Question

If a matrix is given as $2A=\left( \begin{matrix}
  2 & 1 \\
  3 & 2 \\
\end{matrix} \right)$ , then what is ${{A}^{-1}}$ equal to?
 (a). $\left( \begin{matrix}
   2 & -1 \\
   -3 & 2 \\
\end{matrix} \right)$
(b). $\dfrac{1}{2}\left( \begin{matrix}
   2 & -1 \\
   -3 & 2 \\
\end{matrix} \right)$
 (c). $\dfrac{1}{4}\left( \begin{matrix}
   2 & -1 \\
   -3 & 2 \\
\end{matrix} \right)$
 (d). None of the above

Answer 1

Hint: The inverse of a matrix A is the matrix which when multiplied with A gives the identity matrix I, which is given as $I=\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)$ , i.e., If $A\ .\ B\ =\ I\ \Rightarrow \ B\ =\ {{A}^{-1}}$ .
It is somewhat analogous to the multiplicative inverse of a number. Calculation of A is particularly important in solving simultaneous linear equations of several variables. To find ${{A}^{-1}}$ , we require a determinant of A and adjoint of A.

Complete step-by-step solution -
The inverse of the matrix $A$can be calculated as follows:
${{A}^{-1}}=\dfrac{adj(A)}{|A|}$
Where $adj(A)$ is adjoint of the matrix $A$ given by the transpose of the matrix of the cofactors of the elements of $A$ given by $C$such that the matrix element at ith row and jth column, ${{C}_{ij}}$is the cofactor of the matrix element of $A$given by ${{A}_{ij}}$ .
If $2A=\left( \begin{matrix}
   2 & 1 \\
   3 & 2 \\
\end{matrix} \right)$ then $A=\left( \begin{matrix}
   1 & \dfrac{1}{2} \\
   \dfrac{3}{2} & 1 \\
\end{matrix} \right)$
Since, ${{C}_{ij}}$is the co-factor of ${{A}_{ij}}$ , we can write
${{C}_{11}}={{\left( -1 \right)}^{2}}{{A}_{11}}=1$
${{C}_{12}}={{\left( -1 \right)}^{3}}{{A}_{12}}=\dfrac{-3}{2}$
${{C}_{21}}={{\left( -1 \right)}^{3}}{{A}_{21}}=\dfrac{-1}{2}$
${{C}_{22}}={{\left( -1 \right)}^{4}}{{A}_{22}}=1$
Therefore, $C=\left( \begin{matrix}
   1 & -\dfrac{3}{2} \\
   -\dfrac{1}{2} & 1 \\
\end{matrix} \right)$
$adj(A)={{C}^{T}}=\left( \begin{matrix}
   1 & -\dfrac{1}{2} \\
   -\dfrac{3}{2} & 1 \\
\end{matrix} \right)$
Now, we calculate determinant of $A$

$\left| A \right|=\left| \begin{matrix}
   1 & \dfrac{1}{2} \\
   \dfrac{3}{2} & 1 \\
\end{matrix} \right|=\dfrac{1}{4}$
Thus, ${{A}^{-1}}=\ 4.\left( \begin{matrix}
   1 & -\dfrac{1}{2} \\
   -\dfrac{3}{2} & 1 \\
\end{matrix} \right)=4.\dfrac{1}{2}.\left( \begin{matrix}
   2 & -1 \\
   -3 & 2 \\
\end{matrix} \right)=2\left( \begin{matrix}
   2 & -1 \\
   -3 & 1 \\
\end{matrix} \right)$
Hence, the correct answer is (d).

Note: To avoid any mistake it is advisable to multiply the constant term outside the matrix $A$with the matrix. Alternatively, one should remember as a rule that if
 $X=k\ .\ Y$
where X and Y are matrices and k is a constant, then
$|X|\ =\ {{k}^{n}}\ |Y|$ where n is the order of the matrix.
Another source of mistake is the calculation of adjoint matrix $A$. Sometimes, when we are not careful we tend to forget the step where we have to take the transpose of the matrix and use $C$instead as an adjoint. It is a very common source of error.