Question

If A is an invertible matrix, then ${{\left( adjA \right)}^{-1}}$ is equal to (This question has multiple correct options)
A. $adj.\left( {{A}^{-1}} \right)$
B. $\dfrac{A}{\det .A}$
C. $A$
D. $\left( \det .A \right)A$

Answer 1

Hint: To obtain the value of ${{\left( adjA \right)}^{-1}}$ we will use the relation of adjugate with its inverse for an invertible matrix. Firstly we will write the formula to find the inverse of a matrix as it contains adjacent matrices also. Then by using some simplification we will show that ${{\left( adjA \right)}^{-1}}$ is equal to one of the options given. Then by using the formula of adjacent of a matrix we will get that${{\left( adjA \right)}^{-1}}=adj.\left( {{A}^{-1}} \right)$.

Complete step-by-step solution:
As it is given that $A$ is an invertible matrix so we can write inverse of $A$ as below:
${{A}^{-1}}=\dfrac{1}{\det \left( A \right)}adj\left( A \right)$
Now, let us multiply $A$ on both side of the above equation:
$\begin{align}
  & \Rightarrow A{{A}^{-1}}=A\dfrac{1}{\det \left( A \right)}adj\left( A \right) \\
 & \Rightarrow I=\dfrac{A}{\det \left( A \right)}adj\left( A \right) \\
 & \Rightarrow \dfrac{1}{adj\left( A \right)}=\dfrac{A}{\det \left( A \right)} \\
 & \therefore {{\left( adjA \right)}^{-1}}=\dfrac{A}{\det \left( A \right)} \\
\end{align}$
Next we will show that
${{\left( adjA \right)}^{-1}}=adj.\left( {{A}^{-1}} \right)$…….$\left( 1 \right)$
So as we know that ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)$
So that means:
${{\left( adjA \right)}^{-1}}=\dfrac{1}{\left| adjA \right|}adj\left( adjA \right)$
Using $adjA={{A}^{-1}}\left| A \right|$ in above equation we get:
$\begin{align}
  & \Rightarrow {{\left( adjA \right)}^{-1}}=\dfrac{{{\left| A \right|}^{n-2}}A}{{{\left| A \right|}^{n-1}}} \\
 & \Rightarrow {{\left( adjA \right)}^{-1}}=\dfrac{A}{{{\left| A \right|}^{n-1-n+2}}} \\
\end{align}$
$\therefore {{\left( adjA \right)}^{-1}}=\dfrac{A}{\left| A \right|}$…….$\left( 2 \right)$
Next we will solve right side of equation (1)
$\left( adj{{A}^{-1}} \right)={{\left( {{A}^{-1}} \right)}^{-1}}\left| {{A}^{-1}} \right|$
$\Rightarrow \left( adj{{A}^{-1}} \right)=A{{\left| A \right|}^{-1}}$
$\therefore \left( adj{{A}^{-1}} \right)=\dfrac{A}{\left| A \right|}$…….$\left( 3 \right)$
So from equation (2) and (3) ${{\left( adjA \right)}^{-1}}=adj.\left( {{A}^{-1}} \right)$
Hence, option (A) and (B) are correct.

Note: An n-n square matrix $A$ is known as invertible if there exists a matrix $B$ such that it holds the property $AB=BA=I$ where $I$ is an identity matrix. A square matrix which is not invertible is known as a singular matrix. There are various properties of a matrix inverse of an inverse matrix is the matrix itself. Then if a constant is multiplied by a matrix and we find their inverse we can separate the constant and the matrix with inverse power separately. Determinant of an inverse matrix is equal to the determinant of that matrix.