
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
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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 \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}\\{{A_{12}}}&{{A_{22}}}\end{array}} \right]\].
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