--- title: "Positive definite stiffness matrix and Gauss elimination" subject: "Finite Element Method" chapter: 2 key: "semester_8/finite_element_method:matrix-gauss-elimination-stiffness" source: https://www.bctengineer.com/q/semester_8/finite_element_method/matrix-gauss-elimination-stiffness site: BCT Engineering --- # Positive definite stiffness matrix and Gauss elimination Finite Element Method · Chapter 2: Mathematical Background - Asked in: year not given, 6 marks - Video: [Introduction to Gauss Elimination](https://www.youtube.com/watch?v=KChCjzpbnNQ) — Sumit Basu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/matrix-gauss-elimination-stiffness - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-2 ## Question A three-degree-of-freedom structure has the stiffness matrix (in kN/mm) $$ [K] = \begin{bmatrix} 5 & -2 & 0 \\ -2 & 4 & -1 \\ 0 & -1 & 3 \end{bmatrix} $$ and load vector $\{F\} = \{10,\ 0,\ 6\}^T$ kN. (a) Show that $[K]$ is symmetric and positive definite. (b) Solve $[K]\{u\} = \{F\}$ by Gauss elimination and state the displacements in mm. Why must a stiffness matrix be positive definite after boundary conditions are applied? ## Short answer The matrix is symmetric and all leading principal minors (5, 16, 43) are positive, confirming positive definiteness. Gauss elimination gives u1=2.837 mm, u2=2.093 mm, u3=2.698 mm. Positive definiteness after applying boundary conditions ensures positive strain energy for any non-zero displacement and eliminates rigid-body modes, yielding a unique solution. ## Answer #### (a) Symmetry and positive definiteness $K_{12} = K_{21} = -2$, $K_{23} = K_{32} = -1$, $K_{13} = K_{31} = 0$, so $[K]$ is symmetric. For positive definiteness, all leading principal minors must be positive: - $\Delta_1 = 5 > 0$ - $\Delta_2 = 5(4) - (-2)(-2) = 16 > 0$ - $\Delta_3 = \det[K] = 5(12-1) + 2(-6-0) = 55 - 12 = 43 > 0$ All are positive, so $[K]$ is positive definite and a unique solution exists. #### (b) Gauss elimination Augmented matrix: $$ \left[\begin{array}{ccc|c} 5 & -2 & 0 & 10 \\ -2 & 4 & -1 & 0 \\ 0 & -1 & 3 & 6 \end{array}\right] $$ Step 1: $R_2 \leftarrow R_2 + 0.4R_1$ gives $[0,\ 3.2,\ -1\,|\,4]$. Step 2: $R_3 \leftarrow R_3 + \frac{1}{3.2}R_2 = R_3 + 0.3125R_2$ gives $[0,\ 0,\ 2.6875\,|\,7.25]$. Back substitution: $$ \begin{aligned} u_3 &= \frac{7.25}{2.6875} = 2.698\ \text{mm} \\ u_2 &= \frac{4 + 2.698}{3.2} = 2.093\ \text{mm} \\ u_1 &= \frac{10 + 2(2.093)}{5} = 2.837\ \text{mm} \end{aligned} $$ Check: row 3 gives $-2.093 + 3(2.698) = 6.00$ kN. Also the product of pivots $5 \times 3.2 \times 2.6875 = 43 = \det[K]$. **Why positive definite:** strain energy $U = \tfrac{1}{2}\{u\}^T[K]\{u\}$ must be positive for every non-zero displacement. If a rigid-body motion remains (supports missing), $[K]$ is singular and the equations cannot be solved. Positive pivots also guarantee that elimination runs without row interchange. **Answer:** $u_1 = 2.837$ mm, $u_2 = 2.093$ mm, $u_3 = 2.698$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is positive definiteness of [K] checked in this answer? A. All leading principal minors are positive B. The trace is positive C. All off-diagonal terms are positive D. The determinant is zero Answer: A. All leading principal minors (5, 16, 43) are positive, so [K] is positive definite. (Key point: Positive definite test) 2. What is det[K] for the given matrix? A. 55 B. 43 C. 12 D. 16 Answer: B. det = 5(12-1) + 2(-6-0) = 55 - 12 = 43, which also equals the product of pivots 5 x 3.2 x 2.6875. (Key point: Determinant) 3. What is the second pivot after R2 <- R2 + 0.4 R1? A. 4 B. 2.6875 C. 3.2 D. 5 Answer: C. 4 - 0.4(2) = 3.2, giving row [0, 3.2, -1 | 4]. (Key point: Elimination) 4. What are the displacements u1, u2, u3 in mm? A. 2.093, 2.837, 2.698 B. 2.698, 2.093, 2.837 C. 2.837, 2.698, 2.093 D. 2.837, 2.093, 2.698 Answer: D. Back substitution gives u3 = 2.698, u2 = 2.093 and u1 = 2.837 mm. 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