Determinants, Minors and Co-factors

Determinants are unique scalar values associated with square matrices. For a square matrix \( A = [a_{ij}] \) of order \( m \), the determinant is denoted by \( |A| \) or \( \det(A) \).

Determinant of a 2x2 matrix:

Given \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the determinant is

\[ |A| = ad - bc \]

Minors

The minor \( M_{ij} \) of an element \( a_{ij} \) in matrix \( A \) is the determinant of the matrix obtained by deleting the \( i^{th} \) row and \( j^{th} \) column from \( A \).

Co-factors

The cofactor \( C_{ij} \) of element \( a_{ij} \) is defined as

\[ C_{ij} = (-1)^{i+j} M_{ij} \]

where \( M_{ij} \) is the minor of \( a_{ij} \).

Adjoint of a Square Matrix

For a square matrix \( A = [a_{ij}] \), let \( B = [C_{ij}] \) be the matrix of cofactors. The transpose of \( B \), denoted \( B^T \), is called the adjoint of \( A \) and is written as \( \mathrm{adj}(A) \).

For a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint is

\[ \mathrm{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

Worked Example

Find the adjoint of the matrix

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 2 & 0 & 5 \end{bmatrix} \]

Step 1: Find the cofactor matrix \( B = [C_{ij}] \).

Calculate each cofactor:

• \( C_{11} = (-1)^{1+1} \times \det \begin{bmatrix} 3 & 4 \\ 0 & 5 \end{bmatrix} = 1 \times (3 \times 5 - 4 \times 0) = 15 \)
• \( C_{12} = (-1)^{1+2} \times \det \begin{bmatrix} 2 & 4 \\ 2 & 5 \end{bmatrix} = -1 \times (2 \times 5 - 4 \times 2) = -1 \times (10 - 8) = -2 \)
• \( C_{13} = (-1)^{1+3} \times \det \begin{bmatrix} 2 & 3 \\ 2 & 0 \end{bmatrix} = 1 \times (2 \times 0 - 3 \times 2) = -6 \)
• \( C_{21} = (-1)^{2+1} \times \det \begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix} = -1 \times (2 \times 5 - 3 \times 0) = -10 \)
• \( C_{22} = (-1)^{2+2} \times \det \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix} = 1 \times (1 \times 5 - 3 \times 2) = -1 \)
• \( C_{23} = (-1)^{2+3} \times \det \begin{bmatrix} 1 & 2 \\ 2 & 0 \end{bmatrix} = -1 \times (1 \times 0 - 2 \times 2) = 4 \)
• \( C_{31} = (-1)^{3+1} \times \det \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} = 1 \times (2 \times 4 - 3 \times 3) = -1 \)
• \( C_{32} = (-1)^{3+2} \times \det \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = -1 \times (1 \times 4 - 3 \times 2) = 2 \)
• \( C_{33} = (-1)^{3+3} \times \det \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = 1 \times (1 \times 3 - 2 \times 2) = -1 \)

Thus,

\[ B = \begin{bmatrix} 15 & -2 & -6 \\ -10 & -1 & 4 \\ -1 & 2 & -1 \end{bmatrix} \]

Step 2: Transpose \( B \) to get \( \mathrm{adj}(A) \):

\[ \mathrm{adj}(A) = B^T = \begin{bmatrix} 15 & -10 & -1 \\ -2 & -1 & 2 \\ -6 & 4 & -1 \end{bmatrix} \]

Singular and Non-Singular Matrices

A square matrix \( A \) is called singular if its determinant is zero, i.e., \( |A| = 0 \). Otherwise, it is non-singular.

Example of singular matrix:

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 12 \\ 1 & 1 & 3 \end{bmatrix} \]

Calculate \( |A| \):

\[ |A| = 1(15 - 12) - 2(12 - 12) + 3(4 - 5) = 3 - 0 - 3 = 0 \]

Since \( |A| = 0 \), \( A \) is singular.

Example of non-singular matrix:

\[ A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \]

Calculate \( |A| \):

\[ |A| = 0(0 - 1) - 1(0 - 1) + 1(1 - 0) = 0 + 1 + 1 = 2 \neq 0 \]

Since \( |A| \neq 0 \), \( A \) is non-singular.

Summary

• Determinant is a scalar value associated with a square matrix.
• Minors and cofactors are used to compute determinants and adjoints.
• Adjoint matrix is the transpose of the cofactor matrix.
• Singular matrices have zero determinant and are not invertible.
• Non-singular matrices have non-zero determinant and are invertible.

Practice Set

• Level 1 – Easy
• Find the determinant of \( \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix} \).
• Find the minor of element \( a_{12} \) in \( \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \).
• Find the cofactor of element \( a_{21} \) in the above matrix.
• Level 2 – Moderate
• Find the adjoint of \( \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \).
• Determine if the matrix \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \) is singular or non-singular.
• Calculate the determinant of \( \begin{bmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \\ 3 & 1 & 0 \end{bmatrix} \).
• Level 3 – Challenging
• Find the adjoint of \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \).
• Prove that \( A \cdot \mathrm{adj}(A) = |A| I \) for \( A = \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix} \).
• Find the determinant and adjoint of \( \begin{bmatrix} 2 & -1 & 0 \\ 1 & 3 & 4 \\ 0 & 2 & 1 \end{bmatrix} \).

Answer Key

• Level 1
• Determinant = \( 3 \times 5 - 4 \times 2 = 15 - 8 = 7 \)
• Minor of \( a_{12} \) is determinant of \( \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} = 4 \times 9 - 6 \times 7 = 36 - 42 = -6 \)
• Cofactor of \( a_{21} \) is \( (-1)^{2+1} \times \) minor of \( a_{21} \). Minor is determinant of \( \begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix} = 2 \times 9 - 3 \times 8 = 18 - 24 = -6 \). Cofactor = \( -1 \times -6 = 6 \)
• Level 2
• Adjoint of \( \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \) is \( \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \)
• Determinant of given matrix is \( 1(1 \times 0 - 4 \times 6) - 2(0 \times 0 - 4 \times 5) + 3(0 \times 6 - 1 \times 5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1 \neq 0 \), so non-singular.
• Determinant = \( 1(3 \times 0 - 1 \times 2) - 0(-1 \times 0 - 1 \times 3) + 2(-1 \times 1 - 3 \times 3) = 1(0 - 2) - 0(0 - 3) + 2(-1 - 9) = -2 + 0 - 20 = -22 \)
• Level 3
• Adjoint of \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \) is \( \begin{bmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{bmatrix} \)
• For \( A = \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix} \), \( |A| = 1 \times 5 - 3 \times 2 = 5 - 6 = -1 \). \( \mathrm{adj}(A) = \begin{bmatrix} 5 & -3 \\ -2 & 1 \end{bmatrix} \). Then \( A \cdot \mathrm{adj}(A) = \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 5 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} (1)(5)+(3)(-2) & (1)(-3)+(3)(1) \\ (2)(5)+(5)(-2) & (2)(-3)+(5)(1) \end{bmatrix} = \begin{bmatrix} 5 - 6 & -3 + 3 \\ 10 - 10 & -6 + 5 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = |A| I \)
• Determinant of \( \begin{bmatrix} 2 & -1 & 0 \\ 1 & 3 & 4 \\ 0 & 2 & 1 \end{bmatrix} \) is \( 2(3 \times 1 - 4 \times 2) - (-1)(1 \times 1 - 4 \times 0) + 0 = 2(3 - 8) + 1(1 - 0) + 0 = 2(-5) + 1 = -10 + 1 = -9 \). Adjoint can be calculated similarly.

Quick Reference

Term	Definition
Determinant	Scalar value associated with a square matrix
Minor \( M_{ij} \)	Determinant of matrix after deleting \( i^{th} \) row and \( j^{th} \) column
Cofactor \( C_{ij} \)	\( (-1)^{i+j} \times M_{ij} \)
Adjoint \( \mathrm{adj}(A) \)	Transpose of cofactor matrix
Singular Matrix	Matrix with zero determinant, not invertible
Non-Singular Matrix	Matrix with non-zero determinant, invertible

Glossary

• Square Matrix: A matrix with equal number of rows and columns.
• Determinant: A scalar value computed from a square matrix.
• Minor: Determinant of submatrix formed by deleting one row and one column.
• Cofactor: Signed minor used in determinant expansion.
• Adjoint: Transpose of the cofactor matrix.
• Singular Matrix: Matrix with zero determinant.
• Non-Singular Matrix: Matrix with non-zero determinant.

Inverse of a Matrix by Determinant Method

The inverse of a square matrix \( A \) exists only if \( A \) is non-singular, i.e., \( |A| \neq 0 \).

The inverse is given by

\[ A^{-1} = \frac{1}{|A|} \mathrm{adj}(A) \]

Algorithm to find \( A^{-1} \)

1. Calculate \( |A| \).
2. If \( |A| = 0 \), then \( A \) is singular and not invertible.
3. If \( |A| \neq 0 \), calculate the cofactor matrix of \( A \).
4. Find the adjoint \( \mathrm{adj}(A) \) by transposing the cofactor matrix.
5. Calculate \( A^{-1} = \frac{1}{|A|} \mathrm{adj}(A) \).

Worked Example

Find the inverse of \( A = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} \).

Step 1: Calculate determinant

\[ |A| = (-3)(-3) - (2)(5) = 9 - 10 = -1 \]

Step 2: Calculate adjoint

\[ \mathrm{adj}(A) = \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix}^T = \begin{bmatrix} -3 & -5 \\ -2 & -3 \end{bmatrix} \]

Step 3: Calculate inverse

\[ A^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & -5 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix} \]

Practice Set

• Level 1 – Easy
• Find the inverse of \( \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} \).
• Calculate the determinant and inverse of \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
• Level 2 – Moderate
• Find the inverse of \( \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \).
• Verify that \( A \cdot A^{-1} = I \) for \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \).
• Level 3 – Challenging
• Find the inverse of \( \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \) using determinant method.
• Prove that \( A \cdot \mathrm{adj}(A) = |A| I \) for a 3x3 matrix \( A \).

Answer Key

• Level 1
• Inverse of \( \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} \) is \( \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} \).
• Determinant of identity matrix is 1; inverse is the identity matrix itself.
• Level 2
• Inverse of \( \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \) is \( \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \).
• Multiplying \( A \) and \( A^{-1} \) yields the identity matrix.
• Level 3
• Inverse of \( \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \) can be found by calculating \( |A| \), cofactors, adjoint, and then applying the formula.
• Proof involves matrix multiplication and properties of determinants.

Quick Reference

Step	Action
1	Calculate \( |A| \)
2	If \( |A| = 0 \), no inverse exists
3	Calculate cofactor matrix
4	Find adjoint by transposing cofactor matrix
5	Calculate inverse \( A^{-1} = \frac{1}{|A|} \mathrm{adj}(A) \)

Glossary

• Inverse Matrix: Matrix \( A^{-1} \) such that \( A A^{-1} = I \).
• Identity Matrix: Square matrix with 1's on the diagonal and 0's elsewhere.
• Singular Matrix: Matrix with zero determinant, no inverse.
• Non-Singular Matrix: Matrix with non-zero determinant, inverse exists.

Area of Triangle Using Determinants

The area \( \Delta \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by

\[ \Delta = \frac{1}{2} \left| \det \begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix} \right| \]

Since area is positive, take the absolute value.

If \( \Delta = 0 \), the points are collinear.

Equation of a Line Using Determinants

The equation of the line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is

\[ \det \begin{bmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{bmatrix} = 0 \]

Expanding this determinant gives the line equation.

Worked Example

Find the area of the triangle with vertices \( (3, 8), (-4, 2), (5, 1) \).

Step 1: Write the determinant

\[ \Delta = \frac{1}{2} \left| \det \begin{bmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{bmatrix} \right| \]

Step 2: Calculate determinant

\[ = \frac{1}{2} |3(2 \times 1 - 1 \times 1) - 8(-4 \times 1 - 1 \times 5) + 1(-4 \times 1 - 2 \times 5)| \] \[ = \frac{1}{2} |3(2 - 1) - 8(-4 - 5) + 1(-4 - 10)| = \frac{1}{2} |3(1) - 8(-9) + 1(-14)| \] \[ = \frac{1}{2} |3 + 72 - 14| = \frac{1}{2} |61| = 30.5 \]

Practice Set

• Level 1 – Easy
• Find the area of triangle with vertices \( (0,0), (4,0), (0,3) \).
• Check if points \( (1,2), (3,4), (5,6) \) are collinear.
• Level 2 – Moderate
• Find the equation of the line passing through \( (2,3) \) and \( (4,7) \) using determinants.
• Find the area of triangle with vertices \( (1,1), (4,5), (7,2) \).
• Level 3 – Challenging
• Given points \( A(1,3), B(0,0), D(k,0) \), find \( k \) such that area of triangle ABD is 3.
• Prove that three points are collinear if the determinant of their coordinate matrix is zero.

Answer Key

• Level 1
• Area = \( \frac{1}{2} \times 4 \times 3 = 6 \)
• Points are collinear since determinant is zero.
• Level 2
• Equation of line: \( y = 2x - 1 \)
• Area = 9
• Level 3
• \( k = \pm 2 \)
• Proof follows from determinant properties and area formula.

Quick Reference

Formula	Description
\( \Delta = \frac{1}{2} | \det \begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix} | \)	Area of triangle
\( \det \begin{bmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{bmatrix} = 0 \)	Equation of line through two points

Glossary

• Collinear Points: Points lying on the same straight line.
• Determinant: Scalar value used to calculate area and check collinearity.
• Triangle Area: Half the absolute value of the determinant of coordinate matrix.