hw4_solu - Solutions for Selected problems of Homework 4 A....

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: Solutions for Selected problems of Homework 4 A. Agarwal 1016-331 Linear Algebra-1 Spring 2009-10 1. Q24,P.151 Here is the product AB as a combination of rows of B AB = 1 parenleftBig 2 3 0 parenrightBig + parenleftBig 1 1 1 parenrightBig + 2 parenleftBig 1 6 4 parenrightBig 3 parenleftBig 2 3 0 parenrightBig + 1 parenleftBig 1 1 1 parenrightBig + 1 parenleftBig 1 6 4 parenrightBig 2 parenleftBig 2 3 0 parenrightBig + parenleftBig 1 1 1 parenrightBig + 1 parenleftBig 1 6 4 parenrightBig = 4 9 8 6 4 5 5 4 Note that the multiplier red entries are row entries from A . 2. Q30, P.151 We shall see that the lin ear com bi na tions of the rows of B approach that we used to compute AB in the previous problem will be useful here as well. Let the matrix A be written in its row form, A n k = arrowaxisleftaxisshortaxisshort a 1 axisshortaxisshortarrowaxisright arrowaxisleftaxisshortaxisshort a 2 axisshortaxisshortarrowaxisright . . . arrowaxisleftaxisshortaxisshort a n axisshortaxisshortarrowaxisright then, A n k B k p = arrowaxisleftaxisshortaxisshort a 1 axisshortaxisshortarrowaxisright arrowaxisleftaxisshortaxisshort a 2 axisshortaxisshortarrowaxisright . . . arrowaxisleftaxisshortaxisshort a n axisshortaxisshortarrowaxisright B = arrowaxisleftaxisshortaxisshort a 1 B axisshortaxisshortarrowaxisright arrowaxisleftaxisshortaxisshort a 2 B axisshortaxisshortarrowaxisright . . . arrowaxisleftaxisshortaxisshort a n B axisshortaxisshortarrowaxisright Since we are given that the rows of A are linearly dependent , this means the following equation has a non-zero solution (at least one c i negationslash = 0) c 1 a 1 + c 2 a 2 + ... + c n a n = (1) If we right- mul ti ply by ma trix B on both sides of equation (1), we get c 1 a 1 B + c 2 a 2 B + ... + c n a n B = (2) 1 Since a i B is just the i th row of AB , this implies the rows of AB are satisfying equation (2) with a non-zero solution. (remember at least onesolution....
View Full Document

Page1 / 6

hw4_solu - Solutions for Selected problems of Homework 4 A....

This preview shows document pages 1 - 3. Sign up to view the full document.

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