The adjoint and inverse of a matrix are central constructions in linear algebra, particularly for the computation of matrix inverses and the solution of linear systems through matrix methods. These notions are defined analytically using minors, cofactors, determinants, and matrix transposition.

Construction of the Adjoint of a Square Matrix

Let $A = [a_{ij}]_{n \times n}$ be a square matrix of order $n \geq 2$. The process of constructing the adjoint involves the computation of minors, cofactors, and the subsequent transpose of the cofactor matrix.

Minor: The minor $M_{ij}$ of the element $a_{ij}$ is defined as the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column from $A$. For every $i, j$ with $1 \leq i, j \leq n$, $M_{ij} = \det(A_{ij})$, where $A_{ij}$ denotes the $(n-1) \times (n-1)$ submatrix formed as described.

Cofactor: The cofactor $C_{ij}$ of the element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$ for $1 \leq i, j \leq n$. This sign ensures proper expansion by the Laplace formula for determinants.

Consider the matrix $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$. For $a_{11}$, the minor $M_{11}$ is the determinant of the matrix obtained by removing the first row and first column, which yields $a_{22}$. Thus, $M_{11} = a_{22}$. The cofactor $C_{11} = (+1) \cdot a_{22} = a_{22}$. This holds for all other entries by corresponding deletion and sign multiplication.

Cofactor Matrix: The matrix of cofactors is $C = [C_{ij}]_{n \times n}$, where each entry is the cofactor of the corresponding element in $A$.

Transpose: The transpose of a matrix $B = [b_{ij}]_{n \times n}$, denoted $B^T$, is obtained by interchanging its rows and columns, so that the $(i,j)$-th entry becomes the $(j,i)$-th entry.

Adjoint (or Adjugate): The adjoint of $A$, denoted $\operatorname{adj} A$, is defined as the transpose of the cofactor matrix of $A$. That is,

$\operatorname{adj}A = [C_{ji}]_{n \times n}$, where $C_{ji}$ is the cofactor of $a_{ji}$ in $A$.

Explicit Calculation of the Adjoint for $2 \times 2$ and $3 \times 3$ Matrices

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the minor of $a$ is $d$, and the minor of $b$ is $c$. The cofactors are:

$C_{11}$: $+1 \cdot d = d$

$C_{12}$: $-1 \cdot c = -c$

$C_{21}$: $-1 \cdot b = -b$

$C_{22}$: $+1 \cdot a = a$

The cofactor matrix is $\begin{bmatrix} d & -c \\ -b & a \end{bmatrix}$. Thus, the adjoint is its transpose:

$\operatorname{adj}A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

For a $3 \times 3$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, the minor $M_{ij}$ is the determinant of the $2 \times 2$ matrix formed by deleting row $i$ and column $j$, the cofactor $C_{ij}$ is $(-1)^{i+j}M_{ij}$, the cofactor matrix is $[C_{ij}]$, and $\operatorname{adj}A = [C_{ji}]$.

Formal Definition and Matrix Equation Involving the Adjoint

Let $A$ be a square matrix of order $n$. The adjoint satisfies the matrix equation:

$A \cdot \operatorname{adj}A = \operatorname{adj}A \cdot A = \det(A) \cdot I_n$,

where $I_n$ denotes the identity matrix of order $n$. This equation holds for all square matrices and forms the algebraic basis for the construction of the inverse of a matrix.

Analytical Formula for the Inverse of a Square Matrix Using the Adjoint

A square matrix $A$ is invertible if and only if $\det(A) \neq 0$. In this case, the inverse of $A$ is given by

$A^{-1} = \dfrac{1}{\det(A)} \operatorname{adj}A$.

If $\det A = 0$, the matrix is singular and the formula does not apply as division by zero is not defined; such a matrix is non-invertible.

Worked Example: Adjoint and Inverse of a $2 \times 2$ Matrix

Given $A = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}$, compute its adjoint and inverse.

Step 1 (Determinant): Compute $\det A = 4 \times 5 - 3 \times 2 = 20 - 6 = 14$.

Step 2 (Cofactors): Compute the minors and cofactors:

$M_{11} = 5,\quad C_{11} = (+1) \cdot 5 = 5$

$M_{12} = 2, \quad C_{12} = (-1) \cdot 2 = -2$

$M_{21} = 3, \quad C_{21} = (-1) \cdot 3 = -3$

$M_{22} = 4, \quad C_{22} = (+1) \cdot 4 = 4$

Step 3 (Cofactor Matrix): $\begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix}$

Step 4 (Adjoint): Transpose the cofactor matrix:

$\operatorname{adj} A = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}$

Step 5 (Inverse): Substitute into the inverse formula:

$A^{-1} = \dfrac{1}{14} \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}$

Final Result: $A^{-1} = \begin{bmatrix} \dfrac{5}{14} & -\dfrac{3}{14} \\ -\dfrac{1}{7} & \dfrac{2}{7} \end{bmatrix}$

For a more detailed discussion on fundamental matrix operations, refer to Matrix Operations.

Worked Example: Adjoint and Inverse of a $3 \times 3$ Matrix

Given $B = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix}$, find $\operatorname{adj} B$ and $B^{-1}$.

Step 1 (Determinant): Expand by first row:

$\det B = 2 \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} - 1 \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} + 3 \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix}$

$= 2(4 \times 6 - 0 \times 5) - 1(0 \times 6 - 1 \times 5) + 3(0 \times 0 - 1 \times 4)$

$= 2(24 - 0) - 1(0 - 5) + 3(0 - 4)$

$= 2(24) + 5 + 3(-4)$

$= 48 + 5 - 12 = 41$

Step 2 (Cofactors): Compute $C_{ij} = (-1)^{i+j} M_{ij}$ for all $i, j$.

$C_{11} = (+1) \times \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = 1 \times (24-0) = 24$

$C_{12} = (-1) \times \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = -1 \times (0-5) = 5$

$C_{13} = (+1) \times \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = 1 \times (0-4) = -4$

$C_{21} = (-1) \times \begin{vmatrix} 1 & 3 \\ 0 & 6 \end{vmatrix} = -1 \times (1\times6-0\times3) = -1 \times 6 = -6$

$C_{22} = (+1) \times \begin{vmatrix} 2 & 3 \\ 1 & 6 \end{vmatrix} = 1 \times (2\times6-3\times1) = 1 \times (12-3) = 9$

$C_{23} = (-1) \times \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = -1 \times (2\times0-1\times1) = -1 \times (0-1) = 1$

$C_{31} = (+1) \times \begin{vmatrix} 1 & 3 \\ 4 & 5 \end{vmatrix} = 1 \times (1\times5-3\times4) = 1 \times (5-12) = -7$

$C_{32} = (-1) \times \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = -1 \times (2\times5-3\times0) = -1 \times (10-0) = -10$

$C_{33} = (+1) \times \begin{vmatrix} 2 & 1 \\ 0 & 4 \end{vmatrix} = 1 \times (2\times4-1\times0) = 1 \times (8-0) = 8$

Cofactor matrix:

$\begin{bmatrix} 24 & 5 & -4 \\ -6 & 9 & 1 \\ -7 & -10 & 8 \end{bmatrix}$

Step 3 (Adjoint): Transpose the cofactor matrix:

$\operatorname{adj} B = \begin{bmatrix} 24 & -6 & -7 \\ 5 & 9 & -10 \\ -4 & 1 & 8 \end{bmatrix}$

Step 4 (Inverse): $B^{-1} = \dfrac{1}{41} \operatorname{adj} B = \dfrac{1}{41} \begin{bmatrix} 24 & -6 & -7 \\ 5 & 9 & -10 \\ -4 & 1 & 8 \end{bmatrix}$

For further study on the theory of determinants, see Determinants And Their Properties.

Non-Singularity Condition for Invertibility of a Matrix

A matrix $A$ is invertible if and only if $\det A \neq 0$. In this case, there exists a unique $A^{-1}$ such that $A A^{-1} = A^{-1}A = I_n$. If $\det A = 0$, then $A$ is termed singular, and no inverse exists. This non-singularity is essential for the validity of the formula $A^{-1} = \frac{1}{\det A}\operatorname{adj}A$.

For related content covering how determinant properties affect invertibility, consult Properties Of Determinants.

Summary of the Algebraic Procedure for Adjoint and Inverse

The computation of the adjoint of a matrix involves: (i) calculating all $(n-1)\times(n-1)$ minors, (ii) assigning the correct signs to create the cofactor matrix, and (iii) transposing the cofactor matrix. The inverse of a non-singular square matrix $A$ is given by dividing the adjoint by the determinant of $A$. These procedures are fundamental for topics such as matrix equations, linear systems, and higher algebra.

For comprehensive coverage of matrices and their properties, refer also to Matrices And Determinants.