Question

If $A = \left[ {\begin{array}{*{20}{c}}
  3&3&3 \\
  3&3&3 \\
  3&3&3
\end{array}} \right],$ then ${A^4}$ is equal to
A. $27A$
B. $81A$
C. $243A$
D. $729A$

Answer 1

Hint: First find the value of ${A^2}$ by taking the product of matrix A with itself and then use it to find the value of ${A^4}$ that is the required result.

Complete step by step solution:
We have given a matrix having the form:
$A = \left[ {\begin{array}{*{20}{c}}
  3&3&3 \\
  3&3&3 \\
  3&3&3
\end{array}} \right]$
The goal is to find the value of ${A^4}$.
We can express ${A^4}$ as ${A^2} \times {A^2}$, so first, we need to find the value of ${A^2}$.
The value of ${A^2}$ is given as the product of a matrix $A$ with itself. That is,
${A^2} = A \times A$
Substitute the value of the matrix given in the problem:
${A^2} = \left[ {\begin{array}{*{20}{c}}
  3&3&3 \\
  3&3&3 \\
  3&3&3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  3&3&3 \\
  3&3&3 \\
  3&3&3
\end{array}} \right]$
Find the product of matrices,
${A^2} = \left[ {\begin{array}{*{20}{c}}
  {3 \times 3 + 3 \times 3 + 3 \times 3}&{3 \times 3 + 3 \times 3 + 3 \times 3}&{3 \times 3 + 3 \times 3 + 3 \times 3} \\
  {3 \times 3 + 3 \times 3 + 3 \times 3}&{3 \times 3 + 3 \times 3 + 3 \times 3}&{3 \times 3 + 3 \times 3 + 3 \times 3} \\
  {3 \times 3 + 3 \times 3 + 3 \times 3}&{3 \times 3 + 3 \times 3 + 3 \times 3}&{3 \times 3 + 3 \times 3 + 3 \times 3}
\end{array}} \right]$
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
  {27}&{27}&{27} \\
  {27}&{27}&{27} \\
  {27}&{27}&{27}
\end{array}} \right]\]
Now, we have the value of${A^2}$. Use this value to find the value of${A^4}$.
${A^4} = {A^2} \times {A^2}$
Substitute the value of the matrix${A^2}$in the above equation:
${A^4} = \left[ {\begin{array}{*{20}{c}}
  {27}&{27}&{27} \\
  {27}&{27}&{27} \\
  {27}&{27}&{27}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  {27}&{27}&{27} \\
  {27}&{27}&{27} \\
  {27}&{27}&{27}
\end{array}} \right]$
Find the product of the above matrix:
${A^4} = \left[ {\begin{array}{*{20}{c}}
  {27 \times 27 + 27 \times 27 + 27 \times 27}&{27 \times 27 + 27 \times 27 + 27 \times 27}&{27 \times 27 + 27 \times 27 + 27 \times 27} \\
  {27 \times 27 + 27 \times 27 + 27 \times 27}&{27 \times 27 + 27 \times 27 + 27 \times 27}&{27 \times 27 + 27 \times 27 + 27 \times 27} \\
  {27 \times 27 + 27 \times 27 + 27 \times 27}&{27 \times 27 + 27 \times 27 + 27 \times 27}&{27 \times 27 + 27 \times 27 + 27 \times 27}
\end{array}} \right]$
${A^4} = \left[ {\begin{array}{*{20}{c}}
  {3 \times 729}&{3 \times 729}&{3 \times 729} \\
  {3 \times 729}&{3 \times 729}&{3 \times 729} \\
  {3 \times 729}&{3 \times 729}&{3 \times 729}
\end{array}} \right]$
Take out $729$as a common factor from all the elements.
${A^4} = 729\left[ {\begin{array}{*{20}{c}}
  3&3&3 \\
  3&3&3 \\
  3&3&3
\end{array}} \right]$
${A^4} = 729A$

We have obtained the value of ${A^4}$ as$729A$.

Therefore, the option (d) is correct.


Note: Before multiplying the matrix, we have to check, that is it possible to multiply the matrices.
We can say that the matrices are possible to multiply if the number of columns in the first matrix is equal to the number of rows of the other matrix. If this condition does not hold then it is not possible to multiply the matrices.
We can see that we have a matrix $A$ of dimension $3 \times 3$, so its multiplication with itself is possible because the number of rows in the matrix is equal to the number of columns in the matrix.