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Lesson 1: Determinant -- Definition and Cofactor Expansion

Estimated time: 35-45 minutes

Learning Objectives

The 2x2 Determinant

2x2 Determinant: For A = [a b; c d], det(A) = ad - bc.

Worked Example

det([3 1; 2 5]) = 3(5) - 1(2) = 15 - 2 = 13.

det([4 6; 2 3]) = 4(3) - 6(2) = 12 - 12 = 0. (This matrix is singular.)

Minors and Cofactors

Minor M_{ij}: The determinant of the (n-1)x(n-1) matrix obtained by deleting row i and column j from A.

Cofactor C_{ij}: C_{ij} = (-1)^{i+j} M_{ij}. The sign alternates in a checkerboard pattern: + - + / - + - / + - + ...

Example: Sign Pattern for 3x3

[ +   -   + ]
[ -   +   - ]
[ +   -   + ]

C_{11} = +M_{11}, C_{12} = -M_{12}, C_{13} = +M_{13}, etc.

Cofactor Expansion

Cofactor Expansion along row i: det(A) = a_{i1}C_{i1} + a_{i2}C_{i2} + ... + a_{in}C_{in}.

You can also expand along any column. The result is the same regardless of which row or column you choose.

Worked Example: 3x3 Determinant

Find det(A) where A = [2 1 3; 0 -1 2; 1 4 -1].

Expand along row 1:

det(A) = 2 * C_{11} + 1 * C_{12} + 3 * C_{13}

C_{11} = +det([-1 2; 4 -1]) = +( (-1)(-1) - (2)(4) ) = 1 - 8 = -7

C_{12} = -det([0 2; 1 -1]) = -( (0)(-1) - (2)(1) ) = -(-2) = 2

C_{13} = +det([0 -1; 1 4]) = +( (0)(4) - (-1)(1) ) = 0 + 1 = 1

det(A) = 2(-7) + 1(2) + 3(1) = -14 + 2 + 3 = -9.

Strategy: Choose Wisely

Expand along the row or column with the most zeros to minimize computation. Each zero eliminates one cofactor calculation.

Example: Expanding Along a Column with Zeros

For A = [2 1 3; 0 -1 2; 1 4 -1], expand along column 1:

det(A) = 2*C_{11} + 0*C_{21} + 1*C_{31}

The 0 in position (2,1) means we skip that cofactor entirely.

C_{11} = +det([-1 2; 4 -1]) = -7 (computed above)

C_{31} = +det([1 3; -1 2]) = 2 - (-3) = 5

det(A) = 2(-7) + 0 + 1(5) = -14 + 5 = -9. Same answer!

The Determinant and Invertibility

Key Theorem: A square matrix A is invertible if and only if det(A) is not equal to 0.

This connects determinants to everything we learned in Modules 1-2: invertibility, unique solutions, and pivot columns.

Check Your Understanding

1. Compute det([5 2; 3 1]).

Answer: 5(1) - 2(3) = 5 - 6 = -1.

2. What is the cofactor C_{23} of a 3x3 matrix?

Answer: C_{23} = (-1)^{2+3} M_{23} = -M_{23}. Delete row 2 and column 3, take the determinant, then negate.

3. Compute det([1 0 0; 3 2 0; 1 5 4]).

Answer: This is a lower triangular matrix. Expand along row 1: 1*det([2 0; 5 4]) + 0 + 0 = 1*(8-0) = 8. For triangular matrices, det = product of diagonal entries: 1*2*4 = 8.

4. If det(A) = 0, what can you conclude?

Answer: A is NOT invertible (singular). The system Ax = b either has no solution or infinitely many solutions. The columns of A are linearly dependent.

Key Takeaways

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Properties of Determinants.

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