Q.If $A$ is an invertible matrix of order $3 \times 3$, then $|A^{-1}| =$ ________ .

The key idea is that the determinant of the inverse is the reciprocal of the determinant of the original matrix.

Step 1: For any invertible matrix $A$, we have $A A^{-1} = I$.

Step 2: Take determinants on both sides: $|A A^{-1}| = |I|$.

Step 3: Using $|AB| = |A||B|$ and $|I| = 1$, we get $|A| \, |A^{-1}| = 1$. …

The key idea here is a fundamental property linking the determinant of a matrix and its inverse. When a matrix is invertible, its determinant is non-zero, and the inverse matrix "undoes" the original transformation. The determinant measures how much a matrix scales area (or volume), so the inverse must scale by the reciprocal factor.

Let's see why this works step by step.

1. Start with the definition of an inverse.

   If $A$ is invertible, then by definition $A A^{-1} = I$, where $I$ is the identity matrix of the same order (here $3 \times 3$).

2. Take determinants on both sides.

   The determinant of a product equals the product of the determinants. So:

3. Apply the product property. This gives:

4. Recall the determinant of the identity matrix.

   For any $n \times n$ identity matrix, $|I| = 1$. For a $3 \times 3$ identity, this is still $1$.

5. Solve for $|A^{-1}|$. …

Method: Finding the Determinant of a Matrix Inverse from the Determinant Product Rule

Whenever a question asks for $|A^{-1}|$ in terms of $|A|$, don't try to build $A^{-1}$ explicitly — start from the defining equation $AA^{-1}=I$ and take determinants of both sides.

Steps

Step 1: Write down the defining relationship

For any invertible matrix $A$, by definition

This is true regardless of the size of $A$ — the method doesn't depend on the matrix being $3\times3$ specifically.

Step 2: Take the determinant of both sides and apply the product rule

Using $|PQ| = |P|\,|Q|$ for any two square matrices of the same order: …

Common Mistakes

Mistake 1: Confusing this with the $|kA|=k^n|A|$ scaling rule

Why it's wrong: $A^{-1}$ is not "$A$ scaled by a constant" — it's a genuinely different matrix defined by $AA^{-1}=I$, so the order $n$ of the matrix plays no role in $|A^{-1}|$ the way it does for $|kA|$. A student who has just worked a "$|3A|=3^n|A|$"-style question may wrongly expect an exponent here too. Correct approach: derive $|A^{-1}|$ fresh from $AA^{-1}=I$ each time rather than pattern-matching to the scalar-multiplication rule.

Mistake 2: Writing $|A^{-1}| = -|A|$ or $|A^{-1}| = |A|$ …