Q.If 𝑨 and 𝑩 are non-singular matrices of same order with 𝒅𝒆𝒕(𝑨) = 𝟓, then [𝒅𝒆𝒕(𝑩−𝟏𝑨𝑩)]² is equal to
(A) 5
(B) 25
(C) 45
(D) 55

The key idea is that the determinant is invariant under similarity transformations: for any invertible $B$, $\det(B^{-1}AB) = \det(A)$.

Step 1: Simplify the expression inside the determinant.

We have $\det\left( (B^{-1}AB)^2 \right)$.

Step 2: Use the property $\det(X^2) = (\det X)^2$ for any square matrix $X$.

So $\det\left( (B^{-1}AB)^2 \right) = \left[ \det(B^{-1}AB) \right]^2$. …

The key idea here is determinant similarity invariance. When you multiply a matrix on the left by $B^{-1}$ and on the right by $B$, you are performing a similarity transformation. The determinant of a product is the product of determinants, and the determinant of $B^{-1}$ is $1/\det(B)$. So the $B$ and $B^{-1}$ cancel out, leaving only $\det(A)$. This is a powerful shortcut — you never need to know what $A$ or $B$ actually are.

Let’s walk through it step by step.

1. Start with the expression inside the square.

   We need $\det(B^{-1}AB)$. Since $A$ and $B$ are non-singular (determinants are non-zero), all inverses exist.

2. Use the product rule for determinants.

   For any square matrices $X$ and $Y$ of the same order, $\det(XY) = \det(X) \cdot \det(Y)$. Applying this:

3. Recall the determinant of an inverse. For any invertible matrix $B$, $\det(B^{-1}) = \frac{1}{\det(B)}$. So:

4. Cancel $\det(B)$. Since $\det(B) \neq 0$, the $\det(B)$ in numerator and denominator cancel:

5. Now square the result. The problem asks for $\det(B^{-1}AB)^2$, which means: …

Method: Multiplicativity and Inverse Rule to Simplify a Conjugated Determinant

This method solves any question where a determinant of a product/conjugation of matrices (such as $B^{-1}AB$) needs to be reduced to a determinant of a single known matrix, using only the standard CBSE algebraic rules for determinants — no deeper linear-algebra theory required.

Steps

Step 1: Split the product using the multiplicative property

For square matrices of the same order, $\det(XY)=\det(X)\det(Y)$. Apply this repeatedly to break the expression into separate determinant factors:

Step 2: Replace det(B⁻¹) using the inverse-determinant rule

Use $\det(B^{-1})=\dfrac{1}{\det(B)}$ (valid since $B$ is non-singular) to rewrite the expression as …

Common Mistakes

Mistake 1: Assuming the determinant must change because the matrix itself changes under similarity

Why it's wrong: $B^{-1}AB$ is generally a different matrix from $A$ (different entries), so students sometimes assume its determinant must also differ from $\det(A)$. Correct approach: remember similarity transformations always preserve the determinant — $\det(B^{-1}AB)=\det(A)$ regardless of what $B$ is, because $\det(B^{-1})$ and $\det(B)$ always cancel.

Mistake 2: Forgetting that $\det(B^{-1})=1/\det(B)$ before cancelling …