GATE CS 2004 Q27 — Linear Algebra: Matrices, Determinants, System of Linear Equations

We are given that $A, B, C,$ and $D$ are $n \times n$ matrices, each with a non-zero determinant. This means all four matrices are invertible (i.e., their inverses exist).

We are also given:

We want to find the inverse of $B$, denoted as $B^{-1}$. We can solve this using two straightforward methods.

Method 1: Direct Algebraic Manipulation

To isolate $B$, we can pre-multiply and post-multiply both sides of $ABCD = I$ by the respective inverse matrices:

1. Pre-multiply by $A^{-1}$: $$A^{-1}(ABCD) = A^{-1}I \implies BCD = A^{-1}$$
2. Post-multiply by $D^{-1}$: $$(BCD)D^{-1} = A^{-1}D^{-1} \implies BC = A^{-1}D^{-1}$$
3. Post-multiply by $C^{-1}$: $$(BC)C^{-1} = A^{-1}D^{-1}C^{-1} \implies B = A^{-1}D^{-1}C^{-1}$$

Now, we find $B^{-1}$ by taking the inverse of both sides. Recall the reversal property of matrix transposition/inversion, $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$:

Since $(M^{-1})^{-1} = M$ for any invertible matrix $M$, we get:

Method 2: Using the Definition of Inverse

By definition, if $B \cdot X = I$ for square matrices, then $X = B^{-1}$.

Starting with:

Pre-multiply both sides by $A^{-1}$:

Post-multiply both sides by $A$:

Since $B(CDA) = I$, it directly follows that:

Thus, the correct option is B.