{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Linear Algebra Solutions 18

# Linear Algebra Solutions 18 - 12 Also 1 0 A = E3(2)E14...

This preview shows page 1. Sign up to view the full content.

12. A = E 3 (2) E 14 E 42 (3) = E 3 (2) E 14 1 0 0 0 0 1 0 0 0 0 1 0 0 3 0 1 = E 3 (2) 0 3 0 1 0 1 0 0 0 0 1 0 1 0 0 0 = 0 3 0 1 0 1 0 0 0 0 2 0 1 0 0 0 . Also A - 1 = ( E 3 (2) E 14 E 42 (3)) - 1 = ( E 42 (3)) - 1 E - 1 14 ( E 3 (2)) - 1 = E 42 ( - 3) E 14 E 3 (1 / 2) = E 42 ( - 3) E 14 1 0 0 0 0 1 0 0 0 0 1 / 2 0 0 0 0 1 = E 42 ( - 3) 0 0 0 1 0 1 0 0 0 0 1 / 2 0 1 0 0 0 = 0 0 0 1 0 1 0 0 0 0 1 / 2 0 1 - 3 0 0 . 13. (All matrices in this question are over Z 2 .) (a) 1 1 0 1 0 0 1 1 1 1 1 1 1 0 0 1 fl fl fl fl fl fl fl fl 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 0 1 0 0 1 0 0 fl fl fl fl fl fl fl fl 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 1 0 0 0 0 1 0 0 0 1 1 fl fl fl fl fl fl fl fl 1 0 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 fl fl fl fl fl fl fl fl 0 0 0 1 1 0 0 1 1 0 1 0 1 1 1 0
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online