# Linear Algebra with Applications (3rd Edition)

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Math 417 homework 7 solutions Section 6.1 problem 6 Matrix is triangular, so determinant is product of diagonal entries: 6 · 4 · 1 = 24. Nonzero, so matrix invertible. Section 6.1 problem 8 The matrix does not have any zeros or other nice features, so we apply Sarrus’ rule: det = 1 · 1 · 1 + 2 · 1 · 3 + 3 · 1 · 2 - 1 · 1 · 2 - 2 · 1 · 1 - 3 · 1 · 3 = 0 . Determinant 0, so this matrix is not invertible. (One could have observed in the beginning that first plus third column equal two times second column, but that is not quite obvious.) Expansion across column 3: ( - 1) 3+1 · 3 · det 1 1 3 2 + ( - 1) 3+2 · 1 · det 1 2 3 2 + ( - 1) 3+3 · 1 · det 1 2 1 1 = 3 · (1 · 2 - 1 · 3) - (1 · 2 - 2 · 3) + (1 · 1 - 1 · 2) = - 3 - ( - 4) - 1 = 0 . Expansion across row 2: ( - 1) 1+2 · 1 · det 2 3 2 1 + ( - 1) 2+2 · 1 · det 1 3 3 1 + ( - 1) 3+2 · 1 · det 1 2 3 2 = - (2 · 1 - 3 · 2) + (1 · 1 - 3 · 3) - (1 · 2 - 2 · 3) = - ( - 4) - 8 - ( - 4) = 0 . 1 Section 6.2 problem 6 Subtract row 1 from row 2: 1 1 1 1 0 0 3 3 1 - 1 2 - 2 1 - 1 8 - 8 .

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