Matrix Notation
A matrix is a rectangular array of numbers arranged in rows and columns. An m × n matrix has m rows and n columns.
Entry notation
aij = element in row i, column j
A = [aij]m×n
Special Matrix Types
Addition & Scalar Multiplication
Matrix Addition
Add corresponding entries. Requires same dimensions.
If A and B are both m×n:
(A + B)ij = aij + bij
Scalar Multiplication
Multiply every entry by the scalar k.
(kA)ij = k · aij
Example: 3[2, 1; 0, 4] = [6, 3; 0, 12]
Worked Example 1
Compute 2A + B where A = [1, 3; -2, 0] and B = [4, -1; 5, 2]
2A = [2·1, 2·3; 2·(-2), 2·0] = [2, 6; -4, 0]
2A + B = [2+4, 6+(-1); -4+5, 0+2]
= [6, 5; 1, 2]
Matrix Multiplication
(AB)ij = (row i of A) · (column j of B) = Σ aikbkj
Dot product of row i with column j
Worked Example 2
Multiply A = [1, 2; 3, 4] by B = [5, 6; 7, 8]
C11 = 1·5 + 2·7 = 5 + 14 = 19
C12 = 1·6 + 2·8 = 6 + 16 = 22
C21 = 3·5 + 4·7 = 15 + 28 = 43
C22 = 3·6 + 4·8 = 18 + 32 = 50
AB = [19, 22; 43, 50]
AB ≠ BA
Not commutative in general
A(BC) = (AB)C
Associative
A(B + C) = AB + AC
Distributive
Determinants
2×2 Determinant
det[a, b; c, d] = ad − bc
Memorize: multiply down-right diagonal, subtract up-right diagonal.
3×3 — Cofactor Expansion
Expand along the first row:
det(A) = a11M11 − a12M12 + a13M13
where Mij = minor (2×2 det with row i, col j removed)
− + −
+ − +
Worked Example 3
Find det(A) where A = [2, 1, −1; 3, 0, 2; −1, 4, 1]
Expand along row 1:
= 2·det[0, 2; 4, 1] − 1·det[3, 2; −1, 1] + (−1)·det[3, 0; −1, 4]
= 2·(0·1 − 2·4) − 1·(3·1 − 2·(−1)) + (−1)·(3·4 − 0·(−1))
= 2·(−8) − 1·(5) + (−1)·(12)
= −16 − 5 − 12
det(A) = −33
Inverse Matrix
A is invertible ↔ det(A) ≠ 0
If det(A) = 0, the matrix is singular — no inverse exists.
2×2 Inverse Formula
If A = [a, b; c, d], then A⁻¹ = (1/det(A)) · [d, −b; −c, a]
Swap a and d, negate b and c, divide by det(A).
Worked Example 4
Find A⁻¹ for A = [3, 1; 5, 2]
det(A) = 3·2 − 1·5 = 6 − 5 = 1
A⁻¹ = (1/1) · [2, −1; −5, 3]
A⁻¹ = [2, −1; −5, 3]
Verify: A·A⁻¹ = [3·2+1·(−5), 3·(−1)+1·3; 5·2+2·(−5), 5·(−1)+2·3] = [1, 0; 0, 1] = I ✓
Larger Matrices: Row Reduction Method
Write [A | I] and row-reduce. When the left side becomes I, the right side is A⁻¹.
Form augmented matrix [A | I_n]
Apply row operations to reduce A to I
The same operations transform I to A⁻¹
If A cannot be reduced to I, it's singular (no inverse)
Solving Systems with Matrices
Matrix Equation Method
Write system as AX = B
If A is invertible: X = A⁻¹B
2x + y = 7 → A = [2, 1; 5, 3]
5x + 3y = 17 → B = [7; 17]
X = A⁻¹B = [3, -1; -5, 2][7; 17] = [4; -1]
Augmented Matrix (Row Reduction)
Row operations (legal moves):
Ri ↔ Rj (swap rows)
kRi → Ri (scale row)
Ri + kRj → Ri (add multiple)
Goal: [A|B] → [I|X] (reduced row echelon form)
Quick Reference
| Operation | Requirement | Result Dimensions | Commutative? |
|---|---|---|---|
| Addition A + B | A and B same size | m × n | Yes |
| Scalar mult kA | Any matrix | m × n | Yes |
| Multiplication AB | cols(A) = rows(B) | m × p | NO |
| Transpose Aᵀ | Any matrix | n × m | — |
| Inverse A⁻¹ | Square, det ≠ 0 | n × n | Yes (A⁻¹A = I) |
| Determinant det(A) | Square matrix | Scalar | — |
Exam Strategy
Check dimensions first
Before multiplying, verify the inner dimensions match. Write the size next to each matrix. A 3×2 times a 2×4 = a 3×4 result.
Determinant = 0 → no inverse
If you compute det = 0, stop — the matrix is singular and you cannot invert it. The system either has no solution or infinitely many.
2×2 inverse formula
Memorize "swap, negate, divide": swap a↔d, negate b and c, divide everything by ad−bc. Fastest method on exams.
Frequently Asked Questions
When is matrix multiplication defined?
Matrix multiplication AB is defined only when the number of columns in A equals the number of rows in B. If A is m×n and B is n×p, then AB is m×p. Matrix multiplication is generally NOT commutative — AB ≠ BA in most cases.
What does it mean for a matrix to be invertible?
A square matrix A is invertible (non-singular) if there exists a matrix A⁻¹ such that A·A⁻¹ = A⁻¹·A = I (the identity matrix). A matrix is invertible if and only if its determinant is non-zero (det(A) ≠ 0).
How do I solve a system of equations using matrices?
Write the system as AX = B where A is the coefficient matrix, X is the variable column, and B is the constants column. If A is invertible, X = A⁻¹B. Alternatively, write the augmented matrix [A|B] and use row reduction (Gaussian elimination) to get reduced row echelon form.
Related Topics
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