Question

If A is an invertible matrix of order 3$\times $3 such that $\left| \text{A} \right|=5$, find the value of $\left| {{A}^{-1}} \right|$.

Answer 1

Hint: To solve the question, we should know the relation between the determinants of the matrices A and A-1. The relation is
\[\begin{align}
  & {{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \\
 & \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \right| \\
\end{align}\]
We know the property that
$\left| kA \right|={{k}^{n}}\left| A \right|$ where n is the order of the matrix.
\[\left| {{A}^{-1}} \right|=\dfrac{1}{{{\left| A \right|}^{n}}}\left| \left( adj\left( A \right) \right) \right|\]
We know the relation that
$\left| \left( adjA \right) \right|={{\left| A \right|}^{n-1}}$
Using the property, we get that $\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}$, we get the answer.

Complete step-by-step answer:
In the question, we are given the value of $\left| \text{A} \right|=5$ and we are asked to find the value of $\left| {{A}^{-1}} \right|$. To do that, we have to find the relation between the two values $\left| A \right|$ and $\left| {{A}^{-1}} \right|$. From the fundamental properties of inverse of the matrix, we can write that
\[\begin{align}
  & {{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \\
 & \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \right| \\
\end{align}\]
We also know the property of the determinant of the matrix when a constant is multiplied to it. Here, $\left| A \right|$ is a constant, but not a matrix. The relation is
If n is the order of the matrix and determinant of kA is written as
$\left| kA \right|={{k}^{n}}\left| A \right|$.
As the order of matrix A is n, the order of matrix which is adjoint of A is also n. Applying it in the above equation, we get
\[\left| {{A}^{-1}} \right|=\dfrac{1}{{{\left| A \right|}^{n}}}\left| \left( adj\left( A \right) \right) \right|\]
The relation between the determinants of A and adjoint of A is given by
$\left| \left( adjA \right) \right|={{\left| A \right|}^{n-1}}$
Using this in the above relation, we get
\[\begin{align}
  & \left| {{A}^{-1}} \right|=\dfrac{{{\left| A \right|}^{n-1}}}{{{\left| A \right|}^{n}}} \\
 & \left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\to \left( 1 \right) \\
\end{align}\]
In the question, it is given that $\left| \text{A} \right|=5$. Substituting the value of determinant of A in equation-1, we get
\[\left| {{A}^{-1}} \right|=\dfrac{1}{5}\]
$\therefore $The required value is \[\left| {{A}^{-1}} \right|=\dfrac{1}{5}\].

Note: This problem can be done in an alternative way. We know the relation between the product of matrices and its inverse. That is, for a square invertible matrix A of order n$\times $n
$A\times {{A}^{-1}}=I$
Applying determinant on both sides, we get
$\left| A\times {{A}^{-1}} \right|=\left| I \right|$
We know that the determinant of an identity matrix is 1.
$\left| A\times {{A}^{-1}} \right|=1$
From the relation$\left| A\times B \right|=\left| A \right|\times \left| B \right|$, we get
$\left| A \right|\times {{\left| A \right|}^{-1}}=1$
From the above relation, we can write
\[\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\]
Using the value of $\left| \text{A} \right|=5$, we get \[\left| {{A}^{-1}} \right|=\dfrac{1}{5}\] which is the required answer.