
What is the inverse of \[A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\]?
A. \[\left( {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right)\]
B. \[\dfrac{1}{{\left( {ad - bc} \right)}}\left( {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right)\]
C. \[\dfrac{1}{{\left| A \right|}}\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\]
D. \[\left( {\begin{array}{*{20}{c}}b&{ - a}\\d&{ - c}\end{array}} \right)\]
Answer
233.1k+ views
Hint: We will find the cofactors of the given matrix. Using the cofactors, we will find the adjoint of the given matrix. Then we will calculate the determinate of the given matrix. Then put the adjoint matrix and determinant in the formula of the inverse matrix.
Formula used:
The adjoint of the matrix \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\] is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\]where \[{A_{ij}}\] are the cofactors.
The inverse formula of matrix A is \[A = \dfrac{1}{{\left| A \right|}}Adj\,A\].
Complete Step by step solution:
Given matrix is \[A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\]
The cofactors of the matrix are
\[{A_{11}} =(-1)^{1+1} d = d\]
\[{A_{12}} = (-1)^{1+2}c = -c\]
\[{A_{21}} = (-1)^{2+1} b = -b\]
\[{A_{22}} = (-1)^{2+2}a = a\]
The adjoint of the given matrix is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\]\[ = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
Now we will calculate the determinate of the given matrix
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right|\]
\[ \Rightarrow \left| A \right| = ad - bc\]
Now we will substitute the adjoint matrix and determinate of A in the formula \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}Adj\,A\].
\[{A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
Hence option B is the correct option.
Additional information:
The inverse of a matrix exists if and only if the matrix is a non-singular matrix. In other words, if the determinate of a matrix is not equal to zero, then the inverse of the matrix exists.
Note: Students do a mistake to calculate the adjoint matrix. We have to transpose the row and column of the cofactor to find the adjoint of the matrix. But students forgot to transpose the rows and columns that is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\]. The correct formula is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\].
Formula used:
The adjoint of the matrix \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\] is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\]where \[{A_{ij}}\] are the cofactors.
The inverse formula of matrix A is \[A = \dfrac{1}{{\left| A \right|}}Adj\,A\].
Complete Step by step solution:
Given matrix is \[A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\]
The cofactors of the matrix are
\[{A_{11}} =(-1)^{1+1} d = d\]
\[{A_{12}} = (-1)^{1+2}c = -c\]
\[{A_{21}} = (-1)^{2+1} b = -b\]
\[{A_{22}} = (-1)^{2+2}a = a\]
The adjoint of the given matrix is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\]\[ = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
Now we will calculate the determinate of the given matrix
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right|\]
\[ \Rightarrow \left| A \right| = ad - bc\]
Now we will substitute the adjoint matrix and determinate of A in the formula \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}Adj\,A\].
\[{A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
Hence option B is the correct option.
Additional information:
The inverse of a matrix exists if and only if the matrix is a non-singular matrix. In other words, if the determinate of a matrix is not equal to zero, then the inverse of the matrix exists.
Note: Students do a mistake to calculate the adjoint matrix. We have to transpose the row and column of the cofactor to find the adjoint of the matrix. But students forgot to transpose the rows and columns that is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\]. The correct formula is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\].
Recently Updated Pages
JEE Advanced 2026 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced 2021 Chemistry Question Paper 1 with Solutions

JEE Advanced 2022 Physics Question Paper 2 with Solutions

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2021 Chemistry Question Paper 2 with Solutions

JEE Advanced 2022 Maths Question Paper 2 with Solutions

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

Top IIT Colleges in India 2025

Understanding the Mechanisms and Key Differences in SN1 and SN2 Reactions

Difference Between Line Voltage and Phase Voltage

JEE Advanced Previous Year Question Papers [2012 - 2025]

JEE Advanced Syllabus 2026

Other Pages
JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding How a Current Loop Acts as a Magnetic Dipole

Understanding Average and RMS Value in Electrical Circuits

Understanding Collisions: Types and Examples for Students

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

