$
Applying the formula $A \cdot B = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}{b_{11}} + {a_{12}}{b_{21}} + {a_{13}}{b_{31}}}&{{a_{11}}{b_{12}} + {a_{12}}{b_{22}} + {a_{13}}{b_{32}}}&{{a_{11}}{b_{13}} + {a_{12}}{b_{23}} + {a_{13}}{b_{33}}} \\
  {{a_{21}}{b_{11}} + {a_{22}}{b_{21}} + {a_{23}}{b_{31}}}&{{a_{21}}{b_{12}} + {a_{22}}{b_{22}} + {a_{23}}{b_{32}}}&{{a_{21}}{b_{13}} + {a_{22}}{b_{23}} + {a_{23}}{b_{33}}} \\
  {{a_{31}}{b_{11}} + {a_{32}}{b_{21}} + {a_{33}}{b_{31}}}&{{a_{31}}{b_{12}} + {a_{32}}{b_{22}} + {a_{33}}{b_{32}}}&{{a_{31}}{b_{13}} + {a_{32}}{b_{23}} + {a_{33}}{b_{33}}}
\end{array}} \right]$
Hence, by solving this equation for the matrix $A$ multiplied by itself, we get
${A^2} = \left[ {\begin{array}{*{20}{c}}
  {2 \times 2 + 0 \times 0 + 0 \times 0}&{2 \times 0 + 0 \times 2 + 0 \times 0}&{2 \times 0 + 0 \times 0 + 0 \times 2} \\
  {0 \times 2 + 2 \times 0 + 0 \times 0}&{0 \times 0 + 2 \times 2 + 0 \times 0}&{0 \times 0 + 2 \times 0 + 0 \times 2} \\
  {0 \times 2 + 0 \times 0 + 2 \times 0}&{0 \times 0 + 0 \times 2 + 2 \times 0}&{0 \times 0 + 0 \times 0 + 2 \times 2}
\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}
  4&0&0 \\
  0&4&0 \\
  0&0&4
\end{array}} \right] = {2^1}A$
Similarly, ${A^5} = {2^4}A \Rightarrow {A^5} = 16A$ .

Therefore, the correct option is (C).

Note: One can find the following formula for some matrix powers: 1) The exponent's parity. The odd powers might be one way, while the even powers might be another. 2) Differences in the signs. It might be the case, for instance, that the elements of even powers are positive and the elements of odd powers negative, or vice versa. 3) Repetition: Whether or not the same matrix is repeated every predetermined number of powers. 4) The exponent and the components of the matrix have a mathematical relationship.