Linear Algebra with Applications (3rd Edition)

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ² + ( - 1) 3+2 · 1 · det ± 1 2 3 2 ² + ( - 1) 3+3 · 1 · det ± 1 2 1 1 ² = 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 ² + ( - 1) 2+2 · 1 · det ± 1 3 3 1 ² + ( - 1) 3+2 · 1 · det ± 1 2 3 2 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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 01/23/2008 for the course MATH 417 taught by Professor Elling during the Fall '07 term at University of Michigan.

Page1 / 5

hw7sol - Math 417 homework 7 solutions Section 6.1 problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online