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Properties of
Matrix Operations

Discover why matrix multiplication breaks commutativity but preserves associativity and distributivity: the rules that govern how matrices combine.

Precalculus

What Is a Matrix?

A matrix is a rectangular array of numbers arranged in rows and columns. We describe its size as \( m \times n \), where \( m \) is the number of rows and \( n \) is the number of columns.

1 2 3 4 5 6 7 8 9 3 × 3 matrix Row Column

Three Core Operations

Addition
Add corresponding entries: \( (A + B)_{ij} = a_{ij} + b_{ij} \). Matrices must be the same size.
Scalar Multiply
Multiply every entry by a constant: \( (cA)_{ij} = c \cdot a_{ij} \).
Multiplication
Row-by-column dot products: \( (AB)_{ij} = \sum_k a_{ik}\, b_{kj} \). Inner dimensions must match.

Why Do Properties Matter?

When you solve systems of equations, transform graphics in video games, or train machine-learning models, you rearrange matrix expressions constantly. Knowing which rearrangements are valid prevents costly errors.

1
Simplify expressions: factor, regroup, and distribute matrices the same way you would with numbers, except for commutativity.
2
Avoid false steps: assuming \( AB = BA \) leads to wrong answers in proofs and computations.
3
Build to advanced topics: eigenvalues, determinants, and linear transformations all rely on these foundational rules.

The Three Key Properties

✕ Not Commutative
$$ AB \neq BA $$ In general, switching the order of multiplication changes the result.
✓ Associative
$$ (AB)C = A(BC) $$ You can regroup parentheses freely without changing the product.
✓ Distributive
$$ A(B + C) = AB + AC $$ Multiplication distributes over addition from the left (and right).

Why \( AB \neq BA \)

Each entry of a product depends on a specific row from the left matrix and a specific column from the right matrix. Swapping the order feeds different rows into different columns, producing a completely different grid of dot products.

AB 1 2 3 4 · 5 6 7 8 = 19 22 43 50 BA 5 6 7 8 · 1 2 3 4 = 23 34 31 46 ≠

Associativity: Regrouping Is Safe

For any conformable matrices \( A \), \( B \), and \( C \):

A B first · C = A · B C first Order stays A → B → C in both cases

This property lets you choose the most efficient grouping for computation without worrying about getting a different answer.

Distributive Property

Matrix multiplication distributes over addition from both sides:

Left Distribution
$$ A(B + C) = AB + AC $$
Right Distribution
$$ (B + C)A = BA + CA $$

Real Numbers vs Matrices

Property Real Numbers Square Matrices
Commutativity ✓ \( ab = ba \) ✕ \( AB \neq BA \) in general
Associativity ✓ \( (ab)c = a(bc) \) ✓ \( (AB)C = A(BC) \)
Distributivity ✓ \( a(b+c) = ab+ac \) ✓ \( A(B+C) = AB+AC \)
Identity Element ✓ \( a \cdot 1 = a \) ✓ \( AI = IA = A \)
Zero Product \( ab = 0 \Rightarrow a=0 \) or \( b=0 \) ✕ \( AB = O \) possible with \( A,B \neq O \)

How to Apply the Properties

1
Check dimensions: confirm the matrices are conformable (inner dimensions match for multiplication; same size for addition).
2
Preserve order: never swap the positions of matrices around a multiplication sign. \( AB \) and \( BA \) are different expressions.
3
Regroup freely: move parentheses using associativity to simplify computation: \( (AB)C = A(BC) \).
4
Distribute carefully: keep the multiplier on the same side: \( A(B+C) = AB + AC \), not \( BA + CA \).
5
Verify with small examples: if you are unsure about a step, test it with \( 2 \times 2 \) matrices.

Example 1: Non-Commutativity

Let \( A = \begin{bmatrix}1 & 0\\2 & 1\end{bmatrix} \) and \( B = \begin{bmatrix}0 & 1\\3 & 0\end{bmatrix} \). Compute \( AB \) and \( BA \).

\( AB \)
\( \begin{bmatrix}1\cdot0+0\cdot3 & 1\cdot1+0\cdot0\\2\cdot0+1\cdot3 & 2\cdot1+1\cdot0\end{bmatrix} \)

\( = \begin{bmatrix}0 & 1\\3 & 2\end{bmatrix} \)
\( BA \)
\( \begin{bmatrix}0\cdot1+1\cdot2 & 0\cdot0+1\cdot1\\3\cdot1+0\cdot2 & 3\cdot0+0\cdot1\end{bmatrix} \)

\( = \begin{bmatrix}2 & 1\\3 & 0\end{bmatrix} \)

Example 2: Verifying Associativity

Let \( A = \begin{bmatrix}1&2\\0&1\end{bmatrix}\), \( B = \begin{bmatrix}2&0\\1&3\end{bmatrix}\), \( C = \begin{bmatrix}1&1\\0&2\end{bmatrix}\). Show \( (AB)C = A(BC) \).

1
Compute \( AB \): \( \begin{bmatrix}1\cdot2+2\cdot1 & 1\cdot0+2\cdot3\\0\cdot2+1\cdot1 & 0\cdot0+1\cdot3\end{bmatrix} = \begin{bmatrix}4&6\\1&3\end{bmatrix} \)
2
Compute \( (AB)C \): \( \begin{bmatrix}4&6\\1&3\end{bmatrix}\begin{bmatrix}1&1\\0&2\end{bmatrix} = \begin{bmatrix}4&16\\1&7\end{bmatrix} \)
3
Compute \( BC \): \( \begin{bmatrix}2&0\\1&3\end{bmatrix}\begin{bmatrix}1&1\\0&2\end{bmatrix} = \begin{bmatrix}2&2\\1&7\end{bmatrix} \)
4
Compute \( A(BC) \): \( \begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}2&2\\1&7\end{bmatrix} = \begin{bmatrix}4&16\\1&7\end{bmatrix} \) ✓ Same result!

Example 3: Distributive Property

Let \( A = \begin{bmatrix}2&1\\0&3\end{bmatrix}\), \( B = \begin{bmatrix}1&0\\2&1\end{bmatrix}\), \( C = \begin{bmatrix}0&3\\1&0\end{bmatrix}\). Verify \( A(B+C) = AB + AC \).

1
Compute \( B + C \): \( \begin{bmatrix}1+0 & 0+3\\2+1 & 1+0\end{bmatrix} = \begin{bmatrix}1&3\\3&1\end{bmatrix} \)
2
Compute \( A(B+C) \): \( \begin{bmatrix}2&1\\0&3\end{bmatrix}\begin{bmatrix}1&3\\3&1\end{bmatrix} = \begin{bmatrix}5&7\\9&3\end{bmatrix} \)
3
Compute \( AB + AC \): \( \begin{bmatrix}4&1\\6&3\end{bmatrix} + \begin{bmatrix}1&6\\3&0\end{bmatrix} = \begin{bmatrix}5&7\\9&3\end{bmatrix} \) ✓ Match!

The Identity Matrix \( I \)

The identity matrix has 1s on the main diagonal and 0s everywhere else. It is the multiplicative identity for matrices, just as 1 is for real numbers:

1 0 0 1 I₂ (2×2) 1 0 0 0 1 0 0 0 1 I₃ (3×3) AI = A IA = A ✓ Commutes with all matrices!

Common Mistakes to Avoid

✕ Swapping Order
Writing \( AB = BA \) without justification. This is false in general. Always preserve left-to-right order.
✕ Wrong-Side Distribute
Writing \( A(B+C) = BA + CA \). The multiplier \( A \) must stay on the same side of each term.
✕ Cancellation Fallacy
Concluding \( A = B \) from \( CA = CB \). You cannot "cancel" \( C \) unless \( C \) is invertible.

Summary of Properties

Property Formula Status
Commutativity \( AB = BA \) ✕ Fails in general
Associativity \( (AB)C = A(BC) \) ✓ Always holds
Left Distributive \( A(B+C) = AB + AC \) ✓ Always holds
Right Distributive \( (B+C)A = BA + CA \) ✓ Always holds
Identity \( AI = IA = A \) ✓ Always holds
Zero Product \( AB = O \Rightarrow A=O \) or \( B=O \) ✕ Fails in general

Question 1 of 5

Which property does not hold for matrix multiplication in general?

Question 2 of 5

If \( A = \begin{bmatrix}1&2\\3&4\end{bmatrix} \), what is \( AI \) where \( I \) is the \( 2\times2 \) identity matrix?

Question 3 of 5

Which expression is a correct application of the distributive property?

Question 4 of 5

Given \( (AB)C = A(BC) \), which property is being demonstrated?

Question 5 of 5

If \( AB = O \) (the zero matrix), can we conclude that \( A = O \) or \( B = O \)?