1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 10 10.1 Matrix multiplication is distributive [ ]{[ ]{ } { }} [ ]{ } { }LUX D AX B [][]{}[]{} []{}{}LU X L D A X B Therefore, equating like terms, [][]{}[]{}LU X A X []{} {}LD B [][] []LU A 10.2 (a) The coefficient a21 is eliminated by multiplying row 1 by f21 = –3/10 = –0.3 and subtracting the result from row 2. a31 is eliminated by multiplying row 1 by f31 = 1/10 = 0.1 and subtracting the result from row 3. The factors f21 and f31 can be stored in a21 and a31. 10 2 10.3 5.4 1.70.1 0.8 5.1 a32 is eliminated by multiplying row 2 by f32 = 0.8/(–5.4) = –0.14815 and subtracting the result from row 3. The factor f32 can be stored in a32. 10 2 10.3 5.4 1.70.1 0.14815 5.351852 Therefore, the LU decomposition is 100[] 0.3 1 00.1 0.14815 1L 10 2 1[] 0 5.4 1.70 0 5.351852U These two matrices can be multiplied to yield the original system. For example, using MATLAB to perform the multiplication gives >> L=[1 0 0;-.3 1 0;0.1 -.14815 1]; >> U=[10 2 -1;0 -5.4 1.7;0 0 5.351852]; >> L*U ans = 10.0000 2.0000 -1.0000 -3.0000 -6.0000 2.0000 1.0000 1.0000 5.0000 (b) Forward substitution: [L]{D} = {B}2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5.215.6127114815.01.0013.0001321ddd Solving yields d1 = 27, d2 = –53.4, and d3 = –32.1111. Back substitution: 12310 2 1 270 5.4 1.7 53.40 0 5.351852 32.1111xxx 332.111165.351852x 253.4 1.7( 6)85.4x 127 ( 1)( 6) 2(8)0.510x (c) Forward substitution: [L]{D} = {B} 123100 120.3 1 0 180.1 0.14815 1 6ddd Solving yields d1 = 12, d2 = 21.6, and d3 = –4. Back substitution: 12310 2 1 1205.4 1.7 21.60 0 5.351852 4xxx 340.74745.351852x 221.6 1.7( 0.7474)4.235295.4x 112 ( 1)( 0.7474) 2( 4.23529)1.97231810x 10.3 (a) The coefficient a21 is eliminated by multiplying row 1 by f21 = –2/8 = –0.25 and subtracting the result from row 2. a31 is eliminated by multiplying row 1 by f31 = 2/8 = 0.25 and subtracting the result from row 3. The factors f21 and f31 can be stored in a21 and a31. 25.6225.075.0625.01483 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. a32 is eliminated by multiplying row 2 by f32 = –2/6 = –0.33333 and subtracting the result from row 3. The factor f32 can be stored in a32. 5.633333.025.075.0625.0148 Therefore, the LU decomposition is 133333.025.00125.0001][L 5.60075.060148][U Forward substitution: [L]{D} = {B} 7411133333.025.00125.0001321ddd Solving yields d1 = 11, d2 = 6.75, and d3 = 6.5. Back substitution: 5.675.6115.60075.060148321xxx 15.65.63x 16)1(75.075.62x 18)1(4)1)(1(111x (b) The first column of the inverse can be computed by using [L]{D} = {B} 001133333.025.00125.0001321ddd This can be solved for d1 = 1, d2 = 0.25, and d3 = 0.16667. Then, we can implement back substitution 16667.025.015.60075.060148321xxx to yield the first column of the inverse4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 025641.00448718.0099359.0}{X For the second column use {B}T = {0 1 0} which gives {D}T = {0 1 0.33333}. Back substitution then gives {X}T = {0.073718 0.160256 0.051282}. For the third column use {B}T = {0 0 1} which gives {D}T = {0 0 1}. Back substitution then gives {X}T = {0.028846 0.01923 0.153846}. Therefore, the matrix inverse is 153846.0051282.0025641.0019231.0160256.0044872.0028846.0073718.0099359.0][1A We can verify that this is correct by multiplying [A][A]–1 to yield the identity matrix. For example, using MATLAB, >> A=[8 4 -1;-2 5 1;2 -1 6]; >> AI=[0.099359 -0.073718 0.028846; 0.044872 0.160256 -0.019231; -0.025641 0.051282 0.153846]
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