
For the matrix \[A = \left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right]\]which of the following are correct
A. \[{A^3} + 3{A^2} - I = 0\]
B. \[{A^3} - 3{A^2} - I = 0\]
C. \[{A^3} + 2{A^2} - I = 0\]
D. \[{A^3} - {A^2} + I = 0\]
Answer
161.1k+ views
Hint: In the given problem we first find the value of \[{A^2}\]. Similarly, we find the value of \[{A^3}\] by multiplying \[{A^2}\] with \[A\]. We then perform different operations with \[{A^2}\] and \[{A^3}\] in accordance with the given options to get the correct option.
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,
\[A = \left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right]\]
On squaring A, we get \[{A^2}\]
\[{A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&3&1 \\
5&6&2 \\
3&4&1
\end{array}} \right]\]
On multiplying the scalar 3 with \[{A^2}\]we get,
\[3{A^2} = 3\left[ {\begin{array}{*{20}{c}}
2&3&1 \\
5&6&2 \\
3&4&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6&9&3 \\
{15}&{18}&6 \\
9&{12}&3
\end{array}} \right]\]
On multiplying \[{A^2}\] with A we get \[{A^3}\]
\[{A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}
2&3&1 \\
5&6&2 \\
3&4&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
7&9&3 \\
{15}&{19}&6 \\
9&{12}&4
\end{array}} \right]\]
Therefore \[{A^3} - 3{A^2} = \left[ {\begin{array}{*{20}{c}}
7&9&3 \\
{15}&{19}&6 \\
9&{12}&4
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
6&9&3 \\
{15}&{18}&6 \\
9&{12}&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right] = I\]
\[
\\
\Rightarrow {A^3} - 3{A^2} - I = 0 \\
\]
Option B. is the correct answer.
Note: To solve the given problem, one must know to add, subtract and multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix during matrix multiplication.
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,
\[A = \left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right]\]
On squaring A, we get \[{A^2}\]
\[{A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&3&1 \\
5&6&2 \\
3&4&1
\end{array}} \right]\]
On multiplying the scalar 3 with \[{A^2}\]we get,
\[3{A^2} = 3\left[ {\begin{array}{*{20}{c}}
2&3&1 \\
5&6&2 \\
3&4&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6&9&3 \\
{15}&{18}&6 \\
9&{12}&3
\end{array}} \right]\]
On multiplying \[{A^2}\] with A we get \[{A^3}\]
\[{A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}
2&3&1 \\
5&6&2 \\
3&4&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1&0 \\
1&2&1 \\
2&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
7&9&3 \\
{15}&{19}&6 \\
9&{12}&4
\end{array}} \right]\]
Therefore \[{A^3} - 3{A^2} = \left[ {\begin{array}{*{20}{c}}
7&9&3 \\
{15}&{19}&6 \\
9&{12}&4
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
6&9&3 \\
{15}&{18}&6 \\
9&{12}&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right] = I\]
\[
\\
\Rightarrow {A^3} - 3{A^2} - I = 0 \\
\]
Option B. is the correct answer.
Note: To solve the given problem, one must know to add, subtract and multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix during matrix multiplication.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

JEE Advanced 2025 Notes

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

List of Fastest Century in IPL History
