Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# A square 4 ux d 0 d e 00f z 0 triangular matrix has

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Unformatted text preview: 0 00 is r D 2. The special solution to Ax D 0 and U x D 0 is s D . 7; 2; 1/. Since b D .1; 3; 6; 5/ is also the last column of A, a particular solution to Ax D b is .0; 0; 1/ and the complete solution is x D .0; 0; 1/ C c s. (Or use the particular solution x p D .7; 2 ; 0/ with free variable x3 D 0.) &quot; # &quot; # &quot; 1023 2 102 32 1020 1320 5 ! 030 33 ! 0100 2 0 4 9 10 000 36 0001 For b D .1; 0; 0; 0/ elimination leads to U x D .1; 1; 0; 1/ and the fourth equation is 0 D 1. No solution for this b.      1 1 0 10 33 If the complete solution to Ax D is x D C then A D . 3 0 c 30 34 (a) If s D .2; 3; 1; 0/ is the only special solution to Ax D 0, the complete solution is x D c s (line of solution!). The rank of A must be 4 1 D 3. &quot; 10 (b) The fourth variable x4 is not free in s, and R must be 0 1 00 # 20 3 0. 01 (c) Ax D b can be solve for all b, because A and R have full row rank r D 3. 35 For the 1; 2; 1 matrix K (9 by 9) and constant right side b D .10;    ; 10/, the solution x D K 1 b D .45; 80; 105; 120; 125; 120; 105; 80; 45/ rises and falls along the parabola xi D 50i 5i 2 . (A formula for K 1 is later in the text.) 36 If Ax D b and C x D b have the same solutions, A and C have the same shape and the same nullspace (take b D 0). If b D column 1 of A, x D .1; 0; : : : ; 0/ solves Ax D b so it solves C x D b. Then A and C share column 1. Other columns too: A D C ! Problem Set 3.5, page 178 1 &quot; 111 011 001 #&quot; c1 c2 c3 # D 0 gives c3 D c2 D c1 D 0. So those 3 column vectors are &quot; # &quot;# 1112 0 independent. But 0 1 1 3 Œ c  D 0 is solved by c D .1; 1; 4 ; 1/. Then 0014 0 v1 C v2 4v3 C v4 D 0 (dependent). 2 v1 ; v2 ; v3 are independent (the 1’s are in different positions). All six vectors are on the plane .1; 1; 1; 1/  v D 0 so no four of these six vectors can be independent. Solutions to Exercises 37 3 If a D 0 then column 1 D 0; if d D 0 then b.column 1/ a.column 2/ D 0; if f D 0 then all columns end in zero (they are all in the xy plane, they must be dependent). &quot; #&quot; # &quot;# abc x 0 y D 0 gives z D 0 then y D 0 then x D 0. A square 4 Ux D 0 d e 00f z 0 triangular matrix has independent columns (invertible matrix) when its diagonal has no zeros. &quot; #&quot; #&quot; # 123 1 2 3 1 2 3 5 7!0 5 7 : invertible ) independent 5 (a) 3 1 2 ! 0 231 0 1 5 0 0 18=5 columns. &quot; # &quot; # &quot; # &quot;# &quot;# 1 2 3 1 2 3 12 3 1 0 3 1 2!0 7 7!07 7 I A 1 D 0 , columns (b) 2 3 1 0 7 7 00 0 1 0 add to 0. 6 Columns 1, 2, 4 are independent. Also 1, 3, 4 and 2, 3, 4 and others (but not 1, 2, 3). Same column numbers (not same columns!) for A. w2 / D 0. So the # 0 1 1 0 1. difference are dependent and the difference matrix is singular: A D 1 1 1 0 7 The sum v1 v2 C v3 D 0 because .w2 w3 / .w 1 w 3 / C .w 1 &quot; 8 If c1 .w2 C w3 / C c2 .w1 C w3 / C c3 .w1 C w2 / D 0 then .c2 C c3 /w1 C .c1 C c3 /w2 C .c1 C c2 /w3 D 0. Since the w’s are independent, c2 C c3 D c1 C c3 D c1 C c2 D 0. The only solution is c1 D...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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