
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
162.9k+ views
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.
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.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
