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Unformatted text preview: Math 415 — Midterm 1 practice solutions Problem 1: We can start reducing the matrix using Gaussian elimination. We get to 1 0 1 1 0 1 2 1 0 0 3 3 0 0 3 3 and because two rows are equal, we know the matrix is singular and the determinant is 0. Problem 2: We start by writing the augmented matrix 1 1 1 1 0 0 1 2 2 0 1 0 2 1 2 0 0 1 We then reduce the left side to the identity matrix using GaussJordan re duction, and we do the same operations on the right side to get 1 0 0 2 1 0 1 0 2 1 0 0 1 3 1 1 The right side is the inverse we needed. You may want to verify this by multiplying the original matrix with the inverse we found. Problem 3: Let’s call A the original matrix. We start by reducing A to reduced row echelon form using GaussJordan reduction. We obtain U = 1 0 0 6 / 5 0 1 0 3 / 5 0 0 1 1 / 5 0 0 0 There are three leading coefficients in the first three columns of U . The basis....
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 Spring '08
 BERTRANDGUILLOU
 Linear Algebra, Algebra, Determinant, Gaussian Elimination, Row echelon form

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