
If matrix \[A = \left[ {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right]\], then \[{A^{16}} = \]
A. \[\left[ {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
0&1 \\
1&0
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&1
\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right]\]
Answer
162.9k+ views
Hint: To solve this question we will first find the value of \[{A^2}\]. After this we will calculate \[{A^{4}}\] by multiplying \[{A^2}\] with \[{A^2}\]. After calculating the value of \[{A^{4}}\] we will find the value of find \[{A^16}\] by squaring \[{A^{4}}\] that is by multiplying \[{A^{4}}\] with \[{A^{4}}\].
Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]
Complete step by step Solution:
We are given that,
\[A = \left[ {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right]\]
We will first calculate the value of \[{A^2}\] by multiplying \[A\] with \[A\]
We will now evaluate \[{{A}^{4}}\] by calculating \[{{A}^{4}}={{A}^{2}}\times {{A}^{2}}\].
\[\begin{align}
& {{A}^{4}}={{A}^{2}}\times {{A}^{2}} \\
& {{A}^{4}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }-1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\times \left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }-1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right] \\
& {{A}^{4}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]
\end{align}\]
Now we will evaluate \[{{A}^{16}}\] by calculating \[{{A}^{16}}={{A}^{4}}\times {{A}^{4}}\].
\[\begin{align}
& {{A}^{16}}={{A}^{4}}\times {{A}^{4}} \\
& {{A}^{16}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\times \left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right] \\
& {{A}^{16}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]
\end{align}\]
Option D. is the correct answer.
Note:To solve the given problem, one must know to multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix. One must also know to write large numbers in terms of smaller ones for easier simplification.
Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]
Complete step by step Solution:
We are given that,
\[A = \left[ {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right]\]
We will first calculate the value of \[{A^2}\] by multiplying \[A\] with \[A\]
We will now evaluate \[{{A}^{4}}\] by calculating \[{{A}^{4}}={{A}^{2}}\times {{A}^{2}}\].
\[\begin{align}
& {{A}^{4}}={{A}^{2}}\times {{A}^{2}} \\
& {{A}^{4}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }-1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\times \left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }-1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right] \\
& {{A}^{4}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]
\end{align}\]
Now we will evaluate \[{{A}^{16}}\] by calculating \[{{A}^{16}}={{A}^{4}}\times {{A}^{4}}\].
\[\begin{align}
& {{A}^{16}}={{A}^{4}}\times {{A}^{4}} \\
& {{A}^{16}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]\times \left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right] \\
& {{A}^{16}}=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & 1\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]
\end{align}\]
Option D. is the correct answer.
Note:To solve the given problem, one must know to multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix. One must also know to write large numbers in terms of smaller ones for easier simplification.
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
