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Problem on Vector Linear Algebra: Vector Algebra Identity
- 1. Problem on VectorLinear Algebra Daniel Kartawiguna (2024)
- 2. Problem: Prove thefollowing vector identity Prove 𝐀 ∙ 𝐁 × 𝐂 𝐚 ∙ 𝐛 × 𝐜 = 𝐀 ∙ 𝐚 𝐀 ∙ 𝐛 𝐀 ∙ 𝐜 𝐁 ∙ 𝐚 𝐁 ∙ 𝐛 𝐁 ∙ 𝐜 𝐂 ∙ 𝐚 𝐂 ∙ 𝐛 𝐂 ∙ 𝐜
- 3. Solution: • Let • First,we have to compute 𝐁 × 𝐂 𝐀 = 𝐴𝑥𝒊 + 𝐴𝑦𝒋 + 𝐴𝑧𝒌 𝐁 = 𝐵𝑥𝒊 + 𝐵𝑦𝒋 + 𝐵𝑧𝒌 𝐂 = 𝐶𝑥𝒊 + 𝐶𝑦𝒋 + 𝐶𝑧𝒌 𝐚 = 𝑎𝑥𝒊 + 𝑎𝑦𝒋 + 𝑎𝑧𝒌 𝐛 = 𝑏𝑥𝒊 + 𝑏𝑦𝒋 + 𝑏𝑧𝒌 𝐜 = 𝑐𝑥𝒊 + 𝑐𝑦𝒋 + 𝑐𝑧𝒌 𝐁 × 𝐂 = 𝒊 𝒋 𝒌 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧 = 𝐵𝑦 𝐵𝑧 𝐶𝑦 𝐶𝑧 𝒊 + 𝐵𝑥 𝐵𝑧 𝐶𝑥 𝐶𝑧 𝒋 + 𝐵𝑥 𝐵𝑦 𝐶𝑥 𝐶𝑦 𝒌
- 4. • Then, wecan calculate 𝐀∙𝐁×𝐂 𝐀 ∙ 𝐁 × 𝐂 = 𝐴𝑥𝒊 + 𝐴𝑦𝒋 + 𝐴𝑧𝒌 ∙ 𝐵𝑦 𝐵𝑧 𝐶𝑦 𝐶𝑧 𝒊 + 𝐵𝑥 𝐵𝑧 𝐶𝑥 𝐶𝑧 𝒋 + 𝐵𝑥 𝐵𝑦 𝐶𝑥 𝐶𝑦 𝒌 = 𝐵𝑦 𝐵𝑧 𝐶𝑦 𝐶𝑧 𝐴𝑥 + 𝐵𝑥 𝐵𝑧 𝐶𝑥 𝐶𝑧 𝐴𝑦 + 𝐵𝑥 𝐵𝑦 𝐶𝑥 𝐶𝑦 𝐴𝑧 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧 = det 𝑃
- 5. • Thus, thematrix P is 𝑃 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧
- 6. • Second, wecalculated 𝐛 × 𝐜 𝐛 × 𝐜 = 𝒊 𝒋 𝒌 𝑏𝑥 𝑏𝑦 𝑏𝑧 𝑐𝑥 𝑐𝑦 𝑐𝑧 = 𝑏𝑦 𝑏𝑧 𝑐𝑦 𝑐𝑧 𝒊 + 𝑏𝑥 𝑏𝑧 𝑐𝑥 𝑐𝑧 𝒋 + 𝑏𝑥 𝑏𝑦 𝑐𝑥 𝑐𝑦 𝒌
- 7. • Next, wecan calculate 𝐚 ∙ 𝐛 × 𝐜 𝐚 ∙ 𝐛 × 𝐜 = 𝑎𝑥𝒊 + 𝑎𝑦𝒋 + 𝑎𝑧𝒌 ∙ 𝑏𝑦 𝑏𝑧 𝑐𝑦 𝑐𝑧 𝒊 + 𝑏𝑥 𝑏𝑧 𝑐𝑥 𝑐𝑧 𝒋 + 𝑏𝑥 𝑏𝑦 𝑐𝑥 𝑐𝑦 𝒌 = 𝑏𝑦 𝑏𝑧 𝑐𝑦 𝑐𝑧 𝑎𝑥 + 𝑏𝑥 𝑏𝑧 𝑐𝑥 𝑐𝑧 𝑎𝑦 + 𝑏𝑥 𝑏𝑦 𝑐𝑥 𝑐𝑦 𝑎𝑧 = 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 𝑐𝑥 𝑐𝑦 𝑐𝑧
- 8. • Thus, thematrix Q is • If Q is any square matrix, then det(Q)=det(Qt). • The transpose matrix of the Q matrix is 𝑄 = 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 𝑐𝑥 𝑐𝑦 𝑐𝑧 𝑄𝑡 = 𝑎𝑥 𝑏𝑥 𝑐𝑥 𝑎𝑦 𝑏𝑦 𝑐𝑦 𝑎𝑧 𝑏𝑧 𝑐𝑧
- 9. • So that •Then 𝐚 ∙ 𝐛 × 𝐜 = 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 𝑐𝑥 𝑐𝑦 𝑐𝑧 = det 𝑄 = det 𝑄𝑡 = 𝑎𝑥 𝑏𝑥 𝑐𝑥 𝑎𝑦 𝑏𝑦 𝑐𝑦 𝑎𝑧 𝑏𝑧 𝑐𝑧 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧 𝑎𝑥 𝑏𝑥 𝑐𝑥 𝑎𝑦 𝑏𝑦 𝑐𝑦 𝑎𝑧 𝑏𝑧 𝑐𝑧 𝐀 ∙ 𝐁 × 𝐂 𝐚 ∙ 𝐛 × 𝐜 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 𝑐𝑥 𝑐𝑦 𝑐𝑧
- 10. • If Pand Q are squared matrices of equal size, then det(PQ) = det(P)det(Q). • Because det(Q)=det(Qt), then det(PQ)=det(PQt)=det(P)det(Qt) applies. Based on this, we can write 𝑃 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧 𝑄 = 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 𝑐𝑥 𝑐𝑦 𝑐𝑧 𝑄𝑡 = 𝑎𝑥 𝑏𝑥 𝑐𝑥 𝑎𝑦 𝑏𝑦 𝑐𝑦 𝑎𝑧 𝑏𝑧 𝑐𝑧
- 11. • 𝑃𝑄𝑡 canbe written as = 𝐴𝑥𝑎𝑥 + 𝐴𝑦𝑎𝑦 + 𝐴𝑧𝑎𝑧 𝐴𝑥𝑏𝑥 + 𝐴𝑦𝑏𝑦 + 𝐴𝑧𝑏𝑧 𝐴𝑥𝑐𝑥 + 𝐴𝑦𝑐𝑦 + 𝐴𝑧𝑐𝑧 𝐵𝑥𝑎𝑥 + 𝐵𝑦𝑎𝑦 + 𝐵𝑧𝑎𝑧 𝐵𝑥𝑏𝑥 + 𝐵𝑦𝑏𝑦 + 𝐵𝑧𝑏𝑧 𝐵𝑥𝑐𝑥 + 𝐵𝑦𝑐𝑦 + 𝐵𝑧𝑐𝑧 𝐶𝑥𝑎𝑥 + 𝐶𝑦𝑎𝑦 + 𝐶𝑧𝑎𝑧 𝐶𝑥𝑏𝑥 + 𝐶𝑦𝑏𝑦 + 𝐶𝑧𝑏𝑧 𝐶𝑥𝑐𝑥 + 𝐶𝑦𝑐𝑦 + 𝐶𝑧𝑐𝑧 𝑃𝑄𝑡 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐶𝑥 𝐶𝑦 𝐶𝑧 𝑎𝑥 𝑏𝑥 𝑐𝑥 𝑎𝑦 𝑏𝑦 𝑐𝑦 𝑎𝑧 𝑏𝑧 𝑐𝑧
- 12. • If 𝐀∙ 𝐚 = 𝐴𝑥𝑎𝑥 + 𝐴𝑦𝑎𝑦 + 𝐴𝑧𝑎𝑧 𝐀 ∙ 𝐛 = 𝐴𝑥𝑏𝑥 + 𝐴𝑦𝑏𝑦 + 𝐴𝑧𝑏𝑧 𝐀 ∙ 𝐜 = 𝐴𝑥𝑐𝑥 + 𝐴𝑦𝑐𝑦 + 𝐴𝑧𝑐𝑧 𝐁 ∙ 𝐚 = 𝐵𝑥𝑎𝑥 + 𝐵𝑦𝑎𝑦 + 𝐵𝑧𝑎𝑧 𝐁 ∙ 𝐛 = 𝐵𝑥𝑏𝑥 + 𝐵𝑦𝑏𝑦 + 𝐵𝑧𝑏𝑧 𝐁 ∙ 𝐜 = 𝐵𝑥𝑐𝑥 + 𝐵𝑦𝑐𝑦 + 𝐵𝑧𝑐𝑧 𝐂 ∙ 𝐚 = 𝐶𝑥𝑎𝑥 + 𝐶𝑦𝑎𝑦 + 𝐶𝑧𝑎𝑧 𝐂 ∙ 𝐛 = 𝐶𝑥𝑏𝑥 + 𝐶𝑦𝑏𝑦 + 𝐶𝑧𝑏𝑧 𝐂 ∙ 𝐜 = 𝐶𝑥𝑐𝑥 + 𝐶𝑦𝑐𝑦 + 𝐶𝑧𝑐𝑧
- 13. • hence 𝑃𝑄𝑡 = 𝐀∙ 𝐚 𝐀 ∙ 𝐛 𝐀 ∙ 𝐜 𝐁 ∙ 𝐚 𝐁 ∙ 𝐛 𝐁 ∙ 𝐜 𝐂 ∙ 𝐚 𝐂 ∙ 𝐛 𝐂 ∙ 𝐜 det 𝑃𝑄𝑡 = 𝐀 ∙ 𝐚 𝐀 ∙ 𝐛 𝐀 ∙ 𝐜 𝐁 ∙ 𝐚 𝐁 ∙ 𝐛 𝐁 ∙ 𝐜 𝐂 ∙ 𝐚 𝐂 ∙ 𝐛 𝐂 ∙ 𝐜 𝐀 ∙ 𝐁 × 𝐂 𝐚 ∙ 𝐛 × 𝐜 = det 𝑃 det 𝑄 = det 𝑃 det 𝑄𝑡 = det 𝑃𝑄𝑡
- 14. • Thus, itis proved that: 𝐀 ∙ 𝐁 × 𝐂 𝐚 ∙ 𝐛 × 𝐜 = 𝐀 ∙ 𝐚 𝐀 ∙ 𝐛 𝐀 ∙ 𝐜 𝐁 ∙ 𝐚 𝐁 ∙ 𝐛 𝐁 ∙ 𝐜 𝐂 ∙ 𝐚 𝐂 ∙ 𝐛 𝐂 ∙ 𝐜 (qed)
- 15. References Daniel Kartawiguna (2024) daniel.kartawiguna@gmail.com Anton,H., & Kaul, A. (2020). Elementary Linear Algebra. Hoboken, NJ 07030: John Wiley & Sons, Inc. Soemartojo, N. (2011). Kalkulus Lanjutan. Jakarta: Penerbit Universitas Indonesia (UI-Press).