Question

What is the determinant of an inverse matrix?

Answer 1

In the above question, We have to find the determinant of the inverse of that invertible matrix. Let a matrix 'A' of dimension (n x n) is called invertible only under the condition, if there exists another matrix B of the same dimension, such that $AB{\text{ }} = {\text{ }}BA{\text{ }} = {\text{ }}I$ , where I is the identity matrix of the same order. In order to approach the solution, we have to use some basic formulae related to matrices and determinants. The solution is given as follows.

Complete step-by-step answer:
Since we are given an invertible matrix, let a matrix $A$ be an invertible matrix of order $\left( {n \times n} \right)$ .
Now since $A$ is invertible, hence the inverse of $A$ is written as ${A^{ - 1}}$ .
Our aim is to find the determinant of the inverse matrix ${A^{ - 1}}$ , i.e. $det\left( {{A^{ - 1}}} \right)$ .
Since $A$ and ${A^{ - 1}}$ are inverse of each other,
Therefore, we have
$\Rightarrow A{A^{ - 1}} = I$
Where $I$ is the identity matrix of the same order as of $A$ and ${A^{ - 1}}$ .
Now taking determinants of both sides, we have
$\Rightarrow \det \left( {A{A^{ - 1}}} \right) = \det \left( I \right)$
We know the determinant identity that is given by,
$\Rightarrow \det \left( {A \cdot B} \right) = \det A \cdot \det B$
Therefore, using the above identity we can write,
$\Rightarrow \det \left( {A{A^{ - 1}}} \right) = \det A \cdot \det {A^{ - 1}} = \det I$
Or,
$\Rightarrow \det A \cdot \det {A^{ - 1}} = \det I$
Dividing both sides by $\det A$ we can write,
$\Rightarrow \det {A^{ - 1}} = \dfrac{{\det I}}{{\det A}}$
Since we know that the determinant of an identity matrix is $1$ , i.e. $\det I = 1$ ,
Therefore we have,
$\Rightarrow \det {A^{ - 1}} = \dfrac{1}{{\det A}}$
That is the required solution.
Therefore the determinant of an inverse matrix ${A^{ - 1}}$ is $\det {A^{ - 1}} = \dfrac{1}{{\det A}}$ .

Note: There is a condition for a matrix $A$ to be invertible and it is that the determinant of $A$ should not be equal to zero i.e. $A$ is invertible if and only if $\det A \ne 0$ .
Since, if the determinant of $A$ is $0$ then the determinant of the inverse matrix ${A^{ - 1}}$ is not defined.
As $\det {A^{ - 1}} = \dfrac{1}{{\det A}} = \dfrac{1}{0}$ which is undefined. Hence such matrices are not invertible.
A matrix that has no inverse is singular. When the determinant value of a square matrix is exactly zero the matrix is singular, otherwise invertible matrices are known as non-singular which have $\det A \ne 0$ .