Explanation Given information: The given equations are, I A + I B + I C + I D + I E = 0 − 2 I A + 3 I B = 0 3 I B − 3 I C = 6 − 3 I C + I D = 0 − I D + 2 I E = 0 Calculation: Write the given equations in general matrix form [ A ] [ X ] . [ A ] [ X ] = [ 1 1 1 1 1 − 2 3 0 0 0 0 3 − 3 0 0 0 0 − 3 1 0 0 0 0 − 1 2 ] [ 0 0 6 0 0 ] Here, [ A ] is the matrix form of coefficients I A , I B , I C , I D , and I E and [ X ] is the constant. Find the determinant | A | of matrix A . | A | = | 1 1 1 1 1 − 2 3 0 0 0 0 3 − 3 0 0 0 0 − 3 1 0 0 0 0 − 1 2 | Expand matrix A by first column. | A | = 1 | 3 0 0 0 3 − 3 0 0 0 − 3 1 0 0 0 − 1 2 | − ( − 2 ) | 1 1 1 1 3 − 3 0 0 0 − 3 1 0 0 0 − 1 2 | = 1 [ 3 | − 3 0 0 − 3 1 0 0 − 1 2 | ] + 2 [ 1 | − 3 0 0 − 3 1 0 0 − 1 2 | − 3 | 1 1 1 − 3 1 0 0 − 1 2 | ] = 1 [ 3 ( − 3 ) ( 2 − 0 ) ] + 2 { ( − 3 ) ( 2 − 0 ) − 3 [ 1 ( 2 − 0 ) − ( − 3 ) ( 2 + 1 ) ] } = ( − 18 ) + 2 [ ( − 6 ) − 3 ( 2 + 9 ) ] = − 18 − 12 − 66 = − 96 Replace the coefficients of I A (elements of column 1) with constant [ X ] to form matrix [ X A ] . [ X A ] = [ 0 1 1 1 1 0 3 0 0 0 6 3 − 3 0 0 0 0 − 3 1 0 0 0 0 − 1 2 ] Find the determinant | X A | of matrix X A . | X A | = | 0 1 1 1 1 0 3 0 0 0 6 3 − 3 0 0 0 0 − 3 1 0 0 0 0 − 1 2 | Expand matrix X A by first column. | X A | = 0 − 0 + 6 | 1 1 1 1 3 0 0 0 0 − 3 1 0 0 0 − 1 2 | − 0 + 0 = 6 [ − 3 ( | 1 1 1 − 3 1 0 0 − 1 2 | ) ] = 6 { − 3 [ 1 ( 2 − 0 ) − ( − 3 ) ( 2 + 1 ) + 0 ] } = 6 [ − 3 ( 2 + 9 ) ] = − 198 Find the value of current I A using the equation: I A = | X A | | A | Substitute − 198 for | X A | and − 96 for | A | . I A = − 198 − 96 = 33 16 A Therefore, the value of current I A is 33 16 A _ . Replace the coefficients of I B (elements of column 2) with constant [ X ] to form matrix [ X B ] . [ X B ] = [ 1 0 1 1 1 − 2 0 0 0 0 0 6 − 3 0 0 0 0 − 3 1 0 0 0 0 − 1 2 ] Find the determinant | X B | of matrix X B . | X B | = | 1 0 1 1 1 − 2 0 0 0 0 0 6 − 3 0 0 0 0 − 3 1 0 0 0 0 − 1 2 | Expand matrix X B by second column. | X B | = − 0 + 0 − 6 | 1 1 1 1 − 2 0 0 0 0 − 3 1 0 0 0 − 1 2 | + 0 − 0 = − 6 [ − ( − 2 ) ( | 1 1 1 − 3 1 0 0 − 1 2 | ) ] = − 6 { 2 [ 1 ( 2 − 0 ) − ( − 3 ) ( 2 + 1 ) + 0 ] } = − 6 [ 2 ( 2 + 9 ) ] = − 132 Find the value of current I B using the equation: I B = | X B | | A | Substitute − 132 for | X A | and − 96 for | A |

11th Edition

ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 16.6, Problem 33E

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Chapter 16.6, Problem 32E

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Consider the cones K = = {(x1, x2, x3) | € R³ : X3 ≥√√√2x² + 3x² M = = {(21,22,23) (x1, x2, x3) Є R³: x3 > + 2 3 Prove that M = K*. Hint: Adapt the proof from the lecture notes for finding the dual of the Lorentz cone. Alternatively, prove the formula (AL)* = (AT)-¹L*, for any cone LC R³ and any 3 × 3 nonsingular matrix A with real entries, where AL = {Ax = R³ : x € L}, and apply it to the 3-dimensional Lorentz cone with an appropriately chosen matrix A.

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Chapter 16 Solutions

Basic Technical Mathematics

Ch. 16.1 - For matrices A and B, find A + B. Ch. 16.1 - Prob. 2PE Ch. 16.1 - Prob. 3PE Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - In Exercises 3–10, determine the value of the...Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E

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