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Mathematics · Ch 4 — Determinants and Matrices

Properties of Matrix Multiplication

4.6

Properties of Matrix Multiplication

4.6 Properties of Matrix Multiplication

  1. Not commutative. In general AB≠BAAB\ne BA for matrices A,BA,B (§4.5.4 already showed this).
  2. Associative. (AB)C=A(BC)(AB)C=A(BC) whenever the orders are suitable for multiplication.
  3. Distributive over addition. A(B+C)=AB+ACA(B+C)=AB+AC (left distributive law) and (B+C)A=BA+CA(B+C)A=BA+CA (right distributive law).
  4. Multiplicative identity. For a square matrix AA, the identity matrix II of the same order satisfies AI=IA=AAI=IA=A.
  5. Null-matrix absorption. For any matrix AA, there is a null matrix OO (conformable) with AO=OAO=O and OA=OOA=O.
  6. Zero product without a zero factor. The product of two non-zero matrices can itself be the zero matrix: AB=OAB=O is possible even though A≠OA\ne O and B≠OB\ne O — a genuine departure from ordinary number arithmetic, where ab=0⇒a=0ab=0\Rightarrow a=0 or b=0b=0.
  7. Positive integer powers. For a square matrix AA, A2=AA, A3=AAA,…,An=AA⋯A⏟n timesA^2=AA,\ A^3=AAA,\dots,A^n=\underbrace{AA\cdots A}_{n\text{ times}}.

Consequences of the distributive law (for square A,BA,B of the same order)

Because AB≠BAAB\ne BA in general, the familiar numeric expansions pick up extra cross-terms that must be kept separate:

(A+B)2=A2+AB+BA+B2,(A−B)2=A2−AB−BA+B2,(A+B)(A−B)=A2+BA−AB−B2(A+B)^2=A^2+AB+BA+B^2,\qquad (A-B)^2=A^2-AB-BA+B^2,\qquad (A+B)(A-B)=A^2+BA-AB-B^2

(Only when AB=BAAB=BA do these collapse to the familiar A2±2AB+B2A^2\pm2AB+B^2 and A2−B2A^2-B^2 forms.)

Worked Examples

Example 1. Show ABAB is non-singular for suitable matrices A,BA,B.

Step 1: Compute ABAB directly, entry by entry, using the row-times-column rule.

Step 2: Compute ∣AB∣|AB| by cofactor expansion.

Step 3: Since ∣AB∣≠0|AB|\ne0, ABAB is non-singular by definition (§4.4.1, type 14).

Example 2. If A=[133313331]A=\begin{bmatrix}1&3&3\\3&1&3\\3&3&1\end{bmatrix}, prove A2−5AA^2-5A is a scalar matrix.

Step 1: A2=AAA^2=AA; computing entry-by-entry (each diagonal entry of A2A^2 is 1+9+9=191+9+9=19, each off-diagonal entry is 3+3+9=153+3+9=15), so A2=[191515151915151519]A^2=\begin{bmatrix}19&15&15\\15&19&15\\15&15&19\end{bmatrix}.

Step 2: 5A=[515151551515155]5A=\begin{bmatrix}5&15&15\\15&5&15\\15&15&5\end{bmatrix}. …

Misc 4.6Illustration of Property 2 (associativity) with three numeric matrices

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Illustration of Property 4 (identity matrix) with a numeric matrix

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Illustration of Property 6 (zero product from nonzero factors)

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Activity — computing AB−2I for two given 2×2 matrices

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 1 — showing a product matrix AB is non-singular

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 2 — proving a combination A²−5A is a scalar matrix

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 3 — finding an unknown k from a matrix polynomial equation

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 4 — finding unknowns x,y from a chained matrix product

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 5 — finding θ from a matrix-product-equals-scalar equation

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Remark — expanding (A±B)² and (A+B)(A−B) for matrices

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 6 — proving a matrix-power formula by mathematical induction

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 7 — word problem (bookseller's total receipt via matrix multiplication)

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …

Misc 4.6Solved Example 8 — word problem with two data tables multiplied (sports academy fees)

Worked out. Student counts per game per school are arranged in one matrix and the per-student coaching/equipment fee per game in another; multiplying the two matrices gives each school's total coaching and equipment expense, worked out school by school. …