Linear Algebra Solutions 16

# Linear Algebra Solutions 16 - = I n B B 2 B B 2 B 3 = I n-B...

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

5. E 21 (2) 1 2 - 2 - 4 = 1 2 0 0 . Hence, as in the previous question, 1 2 - 2 - 4 is singular. 6. Starting from the equation A 2 - 2 A + 13 I 2 = 0, we deduce A ( A - 2 I 2 ) = - 13 I 2 = ( A - 2 I 2 ) A. Hence AB = BA = I 2 , where B = - 1 13 ( A - 2 I 2 ). Consequently A is non– singular and A - 1 = B . 7. We assume the equation A 3 = 3 A 2 - 3 A + I 3 . (ii) A 4 = A 3 A = (3 A 2 - 3 A + I 3 ) A = 3 A 3 - 3 A 2 + A = 3(3 A 2 - 3 A + I 3 ) - 3 A 2 + A = 6 A 2 - 8 A + 3 I 3 . (iii) A 3 - 3 A 2 + 3 A = I 3 . Hence A ( A 2 - 3 A + 3 I 3 ) = I 3 = ( A 2 - 3 A + 3 I 3 ) A. Hence A is non–singular and A - 1 = A 2 - 3 A + 3 I 3 = - 1 - 3 1 2 4 - 1 0 1 0 . 8. (i) If B 3 = 0 then ( I n - B )( I n + B + B 2 ) = I n ( I n + B + B 2 ) - B ( I n + B + B 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) = ( I n + B + B 2 )-( B + B 2 + B 3 ) = I n-B 3 = I n-0 = I n . Similarly ( I n + B + B 2 )( I n-B ) = I n . Hence A = I n-B is non–singular and A-1 = I n + B + B 2 . It follows that the system AX = b has the unique solution X = A-1 b = ( I n + B + B 2 ) b = b + Bb + B 2 b. 19...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online