Question

If \[A = \left[ {\begin{array}{*{20}{c}}1&2\\3&{ - 5}\end{array}} \right]\], then find the value of \[{A^{ - 1}}\].
A. \[\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 2}\\{ - 3}&1\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}{\dfrac{5}{{11}}}&{\dfrac{2}{{11}}}\\{\dfrac{3}{{11}}}&{\dfrac{{ - 1}}{{11}}}\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}{ - 2}&4\\{ - 3}&6\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}5&2\\3&{ - 1}\end{array}} \right]\]

Answer 1

Hint: A \[2 \times 2\] matrix is given. First, calculate the determinant of the matrix \[A\]. If the value of the determinant is non-zero, then calculate the adjoint matrix of the given matrix and substitute the values in the formula for the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\] to get the required answer. If the value of the determinant is zero, then declare that the inverse matrix does not exist.

Formula used:
The adjoint matrix of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[adj A = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
The determinant of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[\left| A \right| = ad - bc\]
The inverse matrix: \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]

Complete step by step solution:
The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}1&2\\3&{ - 5}\end{array}} \right]\].

Let’s calculate the determinant of the given matrix \[A\].
Apply the formula for the determinant of a \[2 \times 2\] matrix.
We get,
\[\left| A \right| = 1 \times \left( { - 5} \right) - 3 \times 2\]
\[ \Rightarrow \left| A \right| = - 5 - 6\]
\[ \Rightarrow \left| A \right| = - 11\] \[.....\left( 1 \right)\]
Since the determinant is a non-zero value. So, the inverse matrix exists for the given matrix \[A\].

Now find out the adjoint matrix of the given matrix \[A\].
Apply the rule for the adjoint matrix of a \[2 \times 2\] matrix.
We get,
\[adj A = \left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 2}\\{ - 3}&1\end{array}} \right]\] \[.....\left( 2 \right)\]

Substitute the values of the equation \[\left( 1 \right)\] and \[\left( 2 \right)\] in the formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\].
We get,
\[{A^{ - 1}} = \dfrac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 2}\\{ - 3}&1\end{array}} \right]\]
Apply the property of scalar multiplication.
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\dfrac{5}{{11}}}&{\dfrac{2}{{11}}}\\{\dfrac{3}{{11}}}&{\dfrac{{ - 1}}{{11}}}\end{array}} \right]\]
Hence the correct option is B.

Note: Students should keep in mind that if the determinant of a matrix is 0, then \[\dfrac{1}{{det A}}\] is undefined. So, the matrix with a 0 determinant has no inverse. And the adjoint matrix of any matrix is the transpose of its cofactor matrix. But for a \[2 \times 2\] matrix, we don’t need to calculate the cofactor matrix.
So, to find the adjoint matrix of a \[2 \times 2\] matrix:
Interchange the elements of the principal diagonal.
Change the signs of the elements of the other diagonal.