Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant/Proof 1

Theorem

Let $\mathbf A = \sqbrk a_n$ be an invertible square matrix of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $\adj {\mathbf A}$ be the adjugate of $\mathbf A$.

Then:

$\mathbf A^{-1} = \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}$

where $\mathbf A^{-1}$ denotes the inverse of $\mathbf A$

Proof

Let $\mathbf I_n$ denote the unit matrix of order $n$.

\(\ds \map \det {\mathbf A} \cdot \mathbf I_n\)

\(=\)

\(\ds \mathbf A \cdot \adj {\mathbf A}\)

Matrix Product with Adjugate Matrix

\(\ds \map \det {\mathbf A} \cdot \mathbf A^{-1} \cdot \mathbf I_n\)

\(=\)

\(\ds \mathbf A^{-1} \cdot \mathbf A \cdot \adj {\mathbf A}\)

\(\ds \map \det {\mathbf A} \cdot \mathbf A^{-1} \cdot \mathbf I_n\)

\(=\)

\(\ds \mathbf I_n \cdot \adj {\mathbf A}\)

Definition of Inverse Matrix

\(\ds \map \det {\mathbf A} \cdot \mathbf A^{-1}\)

\(=\)

\(\ds \adj {\mathbf A}\)

Unit Matrix is Identity for Matrix Multiplication

\(\ds \mathbf A^{-1}\)

\(=\)

\(\ds \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}\)

$\blacksquare$