Question

If A is a square matrix such that ${{A}^{3}}=I$ then ${{A}^{-1}}$ is equal to
(A) $I$
(B) $A$
(C) ${{A}^{2}}$
(D) None of these

Answer 1

Hint: We solve this problem by first considering the given equation then multiplying it with the matrix ${{A}^{-1}}$. Then we use the properties of matrices, $A\left( BC \right)=\left( AB \right)C$, ${{A}^{-1}}A=I$ and $AI=IA=A$. Then we use them and simplify the obtained equation to find the value of the inverse of A, that is ${{A}^{-1}}$.

Complete step-by-step answer:
We are given that A is a square matrix.
We are also given that ${{A}^{3}}=I$, where $I$ is an identity matrix.
Now let us multiply the given equation with the inverse of A, that is ${{A}^{-1}}$. So, multiplying with ${{A}^{-1}}$ we get,
${{A}^{-1}}{{A}^{3}}={{A}^{-1}}I.........\left( 1 \right)$
Now let us consider the left-hand side of the equation (1), that is ${{A}^{-1}}{{A}^{3}}$.
Now let us consider a property of matrices.
$A\left( BC \right)=\left( AB \right)C$
Using this property, we can write ${{A}^{-1}}{{A}^{3}}$ as,
$\Rightarrow {{A}^{-1}}{{A}^{3}}={{A}^{-1}}A{{A}^{2}}=\left( {{A}^{-1}}A \right){{A}^{2}}$
Now, let us consider another property of matrices.
${{A}^{-1}}A=I$
Using the above property, we can write ${{A}^{-1}}{{A}^{3}}$ as,
$\begin{align}
  & \Rightarrow {{A}^{-1}}{{A}^{3}}=\left( {{A}^{-1}}A \right){{A}^{2}} \\
 & \Rightarrow {{A}^{-1}}{{A}^{3}}=\left( I \right){{A}^{2}} \\
\end{align}$
Now let us consider another property of matrices that when a matrix is multiplied to identity matrix, we get the same matrix, that is
$AI=IA=A$
So, we get that
$\Rightarrow {{A}^{-1}}{{A}^{3}}={{A}^{2}}.......\left( 2 \right)$
Now, let us consider the right-hand side of the equation (1), that is ${{A}^{-1}}I$.
Using the same above property we can write it as,
$\Rightarrow {{A}^{-1}}I={{A}^{-1}}...........\left( 3 \right)$
So, substituting the values obtained in equation (2) and equation (3) in the equation (1) we get,
$\begin{align}
  & \Rightarrow {{A}^{-1}}{{A}^{3}}={{A}^{-1}}I \\
 & \Rightarrow {{A}^{2}}={{A}^{-1}} \\
\end{align}$
So, we get the value of inverse of A as,
$\Rightarrow {{A}^{-1}}={{A}^{2}}$
Hence the answer is Option C.

Note: We can also solve this problem in another method.
Let us consider a property of matrices.
${{A}^{-1}}A=I$
Now let us multiply the above equation with ${{A}^{2}}$ on both sides. Then we get,
$\begin{align}
  & \Rightarrow {{A}^{-1}}A{{A}^{2}}=I{{A}^{2}} \\
 & \Rightarrow {{A}^{-1}}{{A}^{3}}={{A}^{2}} \\
\end{align}$
As we are given that ${{A}^{3}}=I$, let us substitute it in the above equation.
$\begin{align}
  & \Rightarrow {{A}^{-1}}\left( I \right)={{A}^{2}} \\
 & \Rightarrow {{A}^{-1}}={{A}^{2}} \\
\end{align}$
Hence the answer is Option C.