
Let \[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\5&2&0\\{ - 1}&6&1\end{array}} \right]\] then the adjoint of \[{\rm{A}}\] is
A. \[\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]\]
В. \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\{ - 5}&{ - 2}&0\\1&{ - 6}&1\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\{ - 5}&{ - 2}&0\\1&{ - 6}&{ - 1}\end{array}} \right]\]
D. None of these
Answer
163.2k+ views
Hint:
We must first find the adjoint of a matrix by taking the cofactor of each element in the matrix and then taking the matrix's transpose. The adjoint of a matrix \[A = {(a_{ij})_{n \times n}}\] is defined as the transpose of the matrix \[{(A_{ij})_{n \times n}}\] where \[(A_{ij})\] is the element's cofactor \[(a_{ij})\]
Formula Used:
Cofactor can be found using formula:
\[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\]
Complete Step by Step Solution:
We have been provided in the question the \[3 \times 3\] matrix ‘A’
The \[3 \times 3\] matrix of A is
\[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\5&2&0\\{ - 1}&6&1\end{array}} \right]\]
First we have to determine the cofactor of the matrix
We have been already known that the cofactor matrix is made up of all the cofactors of the provided matrix, which are determined using the formula \[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\] where minor is \[{M_{ij}}\]
\[{C_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}2&0\\6&1\end{array}} \right| = 2\]
\[{C_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}5&0\\{ - 1}&1\end{array}} \right| = - 5\]
\[{C_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}5&2\\{ - 1}&6\end{array}} \right| = 32\]
\[{C_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{*{20}{c}}0&0\\6&1\end{array}} \right| = 0\]
\[{C_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right| = 1\]
\[{C_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{*{20}{c}}1&0\\{ - 1}&6\end{array}} \right| = - 6\]
\[{C_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{*{20}{c}}0&0\\2&0\end{array}} \right| = 0\]
\[{C_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{*{20}{c}}1&0\\5&0\end{array}} \right| = 0\]
\[{C_{33}} = {( - 1)^{3 - 3}}\left| {\begin{array}{*{20}{l}}1&0\\5&2\end{array}} \right| = 2\]
The cofactor matrix is \[\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]\]
\[ \Rightarrow {\mathop{\rm adj}\nolimits} (A) = {\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]^T}\]
Now, we have to determine the transpose of co-factor matrix
Here, rows becomes columns and columns becomes rows, we have
The matrix’s transpose is
\[ = \left[ {\begin{array}{*{20}{c}}2&0&0\\{ - 5}&1&0\\{32}&{ - 6}&2\end{array}} \right]\]
Therefore, the adjoint of \[{\rm{A}}\] is \[\left[ {\begin{array}{*{20}{c}}2&0&0\\{ - 5}&1&0\\{32}&{ - 6}&2\end{array}} \right]\]
Hence, the option D is correct
Note: In this type of question, we must first get the cofactor of each element, and then use transpose to find the adjoint of the provided matrix. Students must be mindful of signs when calculating cofactors and transposing the matrix. When the adjoint matrix is evaluated, the value must be the transpose of the obtained value. Another consideration is that anytime the adjoint matrix is assessed, we must ensure that the cofactor and minor of that matrix have the correct sign.
We must first find the adjoint of a matrix by taking the cofactor of each element in the matrix and then taking the matrix's transpose. The adjoint of a matrix \[A = {(a_{ij})_{n \times n}}\] is defined as the transpose of the matrix \[{(A_{ij})_{n \times n}}\] where \[(A_{ij})\] is the element's cofactor \[(a_{ij})\]
Formula Used:
Cofactor can be found using formula:
\[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\]
Complete Step by Step Solution:
We have been provided in the question the \[3 \times 3\] matrix ‘A’
The \[3 \times 3\] matrix of A is
\[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\5&2&0\\{ - 1}&6&1\end{array}} \right]\]
First we have to determine the cofactor of the matrix
We have been already known that the cofactor matrix is made up of all the cofactors of the provided matrix, which are determined using the formula \[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\] where minor is \[{M_{ij}}\]
\[{C_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}2&0\\6&1\end{array}} \right| = 2\]
\[{C_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}5&0\\{ - 1}&1\end{array}} \right| = - 5\]
\[{C_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}5&2\\{ - 1}&6\end{array}} \right| = 32\]
\[{C_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{*{20}{c}}0&0\\6&1\end{array}} \right| = 0\]
\[{C_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right| = 1\]
\[{C_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{*{20}{c}}1&0\\{ - 1}&6\end{array}} \right| = - 6\]
\[{C_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{*{20}{c}}0&0\\2&0\end{array}} \right| = 0\]
\[{C_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{*{20}{c}}1&0\\5&0\end{array}} \right| = 0\]
\[{C_{33}} = {( - 1)^{3 - 3}}\left| {\begin{array}{*{20}{l}}1&0\\5&2\end{array}} \right| = 2\]
The cofactor matrix is \[\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]\]
\[ \Rightarrow {\mathop{\rm adj}\nolimits} (A) = {\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]^T}\]
Now, we have to determine the transpose of co-factor matrix
Here, rows becomes columns and columns becomes rows, we have
The matrix’s transpose is
\[ = \left[ {\begin{array}{*{20}{c}}2&0&0\\{ - 5}&1&0\\{32}&{ - 6}&2\end{array}} \right]\]
Therefore, the adjoint of \[{\rm{A}}\] is \[\left[ {\begin{array}{*{20}{c}}2&0&0\\{ - 5}&1&0\\{32}&{ - 6}&2\end{array}} \right]\]
Hence, the option D is correct
Note: In this type of question, we must first get the cofactor of each element, and then use transpose to find the adjoint of the provided matrix. Students must be mindful of signs when calculating cofactors and transposing the matrix. When the adjoint matrix is evaluated, the value must be the transpose of the obtained value. Another consideration is that anytime the adjoint matrix is assessed, we must ensure that the cofactor and minor of that matrix have the correct sign.
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

JEE Electricity and Magnetism 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

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

Instantaneous Velocity - Formula based Examples for JEE

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

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
