Q.$|A^{-1}| \neq |A|^{-1}$, where $A$ is a non-singular matrix.

The key idea is the inverse of a product property: for any two non-singular matrices $A$ and $B$, $(AB)^{-1} = B^{-1}A^{-1}$. The determinant of a product is the product of determinants: $|AB| = |A||B|$.

Take determinants on both sides of $(AB)^{-1} = B^{-1}A^{-1}$:

But also $|(AB)^{-1}| = \frac{1}{|AB|} = \frac{1}{|A||B|}$.

Equating the two expressions gives:

$|A^{-1}||B^{-1}| = \frac{1}{|A||B|}.$ …

The claim in the question is a common trap. Let’s see why it’s wrong — and why the correct equality holds.

1. The core idea: determinant of a product

For any two square matrices $A$ and $B$ of the same order, we have the fundamental property:

This is not a coincidence — it reflects how determinants measure scaling of volume under linear transformations. When you apply $A$ then $B$, the total volume scaling factor is the product of the individual scaling factors.

2. Apply this to $A$ and $A^{-1}$

Since $A$ is non-singular, its inverse $A^{-1}$ exists and satisfies:

Take determinants on both sides:

Using the product property on the left:

3. What is $|I|$?

The identity matrix $I$ has 1’s on the diagonal and 0’s elsewhere. Its determinant is $1$ (the product of the diagonal entries). So:

4. Solve for $|A^{-1}|$

Since $|A| \neq 0$ (that’s what non-singular means), we can divide:

This is an exact equality, not an inequality.

Method: Deriving the Relationship Between $|A^{-1}|$ and $|A|^{-1}$

Use this method for any question about the determinant of an inverse matrix — derive the relationship from the defining equation $AA^{-1}=I$ rather than memorising it blindly.

Steps

Step 1: Start from the definition of the inverse

This is what "inverse" means — it is always your starting point for any inverse-determinant question.

Step 2: Take the determinant of both sides

Step 3: Apply the determinant-of-a-product rule

using $|XY| = |X||Y|$ for square matrices of the same order.

Step 4: Evaluate $|I|$ and solve

The identity matrix has determinant $1$ (product of its diagonal $1$'s), so:

$|A|\cdot|A^{-1}| = 1 \quad\Longrightarrow\quad |A^{-1}| = \frac{1}{|A|} = |A|^{-1}$ …

Common Mistakes

Mistake 1: Confusing $|A^{-1}|$ (determinant of the inverse matrix) with $(|A|)^{-1}$ (reciprocal of the determinant)

Why it's wrong: the notation looks similar, and a student may think these are unrelated quantities being compared, when in fact the whole point of the identity is that they are provably EQUAL. Correct approach: derive the relationship from $AA^{-1}=I$ each time rather than guessing — it always gives $|A^{-1}|=|A|^{-1}$.

Mistake 2: Accepting the given inequality at face value instead of deriving the true relationship …