Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant
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 1
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 | |||||||||||
\(\ds \mathbf A^{-1}\) | \(=\) | \(\ds \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}\) |
$\blacksquare$
Proof 2
Let:
- $\mathbf A = \begin {bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end {bmatrix}$
- $\mathbf A^{-1} = \begin {bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nn} \end {bmatrix}$
Let $\tuple {\mathbf e_1, \mathbf e_2, \cdots, \mathbf e_n}$ be the standard ordered basis of $\R^n$.
Let $T: \R^n \to \R^n, \mathbf x \mapsto \map T {\mathbf x}$ be a linear transformation.
From Linear Transformation as Matrix Product, let:
\(\ds \map T {\mathbf x}\) | \(=\) | \(\ds \mathbf A^{-1}\mathbf x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf A^{-1}\) | \(=\) | \(\ds \begin {bmatrix} \map T {\mathbf e_1} & \map T {\mathbf e_2} & \cdots & \map T {\mathbf e_n} \end {bmatrix}\) |
Let $p, q \in \set {1, \dots, n}$.
Let $\mathbf I_n$ be the unit matrix of order $n$.
\(\ds \map T {\mathbf e_q}\) | \(=\) | \(\ds \mathbf A^{-1} \mathbf e_q\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \begin {bmatrix} b_{1q} \\ \vdots \\ b_{nq} \end {bmatrix}\) | \(=\) | \(\ds \mathbf A^{-1} \mathbf e_q\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf A \begin {bmatrix} b_{1q} \\ \vdots \\ b_{nq} \end {bmatrix}\) | \(=\) | \(\ds \mathbf A \mathbf A^{-1} \mathbf e_q\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \mathbf I_n \mathbf e_q\) | Definition of Inverse Matrix | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \mathbf e_q\) | Unit Matrix is Identity for Matrix Multiplication |
Let $\mathbf {A_p}$ be the matrix obtained by replacing the $p$th column of $\mathbf A$ with $\mathbf e_q$.
Let $C_{q p}$ be the cofactor of $a_{q p}$ in $\map \det {\mathbf A_p}$.
\(\ds b_{pq}\) | \(=\) | \(\ds \frac {\map \det {\mathbf A_p} } {\map \det {\mathbf A} }\) | Cramer's Rule | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \frac {\begin {vmatrix} a_{1 1} & \cdots & a_{1, p - 1} & 0 & a_{1, p + 1} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{q - 1, 1} & \cdots & a_{q - 1, p - 1} & 0 & a_{q - 1, p + 1} & \cdots & a_{q - 1, n} \\ a_{q 1} & \cdots & a_{q, p - 1} & 1 & a_{q, p + 1} & \cdots & a_{q n} \\ a_{q + 1, 1} & \cdots & a_{q + 1, p - 1} & 0 & a_{q + 1, p + 1} & \cdots & a_{q + 1, n} \\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n, p - 1} & 0 & a_{n, p + 1} & \cdots & a_{n n} \end {vmatrix} } {\map \det {\mathbf A} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \frac {\ds \sum_{k \mathop = 1}^n a_{k p} C_{k p } } {\map \det {\mathbf A} }\) | Expansion Theorem for Determinants | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \frac {0 \cdot C_{1 p} + 0 \cdot C_{2 p} + \cdots + 1 \cdot C_{q p} + \cdots + 0 \cdot C_{n p} } {\map \det {\mathbf A} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \frac {C_{q p} } {\map \det {\mathbf A} }\) |
Hence:
\(\ds \mathbf A^{-1}\) | \(=\) | \(\ds \begin {bmatrix} \dfrac {C_{1 1} } {\map \det {\mathbf A} } & \cdots & \dfrac {C_{n 1} } {\map \det {\mathbf A} } \\ \vdots & \ddots & \vdots \\ \dfrac {C_{1 n} } {\map \det {\mathbf A} } & \cdots & \dfrac {C_{n n} } {\map \det {\mathbf A} } \end {bmatrix}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \dfrac 1 {\det {\mathbf A} } \begin {bmatrix} C_{1 1} & \cdots & C_{n 1} \\ \vdots & \ddots & \vdots \\ C_{1 n} & \cdots & C_{n n} \end {bmatrix}\) | Definition of Matrix Scalar Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}\) | Definition of Adjugate Matrix |
$\blacksquare$
Examples
Order $2$ Matrix
- $\begin {pmatrix} a & b \\ c & d \end {pmatrix}^{-1} = \dfrac 1 {a d - b c} \begin {pmatrix} d & -b \\ -c & a \end {pmatrix}$
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.7$ Minors and cofactors
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): adjoint