
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
162.3k+ 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
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

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

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

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

NIT Cutoff Percentile for 2025

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: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
