STA3710 Assignment 4 (COMPLETE ANSWERS) 2025 (894289) - Due 9 September 2025

STA3710
ASSIGNMENT 4 2025

UNIQUE NO. 894289
DUE DATE: 9 SEPTEMBER 2025

, Question 2.1

Given

2 −1
0 −2 1
𝐴 = ൭−1 4൱ , 𝐵=ቀ ቁ ,
2 0 1 2×3
0 6 3×2

3 1 3
𝐶=ቀ ቁ , 𝐷=ቀቁ .
1 2 2×2 1 2×1

2.1.1 Show (𝐴 ⊗ 𝐵)′ = 𝐴′ ⊗ 𝐵′ .

Proof (general). For any matrices 𝐴 and 𝐵 the Kronecker product satisfies the
bilinearity property and the transpose distributes with reversal of factor transposes.
Concretely, if 𝐴 = (𝑎𝑖𝑗 ) and 𝐵 arbitrary, the (𝑖, 𝑗)-block of 𝐴 ⊗ 𝐵 is 𝑎𝑖𝑗 𝐵. Taking
transpose we get the (𝑗, 𝑖)-block equal to (𝑎𝑖𝑗 𝐵)′ = 𝑎𝑖𝑗 𝐵 ′ = (𝐴′ )𝑗𝑖 𝐵 ′ . Hence the block
structure of (𝐴 ⊗ 𝐵)′ is exactly that of 𝐴′ ⊗ 𝐵 ′ . Therefore

(𝐴 ⊗ 𝐵)′ = 𝐴′ ⊗ 𝐵 ′ .

(This is a standard Kronecker product identity.) ▫

Quick numeric check (with your A,B): compute both sides and compare — they are
equal. (You can compute explicitly if you want; the identity holds elementwise.)

2.1.2 Show (𝐴 ⊗ 𝐵)(𝐶 ⊗ 𝐷) = (𝐴𝐶) ⊗ (𝐵𝐷) (when products are conformable).

Statement and requirement. The identity

(𝐴 ⊗ 𝐵)(𝐶 ⊗ 𝐷) = (𝐴𝐶) ⊗ (𝐵𝐷)

holds whenever the individual products 𝐴𝐶 and 𝐵𝐷 are defined (i.e. the inner
dimensions match: if 𝐴 is 𝑚 × 𝑛 and 𝐶 is 𝑛 × 𝑝, and 𝐵 is 𝑟 × 𝑠 and 𝐷 is 𝑠 × 𝑡). Under
those conformability conditions the Kronecker product obeys the mixed-product
property.