{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

201f04mt2[1]

# 201f04mt2[1] - B U Department of Mathematics Math 201...

This preview shows pages 1–3. Sign up to view the full content.

B U Department of Mathematics Math 201 Matrix Theory Fall 2004 Second Midterm This archive is a property of Bo˘ gazi¸ ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. (a) Let A = a b c d e f g h k and det A = - 3. Find the value of the determinant: det B = a b c g + 2 a h + 2 b k + 2 c 3 d 3 e 3 f . Solution: det A = - a b c g h k d e f (interchanging 2 nd and 3 rd rows) = - a b c g + 2 a h + 2 b k + 2 c d e f (adding twice the 1 st row to the 2 nd row) = - 1 3 a b c g + 2 a h + 2 b k + 2 c 3 d 3 e 3 f (multiplying the 3 rd row by 3) = - 1 3 det B . Hence det B = - 3 · ( - 3) = 9 .

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

View Full Document
(b) Find the n × n determinant: n 0 n - 1 . . . 2 0 1 . Solution: Interchange 1 st and n th rows, 2 nd and ( n - 1) st rows, in general, k th and ( n - k + 1) st rows to obtain the diagonal matrix 1 2 . . . n - 1 n provided k < n - k + 1 i.e. 2 k < n + 1. If n is odd then the number s of swappings is equal to n +1 2 - 1 = n - 1 2 . If n is even then s = n 2 . Hence, n 0 n - 1 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}