Question

Find the inverse of the matrix given below
\[\left| {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 1}&2
\end{array}} \right|\]

Answer 1

 Hint: Inverse of the matrix is the relation of the adjoint and determinant value of the matrix, which is represented as\[{A^{ - 1}} = \dfrac{{adj(A)}}{{\det (A)}}\] where the determinant and adjoint are calculated individually and are kept in the formula. The given matrix is a \[2 \times 2\]matrix that contains four elements in it.

Complete step-by-step answer:
A square matrix is the one that has the same number of rows as well as columns. The standard $2 \times 2$ square matrix is\[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right|\].
Compare the given matrix \[A = \left| {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 1}&2
\end{array}} \right|\] with the standard $2 \times 2$ matrix\[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right|\].
By equating, we get the value as\[{\text{a = 3, b = - 5, c = - 1, d = 2}}\].
A determinant of a matrix is the scalar value of a matrix that can be computed from elements of a matrix in $\left( {ad - bc} \right)$ form. The matrix which has its determinant value as 0 is known as Singular or Non-invertible matrix.
First, we find the determinant value of the matrix,
Let \[\det ({\text{A}})\]represents the determinant value of the matrix \[A = \left| {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 1}&2
\end{array}} \right|\]
\[
  \det A = \left[ {\left( {a \times d} \right) - \left( {b \times c} \right)} \right] \\
   = \left[ {\left( {3 \times 2} \right) - \left( {\left( { - 1} \right) \times \left( { - 5} \right)} \right)} \right] \\
   = \left[ {6 - 5} \right] \\
   = 1 \\
 \]
Now the next step for the inverse of the matrix is to find the adjoint value of the matrix
\[A = \left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right|\]
In general, adjoint of the matrix is the transpose of the cofactor of the matrix. To find the adjoint of a $2 \times 2$ square matrix, swap the positions of a and d, and put a negative sign in front of b and c in the matrix
\[adjA = {\left| {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right|^{}}\]
We get \[a = 2,b = 5,c = 1,d = 3\]
So we can write adjoint of the matrix \[A = \left| {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 1}&2
\end{array}} \right|\] as, \[adjA = {\left| {\begin{array}{*{20}{c}}
  2&5 \\
  1&3
\end{array}} \right|^{}}\]
Finally, substitute \[adjA = {\left| {\begin{array}{*{20}{c}}
  2&5 \\
  1&3
\end{array}} \right|^{}}\] and $\det (A) = 1$ in the formula \[{A^{ - 1}} = \dfrac{{adj(A)}}{{\det (A)}}\] to determine the inverse of the matrix.
\[
  {A^{ - 1}} = \dfrac{{adj(A)}}{{\det (A)}} \\
   = \dfrac{1}{{\left| 1 \right|}}\left| {\begin{array}{*{20}{c}}
  2&5 \\
  1&3
\end{array}} \right| \\
   = \left| {\begin{array}{*{20}{c}}
  {\dfrac{2}{1}}&{\dfrac{5}{1}} \\
  {\dfrac{1}{1}}&{\dfrac{3}{1}}
\end{array}} \right| \\
   = \left| {\begin{array}{*{20}{c}}
  2&5 \\
  1&3
\end{array}} \right| \\
 \]
Hence, the inverse of the matrix \[A = \left| {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 1}&2
\end{array}} \right|\] is \[{A^{ - 1}} = \left| {\begin{array}{*{20}{c}}
  2&5 \\
  1&3
\end{array}} \right|\]

Note: Inverse of a matrix A is applicable only when it satisfies the condition \[A \times {A^{ - 1}} = {A^{ - 1}} \times A = I\] where \[I\] represents the singular matrix.