--- title: "Matrix inverse by Gauss elimination" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:inverse-gauss-elimination" source: https://www.bctengineer.com/q/semester_4/numerical_method/inverse-gauss-elimination site: BCT Engineering --- # Matrix inverse by Gauss elimination Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2068 Magh, 8 marks - Video: [Inverse of a matrix by Gauss Elimination Method](https://www.youtube.com/watch?v=8hQ7oVAiOkw) — Dr Samir Dey Classes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/inverse-gauss-elimination - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Find the inverse of the matrix A = [4 3 -1; 1 1 1; 3 5 3] using Gauss elimination. ## Short answer Form the augmented matrix [A|I] and apply Gauss-Jordan elimination to transform A to I; the right side becomes A⁻¹ = (1/10)[2 14 -4; 0 -15 5; -2 11 -1]. ## Answer Write the augmented matrix [A | I] and reduce A to I by row operations; the right half then becomes A⁻¹. (Gauss–Jordan form of Gauss elimination.) ```text [A | I] [ 4 3 -1 | 1 0 0 ] [ 1 1 1 | 0 1 0 ] [ 3 5 3 | 0 0 1 ] ``` **R1 → R1/4 (pivot 4 becomes 1)** ```text [ 1 3/4 -1/4 | 1/4 0 0 ] [ 1 1 1 | 0 1 0 ] [ 3 5 3 | 0 0 1 ] ``` **R2 → R2 − 1·R1, R3 → R3 − 3·R1** ```text [ 1 3/4 -1/4 | 1/4 0 0 ] [ 0 1/4 5/4 | -1/4 1 0 ] [ 0 11/4 15/4 | -3/4 0 1 ] ``` **R2 → 4·R2 (second pivot 1/4 becomes 1)** ```text [ 1 3/4 -1/4 | 1/4 0 0 ] [ 0 1 5 | -1 4 0 ] [ 0 11/4 15/4 | -3/4 0 1 ] ``` **R1 → R1 − (3/4)·R2, R3 → R3 − (11/4)·R2** ```text [ 1 0 -4 | 1 -3 0 ] [ 0 1 5 | -1 4 0 ] [ 0 0 -10 | 2 -11 1 ] ``` **R3 → R3/(−10) (third pivot becomes 1)** ```text [ 1 0 -4 | 1 -3 0 ] [ 0 1 5 | -1 4 0 ] [ 0 0 1 | -1/5 11/10 -1/10 ] ``` **R1 → R1 + 4·R3, R2 → R2 − 5·R3** ```text [ 1 0 0 | 1/5 7/5 -2/5 ] [ 0 1 0 | 0 -3/2 1/2 ] [ 0 0 1 | -1/5 11/10 -1/10 ] ``` Hence ```text 1 [ 2 14 −4 ] A⁻¹ = ──── [ 0 −15 5 ] = [ 0.2 1.4 −0.4 ] 10 [ −2 11 −1 ] [ 0.0 −1.5 0.5 ] [ −0.2 1.1 −0.1 ] ``` Check: first row of A times first column of A⁻¹ = 4(0.2) + 3(0) + (−1)(−0.2) = 1, and row 1 × column 2 = 4(1.4) + 3(−1.5) + (−1)(1.1) = 0, as required (det A = −10). **Answer: A⁻¹ = (1/10)[2 14 −4; 0 −15 5; −2 11 −1]** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To invert A with this method, which augmented matrix is reduced? A. [A | I], reducing A to I so the right half becomes A^-1 B. [A | b], reducing to upper triangular form only C. [I | A], reducing I to zero D. [A | A], reducing both halves equally Answer: A. Row operations that turn A into I turn I into A^-1. (Key point: Method) 2. After R1 → R1/4, R2 → R2 - R1, what is row 2? A. [0, 1/4, 5/4 | 1/4, 1, 0] B. [0, 1/4, 5/4 | -1/4, 1, 0] C. [0, 1, 5 | -1, 4, 0] D. [0, 1/4, 3/4 | -1/4, 1, 0] Answer: B. Row 2 = [1,1,1|0,1,0] - [1,3/4,-1/4|1/4,0,0] = [0,1/4,5/4|-1/4,1,0]. (Key point: Row operation result) 3. The answer states det A = -10. What does it show? A. A has no inverse B. A is symmetric C. A is invertible, and the third pivot -10 appears in the reduction D. A^-1 has determinant -10 Answer: C. A nonzero determinant means the inverse exists; the pivot R3 → R3/(-10) divides by this value. (Key point: Determinant) 4. What is the middle row of A^-1? A. [0, -1.5, 0.5] B. [0, 1.5, -0.5] C. [0.2, 1.4, -0.4] D. [-0.2, 1.1, -0.1] Answer: A. A^-1 = (1/10)[2 14 -4; 0 -15 5; -2 11 -1], so the middle row is [0, -1.5, 0.5]. 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