1. **Problem Statement:** Prove that if $A$ is a non-singular matrix of order $n$, then $\operatorname{adj}(A) = \det(A) A^{-1}$. 2. **Recall Definitions and Properties:** - The adjoint (or adjugate) matrix $\operatorname{adj}(A)$ is the transpose of the cofactor matrix of $A$. - For any square matrix $A$, the following identity holds: $$A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = \det(A) I_n,$$ where $I_n$ is the identity matrix of order $n$. - Since $A$ is non-singular, $\det(A) \neq 0$ and $A^{-1}$ exists. 3. **Proof Steps:** - Starting from the identity: $$A \cdot \operatorname{adj}(A) = \det(A) I_n,$$ - Multiply both sides on the left by $A^{-1}$: $$A^{-1} A \cdot \operatorname{adj}(A) = A^{-1} \det(A) I_n,$$ which simplifies to: $$I_n \cdot \operatorname{adj}(A) = \det(A) A^{-1},$$ - Therefore: $$\operatorname{adj}(A) = \det(A) A^{-1}.$$ 4. **Explanation:** This shows that the adjoint matrix of a non-singular matrix $A$ can be expressed as the product of its determinant and its inverse. This is a fundamental result in linear algebra linking the adjoint, determinant, and inverse of a matrix. **Final answer:** $$\boxed{\operatorname{adj}(A) = \det(A) A^{-1}}$$