Q.If $A$ is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to (A) $\det(A)$ (B) $\dfrac{1}{\det(A)}$ (C) $1$ (D) $0$

Concept: Determinant of an inverse matrix.

For any invertible matrix $A$, we know $A A^{-1} = I$. Taking determinants on both sides:

Since $\det(AB) = \det(A) \det(B)$ and $\det(I) = 1$, we get:

Therefore:

$\det(A^{-1}) = \frac{1}{\det(A)}$ …

The key here is a fundamental property connecting the determinant of a matrix and its inverse. If you understand why this property holds, you never need to memorise it — it follows directly from the definition of an inverse.

The core idea: For any invertible matrix $A$, we have $A A^{-1} = I$, where $I$ is the identity matrix. Taking determinants on both sides gives $\det(A A^{-1}) = \det(I)$. Since the determinant of a product is the product of the determinants, and $\det(I) = 1$, we get $\det(A) \cdot \det(A^{-1}) = 1$. Rearranging gives $\det(A^{-1}) = \frac{1}{\det(A)}$.

This reasoning works for any square matrix, not just order 2. The order being 2 is just a detail — the property is universal.

Let’s walk through it step by step.

1. Start with the definition of an inverse. If $A$ is invertible, there exists a matrix $A^{-1}$ such that

where $I$ is the identity matrix of the same order (here, $2 \times 2$).

2. Take the determinant of both sides. The determinant is a function that respects multiplication: for any two square matrices $X$ and $Y$ of the same order,

Applying this to our equation:

3. Use the product property. The left side becomes $\det(A) \cdot \det(A^{-1})$. The right side is $\det(I)$. For a $2 \times 2$ identity matrix,

and its determinant is $1 \cdot 1 - 0 \cdot 0 = 1$. So we have:

4. Solve for $\det(A^{-1})$. …

Method: Determinant of an Inverse Matrix

This method derives the relationship between $\det(A^{-1})$ and $\det(A)$ from the defining property of an inverse, so it applies to a matrix of any order, not just order 2.

Steps

Step 1: Start from the defining relation $AA^{-1}=I$

Every invertible matrix satisfies this by definition.

Step 2: Take the determinant of both sides

Use $\det(XY)=\det(X)\det(Y)$ on the left, and $\det(I)=1$ on the right:

Step 3: Solve for $\det(A^{-1})$ …

Common Mistakes

Mistake 1: Choosing $\det(A)$ itself instead of its reciprocal

Why it's wrong: option (A) tempts students who recall "there's a relationship between $\det A$ and $\det A^{-1}$" but misremember which way it goes — the correct relationship, from $\det(A)\det(A^{-1})=\det(I)=1$, is a reciprocal, not equality. Correct approach: re-derive it from $AA^{-1}=I$ and the product rule for determinants each time, rather than recalling the answer from memory.

Mistake 2: Assuming a sign change is involved …