{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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 Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

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 bookmark
Ask a homework question - tutors are online