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Question: If A is a square matrix such that \({{A}^{3}}=I\) then \({{A}^{-1}}\) is equal to (A) \(I\) (B) ...

If A is a square matrix such that A3=I{{A}^{3}}=I then A−1{{A}^{-1}} is equal to
(A) II
(B) AA
(C) A2{{A}^{2}}
(D) None of these

Explanation

Solution

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

Complete step-by-step answer :
We are given that A is a square matrix.
We are also given that A3=I{{A}^{3}}=I, where II is an identity matrix.
Now let us multiply the given equation with the inverse of A, that is A−1{{A}^{-1}}. So, multiplying with A−1{{A}^{-1}} we get,
A−1A3=A−1I.........(1){{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−1A3{{A}^{-1}}{{A}^{3}}.
Now let us consider a property of matrices.
A(BC)=(AB)CA\left( BC \right)=\left( AB \right)C
Using this property, we can write A−1A3{{A}^{-1}}{{A}^{3}} as,
⇒A−1A3=A−1AA2=(A−1A)A2\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−1A=I{{A}^{-1}}A=I
Using the above property, we can write A−1A3{{A}^{-1}}{{A}^{3}} as,
⇒A−1A3=(A−1A)A2 ⇒A−1A3=(I)A2 \begin{aligned} & \Rightarrow {{A}^{-1}}{{A}^{3}}=\left( {{A}^{-1}}A \right){{A}^{2}} \\\ & \Rightarrow {{A}^{-1}}{{A}^{3}}=\left( I \right){{A}^{2}} \\\ \end{aligned}
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=AAI=IA=A
So, we get that
⇒A−1A3=A2.......(2)\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−1I{{A}^{-1}}I.
Using the same above property we can write it as,
⇒A−1I=A−1...........(3)\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,
⇒A−1A3=A−1I ⇒A2=A−1 \begin{aligned} & \Rightarrow {{A}^{-1}}{{A}^{3}}={{A}^{-1}}I \\\ & \Rightarrow {{A}^{2}}={{A}^{-1}} \\\ \end{aligned}
So, we get the value of inverse of A as,
⇒A−1=A2\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−1A=I{{A}^{-1}}A=I
Now let us multiply the above equation with A2{{A}^{2}} on both sides. Then we get,
⇒A−1AA2=IA2 ⇒A−1A3=A2 \begin{aligned} & \Rightarrow {{A}^{-1}}A{{A}^{2}}=I{{A}^{2}} \\\ & \Rightarrow {{A}^{-1}}{{A}^{3}}={{A}^{2}} \\\ \end{aligned}
As we are given that A3=I{{A}^{3}}=I, let us substitute it in the above equation.
⇒A−1(I)=A2 ⇒A−1=A2 \begin{aligned} & \Rightarrow {{A}^{-1}}\left( I \right)={{A}^{2}} \\\ & \Rightarrow {{A}^{-1}}={{A}^{2}} \\\ \end{aligned}
Hence the answer is Option C.