Adjoint and Inverse of a Square Matrix

1. Adjoint of a Square Matrix

The adjoint of a square matrix is a crucial concept that helps us find the inverse of a matrix. It's closely related to the transpose of the cofactor matrix. Let's break down how to find it.

1.1 Minors of an Element

Before we get to the adjoint, we need to understand minors. The minor of an element $a_{ij}$ in a square matrix $A$ is the determinant of the submatrix formed by deleting the $i$-th row and $j$-th column of $A$. We denote the minor of $a_{ij}$ as $M_{ij}$.

Example: Consider the matrix $ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} $ To find the minor of the element $a_{11}$ (which is 1), we delete the 1st row and 1st column: $ M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = (5 \times 9) - (6 \times 8) = 45 - 48 = -3 $ Similarly, to find the minor of $a_{23}$ (which is 6), we delete the 2nd row and 3rd column: $ M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = (1 \times 8) - (2 \times 7) = 8 - 14 = -6 $

1.2 Cofactors of an Element

The cofactor of an element $a_{ij}$, denoted by $C_{ij}$, is obtained by multiplying its corresponding minor $M_{ij}$ by $(-1)^{i+j}$. The sign pattern for cofactors is often remembered as: $ \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} $ So, $C_{ij} = (-1)^{i+j} M_{ij}$.

Example (continuing with matrix A): $ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} $ We already found $M_{11} = -3$. The cofactor $C_{11}$ is: $ C_{11} = (-1)^{1+1} M_{11} = (-1)^2 (-3) = 1 \times (-3) = -3 $ We found $M_{23} = -6$. The cofactor $C_{23}$ is: $ C_{23} = (-1)^{2+3} M_{23} = (-1)^5 (-6) = (-1) \times (-6) = 6 $ Let's find a few more: $ M_{12} = \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = (4 \times 9) - (6 \times 7) = 36 - 42 = -6 $ $ C_{12} = (-1)^{1+2} M_{12} = (-1)^3 (-6) = (-1) \times (-6) = 6 $ $ M_{22} = \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = (1 \times 9) - (3 \times 7) = 9 - 21 = -12 $ $ C_{22} = (-1)^{2+2} M_{22} = (-1)^4 (-12) = 1 \times (-12) = -12 $

1.3 Cofactor Matrix

The cofactor matrix of $A$, denoted by $C$, is a matrix where each element $a_{ij}$ is replaced by its cofactor $C_{ij}$.

Example (for matrix A): We need to find all cofactors. $C_{11} = -3$ $C_{12} = 6$ $C_{13} = (-1)^{1+3} \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = 1 \times (32 - 35) = -3$ $C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix} = -1 \times (18 - 24) = -1 \times (-6) = 6$ $C_{22} = -12$ $C_{23} = 6$ $C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} = 1 \times (12 - 15) = -3$ $C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} = -1 \times (6 - 12) = -1 \times (-6) = 6$ $C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = 1 \times (5 - 8) = -3$ So, the cofactor matrix $C$ is: $ C = \begin{pmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{pmatrix} $

1.4 Adjoint of a Matrix

The adjoint of a square matrix $A$, denoted by $adj(A)$, is the transpose of its cofactor matrix $C$. $ adj(A) = C^T $

Example (for matrix A): Given the cofactor matrix $ C = \begin{pmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{pmatrix} $, its transpose is: $ adj(A) = C^T = \begin{pmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{pmatrix}^T = \begin{pmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{pmatrix} $ In this particular case, the cofactor matrix is symmetric, so its transpose is itself.

Another Example: Let $ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} $. Minors: $M_{11} = 4$, $M_{12} = 3$ $M_{21} = 2$, $M_{22} = 1$ Cofactors: $C_{11} = (-1)^{1+1} M_{11} = 1 \times 4 = 4$ $C_{12} = (-1)^{1+2} M_{12} = -1 \times 3 = -3$ $C_{21} = (-1)^{2+1} M_{21} = -1 \times 2 = -2$ $C_{22} = (-1)^{2+2} M_{22} = 1 \times 1 = 1$ Cofactor Matrix: $ C = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix} $ Adjoint Matrix: $ adj(A) = C^T = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} $

1.5 Properties of Adjoint

For a square matrix $A$ of order $n$:

• $ A \cdot adj(A) = adj(A) \cdot A = |A| I_n $, where $|A|$ is the determinant of $A$ and $I_n$ is the identity matrix of order $n$. This is a very important property.
• If $|A| \neq 0$, then $ adj(A) = |A| A^{-1} $.
• $ adj(I_n) = I_n $.
• $ adj(A^T) = (adj(A))^T $.
• $ adj(AB) = adj(B) adj(A) $.
• $ |adj(A)| = |A|^{n-1} $.
• For a scalar $k$, $ adj(kA) = k^{n-1} adj(A) $.
• If $A$ is a non-singular matrix, then $ adj(A^{-1}) = (adj(A))^{-1} = \frac{1}{|A|} A $.

2. Inverse of a Square Matrix

The inverse of a square matrix $A$, denoted by $A^{-1}$, is a matrix such that when multiplied by $A$, it results in the identity matrix $I$. That is, $A \cdot A^{-1} = A^{-1} \cdot A = I$.

A matrix has an inverse if and only if its determinant is non-zero. Such a matrix is called a non-singular matrix. If the determinant is zero, the matrix is called a singular matrix, and it does not have an inverse.

2.1 Formula for Inverse using Adjoint

The formula for finding the inverse of a square matrix $A$ of order $n$ is: $ A^{-1} = \frac{1}{|A|} adj(A) $ This formula is valid only if $|A| \neq 0$.

2.2 Steps to Find the Inverse

1. Calculate the determinant of the matrix, $|A|$.
2. If $|A| = 0$, the matrix is singular, and its inverse does not exist.
3. If $|A| \neq 0$, proceed to find the adjoint of the matrix, $adj(A)$.
4. Divide the adjoint matrix by the determinant: $ A^{-1} = \frac{1}{|A|} adj(A) $.

2.3 Examples

Example 1: Find the inverse of $ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} $. Step 1: Calculate the determinant. $ |A| = (1 \times 4) - (2 \times 3) = 4 - 6 = -2 $. Step 2: Since $|A| = -2 \neq 0$, the inverse exists. Step 3: Find the adjoint. Using the shortcut for 2x2 matrices: $ adj(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} $. Step 4: Calculate the inverse. $ A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{4}{-2} & \frac{-2}{-2} \\ \frac{-3}{-2} & \frac{1}{-2} \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} $.

Example 2: Find the inverse of $ B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} $. Step 1: Calculate the determinant. Expand along the first row: $ |B| = 1 \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2 \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3 \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} $ $ |B| = 1((1 \times 0) - (4 \times 6)) - 2((0 \times 0) - (4 \times 5)) + 3((0 \times 6) - (1 \times 5)) $ $ |B| = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) $ $ |B| = -24 - 2(-20) + 3(-5) $ $ |B| = -24 + 40 - 15 = 1 $. Step 2: Since $|B| = 1 \neq 0$, the inverse exists. Step 3: Find the adjoint. We need the cofactors: $C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = 1(0 - 24) = -24$ $C_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = -1(0 - 20) = 20$ $C_{13} = (-1)^{1+3} \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = 1(0 - 5) = -5$ $C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} = -1(0 - 18) = 18$ $C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} = 1(0 - 15) = -15$ $C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} = -1(6 - 10) = 4$ $C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 1(8 - 3) = 5$ $C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -1(4 - 0) = -4$ $C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1(1 - 0) = 1$ Cofactor Matrix: $ C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix} $ Adjoint Matrix: $ adj(B) = C^T = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} $ Step 4: Calculate the inverse. $ B^{-1} = \frac{1}{|B|} adj(B) = \frac{1}{1} \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} $.

2.4 Properties of Inverse

For square matrices $A$ and $B$ of the same order, and a scalar $k$:

• $(A^{-1})^{-1} = A$.
• $(A^T)^{-1} = (A^{-1})^T$.
• $(AB)^{-1} = B^{-1} A^{-1}$. (Note the reversal of order).
• $(kA)^{-1} = \frac{1}{k} A^{-1}$, provided $k \neq 0$.
• $A \cdot A^{-1} = A^{-1} \cdot A = I$.
• If $A$ is a non-singular matrix, then $A^{-1}$ is also non-singular.
• $|A^{-1}| = \frac{1}{|A|}$.

2.5 Solving Systems of Linear Equations using Inverse

A system of linear equations can be represented in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix. If $A$ is a non-singular matrix (i.e., $|A| \neq 0$), then the unique solution to this system is given by: $ X = A^{-1} B $

Example: Solve the system of equations: $ x + 2y + 3z = 6 $ $ 2x - y + z = 2 $ $ x + 3y - z = 2 $ In matrix form, this is $AX = B$: $ A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ 2 \\ 2 \end{pmatrix} $ First, find the determinant of $A$: $|A| = 1(1-3) - 2(-2-1) + 3(6-(-1)) = 1(-2) - 2(-3) + 3(7) = -2 + 6 + 21 = 25$. Since $|A| = 25 \neq 0$, a unique solution exists. Next, find $A^{-1}$. This involves finding the adjoint of $A$ and dividing by $|A|$. Let's find the cofactors: $C_{11} = -4$, $C_{12} = 3$, $C_{13} = 7$ $C_{21} = 11$, $C_{22} = -4$, $C_{23} = -1$ $C_{31} = 5$, $C_{32} = 5$, $C_{33} = -5$ Cofactor Matrix: $ C = \begin{pmatrix} -4 & 3 & 7 \\ 11 & -4 & -1 \\ 5 & 5 & -5 \end{pmatrix} $ Adjoint Matrix: $ adj(A) = C^T = \begin{pmatrix} -4 & 11 & 5 \\ 3 & -4 & 5 \\ 7 & -1 & -5 \end{pmatrix} $ Inverse Matrix: $ A^{-1} = \frac{1}{25} \begin{pmatrix} -4 & 11 & 5 \\ 3 & -4 & 5 \\ 7 & -1 & -5 \end{pmatrix} $ Now, solve for $X$: $ X = A^{-1} B = \frac{1}{25} \begin{pmatrix} -4 & 11 & 5 \\ 3 & -4 & 5 \\ 7 & -1 & -5 \end{pmatrix} \begin{pmatrix} 6 \\ 2 \\ 2 \end{pmatrix} $ $ X = \frac{1}{25} \begin{pmatrix} (-4 \times 6) + (11 \times 2) + (5 \times 2) \\ (3 \times 6) + (-4 \times 2) + (5 \times 2) \\ (7 \times 6) + (-1 \times 2) + (-5 \times 2) \end{pmatrix} $ $ X = \frac{1}{25} \begin{pmatrix} -24 + 22 + 10 \\ 18 - 8 + 10 \\ 42 - 2 - 10 \end{pmatrix} = \frac{1}{25} \begin{pmatrix} 8 \\ 20 \\ 30 \end{pmatrix} = \begin{pmatrix} 8/25 \\ 20/25 \\ 30/25 \end{pmatrix} = \begin{pmatrix} 8/25 \\ 4/5 \\ 6/5 \end{pmatrix} $ So, $x = 8/25$, $y = 4/5$, and $z = 6/5$.