GATE CS 2019 Q9 — Linear Algebra: Matrices, Determinants, System of Linear Equations

A fundamental theorem in linear algebra establishes the equivalence between the invertibility of a square matrix and its determinant.

Let $X$ be an $n \times n$ square matrix.

• Statement I $\implies$ Statement II:
  If $X$ is invertible, there exists a matrix $X^{-1}$ such that $X X^{-1} = I$, where $I$ is the identity matrix. Taking the determinant on both sides:
  $$\det(X X^{-1}) = \det(I) \implies \det(X) \cdot \det(X^{-1}) = 1$$
  Since the product of their determinants is $1$, the determinant of $X$ cannot be zero (i.e., $\det(X) \neq 0$).
• Statement II $\implies$ Statement I:
  If $\det(X) \neq 0$, we can construct the inverse of $X$ using its adjugate matrix:
  $$X^{-1} = \frac{1}{\det(X)} \text{adj}(X)$$
  Since $\det(X) \neq 0$, the division is well-defined, and thus $X$ is invertible.

Since both implications hold, I and II are equivalent statements.