
The inverse of \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\] is
A. \[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&3\\0&1&{ - 2}\\0&0&0\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&1\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&2\\0&0&1\end{array}} \right]\]
D. None of these
Answer
161.7k+ views
Hint:We have to find \[{A^{ - 1}}\]of this 3×3 matrix\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\], so for that firstly we will solve adjA and the formula of this is mentioned below and then |A|which has also the formula below. And then dividing both of them will get our required solution. As for finding \[{A^{ - 1}}\] the formula suggests\[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\].
Formula Used:\[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\]
If A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\]
Then signs of adjA for this matrix will be \[\left[ {\begin{array}{*{20}{c}} + & - & + \\ - & + & - \\ + & - & + \end{array}} \right]\]as in \[{a_{11}}\]the sum of this two numbers are 2 which is even so we will put + sign and like that for \[{a_{12}}\]The sum of these two numbers are 3 which is odd so we will put – sign. Like that it will be applicable for all elements in the matrix.
Now after knowing the signs of each element we will find cofactors of each eleme\[adjA = \left[ {\begin{array}{*{20}{c}}{{a_{22}}{a_{33}} - {a_{23}}{a_{32}}}&{{a_{21}}{a_{33}} - {a_{23}}{a_{31}}}&{{a_{21}}{a_{32}} - {a_{22}}{a_{31}}}\\{{a_{12}}{a_{33}} - {a_{13}}{a_{32}}}&{{a_{11}}{a_{33}} - {a_{13}}{a_{31}}}&{{a_{11}}{a_{32}} - {a_{12}}{a_{31}}}\\{{a_{12}}{a_{23}} - {a_{13}}{a_{22}}}&{{a_{11}}{a_{23}} - {a_{13}}{a_{21}}}&{{a_{11}}{a_{22}} - {a_{12}}{a_{21}}}\end{array}} \right]\]nt in this matrix.
\[adjA = \left[ {\begin{array}{*{20}{c}}{{a_{22}}{a_{33}} - {a_{23}}{a_{32}}}&{{a_{21}}{a_{33}} - {a_{23}}{a_{31}}}&{{a_{21}}{a_{32}} - {a_{22}}{a_{31}}}\\{{a_{12}}{a_{33}} - {a_{13}}{a_{32}}}&{{a_{11}}{a_{33}} - {a_{13}}{a_{31}}}&{{a_{11}}{a_{32}} - {a_{12}}{a_{31}}}\\{{a_{12}}{a_{23}} - {a_{13}}{a_{22}}}&{{a_{11}}{a_{23}} - {a_{13}}{a_{21}}}&{{a_{11}}{a_{22}} - {a_{12}}{a_{21}}}\end{array}} \right]\]
And if A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\]
So for finding |A| we will hide that row and that column of that element and make a determinant of the other 2×2 matrix which will be left.
Then |A|=\[{a_{11}}({a_{22}}{a_{33}} - {a_{23}}{a_{32}}) - {a_{12}}({a_{21}}{a_{33}} - {a_{23}}{a_{31}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{22}}{a_{31}})\]
Complete step by step solution:we are given matrix= \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\]and we have to find \[{A^{ - 1}}\]
so first of all as the formula suggests we have to find adjA
so adjA of this matrix\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\] is
\[adjA = \left[ {\begin{array}{*{20}{c}}{1(1) - 2(0)}&{0(1) - 2(0)}&{0(0) - 1(0)}\\{2(1) - 3(0)}&{1(1) - 3(0)}&{1(0) - 2(0)}\\{2(2) - 3(1)}&{1(2) - 3(0)}&{1(1) - 2(0)}\end{array}} \right]\]
\[adjA = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\1&2&1\end{array}} \right]\]
Now as we have done above that the signs of the adj will change as per the formula
\[adjA = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
Now |A| of matrix \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\]is
\[\begin{array}{l}\left| A \right| = 1(1 - 0) - 2(0 - 0) + 3(0 - 0)\\\left| A \right| = 1(1) - 0 + 0\\\left| A \right| = 1 - 0\\\left| A \right| = 1\end{array}\]
Now as \[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\]
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
Now by taking transpose of this matrix we will get
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
So, Option ‘B’ is correct
Note:We have to remember that if our final answer is not matching with the options we can try taking the transpose of that matrix and then we can get our required solution. \[{A^{ - 1}}\] exists only when\[\left| A \right| \ne 0\]. We have to remember the formula for\[{A^{ - 1}}\]. Sometimes students make mistakes while solving adjA and |A| for a 3×3 matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.
Formula Used:\[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\]
If A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\]
Then signs of adjA for this matrix will be \[\left[ {\begin{array}{*{20}{c}} + & - & + \\ - & + & - \\ + & - & + \end{array}} \right]\]as in \[{a_{11}}\]the sum of this two numbers are 2 which is even so we will put + sign and like that for \[{a_{12}}\]The sum of these two numbers are 3 which is odd so we will put – sign. Like that it will be applicable for all elements in the matrix.
Now after knowing the signs of each element we will find cofactors of each eleme\[adjA = \left[ {\begin{array}{*{20}{c}}{{a_{22}}{a_{33}} - {a_{23}}{a_{32}}}&{{a_{21}}{a_{33}} - {a_{23}}{a_{31}}}&{{a_{21}}{a_{32}} - {a_{22}}{a_{31}}}\\{{a_{12}}{a_{33}} - {a_{13}}{a_{32}}}&{{a_{11}}{a_{33}} - {a_{13}}{a_{31}}}&{{a_{11}}{a_{32}} - {a_{12}}{a_{31}}}\\{{a_{12}}{a_{23}} - {a_{13}}{a_{22}}}&{{a_{11}}{a_{23}} - {a_{13}}{a_{21}}}&{{a_{11}}{a_{22}} - {a_{12}}{a_{21}}}\end{array}} \right]\]nt in this matrix.
\[adjA = \left[ {\begin{array}{*{20}{c}}{{a_{22}}{a_{33}} - {a_{23}}{a_{32}}}&{{a_{21}}{a_{33}} - {a_{23}}{a_{31}}}&{{a_{21}}{a_{32}} - {a_{22}}{a_{31}}}\\{{a_{12}}{a_{33}} - {a_{13}}{a_{32}}}&{{a_{11}}{a_{33}} - {a_{13}}{a_{31}}}&{{a_{11}}{a_{32}} - {a_{12}}{a_{31}}}\\{{a_{12}}{a_{23}} - {a_{13}}{a_{22}}}&{{a_{11}}{a_{23}} - {a_{13}}{a_{21}}}&{{a_{11}}{a_{22}} - {a_{12}}{a_{21}}}\end{array}} \right]\]
And if A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\]
So for finding |A| we will hide that row and that column of that element and make a determinant of the other 2×2 matrix which will be left.
Then |A|=\[{a_{11}}({a_{22}}{a_{33}} - {a_{23}}{a_{32}}) - {a_{12}}({a_{21}}{a_{33}} - {a_{23}}{a_{31}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{22}}{a_{31}})\]
Complete step by step solution:we are given matrix= \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\]and we have to find \[{A^{ - 1}}\]
so first of all as the formula suggests we have to find adjA
so adjA of this matrix\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\] is
\[adjA = \left[ {\begin{array}{*{20}{c}}{1(1) - 2(0)}&{0(1) - 2(0)}&{0(0) - 1(0)}\\{2(1) - 3(0)}&{1(1) - 3(0)}&{1(0) - 2(0)}\\{2(2) - 3(1)}&{1(2) - 3(0)}&{1(1) - 2(0)}\end{array}} \right]\]
\[adjA = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\1&2&1\end{array}} \right]\]
Now as we have done above that the signs of the adj will change as per the formula
\[adjA = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
Now |A| of matrix \[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]\]is
\[\begin{array}{l}\left| A \right| = 1(1 - 0) - 2(0 - 0) + 3(0 - 0)\\\left| A \right| = 1(1) - 0 + 0\\\left| A \right| = 1 - 0\\\left| A \right| = 1\end{array}\]
Now as \[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\]
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
Now by taking transpose of this matrix we will get
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\1&{ - 2}&1\end{array}} \right]\]
So, Option ‘B’ is correct
Note:We have to remember that if our final answer is not matching with the options we can try taking the transpose of that matrix and then we can get our required solution. \[{A^{ - 1}}\] exists only when\[\left| A \right| \ne 0\]. We have to remember the formula for\[{A^{ - 1}}\]. Sometimes students make mistakes while solving adjA and |A| for a 3×3 matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.
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 Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE 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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

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