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Find the vector form of the general solution of the given linear system Ax = b ; then use that result to find the vector form of the general solution of Ax = 0. + x2 + 2x3 + X3 -2 2x1 + x2 + 3x3 4 O The general solution of Ax = b is (x1, x2, x3) = (– 2, 8, 0) + s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2, x3) = s(– 1, – 1, 1). O The general solution of Ax = b is (x1, x2, x3) = (- 2, 8, 0) + s( – 1, - 1, 1); and thejẩjneral ẩjolution ofjẩ = 0 isẩ(X1, X2, X3) = (- 2, 8, 0). O The general solution of Ax = b is (x1, X2, x3) = s(- 2, 8, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x|, x2, x3) = s(- 2, 8, 0). O The general solution of Ax = b is(x1, x2, x3) = s(- 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = (– 2, 8, 0) + s( – 1, – 1, 1). O The general solution of Ax b is (x1,x2, x3) = s(- 2, 8, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x1, X2, x3) = s(- 1, - 1, 1).

Find the vector form of the general solution of the given linear system Ax = b ; then use that result to find the vector form of the general solution of Ax = 0. + x2 + 2x3 + X3 -2 2x1 + x2 + 3x3 4 O The general solution of Ax = b is (x1, x2, x3) = (– 2, 8, 0) + s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2, x3) = s(– 1, – 1, 1). O The general solution of Ax = b is (x1, x2, x3) = (- 2, 8, 0) + s( – 1, - 1, 1); and thejẩjneral ẩjolution ofjẩ = 0 isẩ(X1, X2, X3) = (- 2, 8, 0). O The general solution of Ax = b is (x1, X2, x3) = s(- 2, 8, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x|, x2, x3) = s(- 2, 8, 0). O The general solution of Ax = b is(x1, x2, x3) = s(- 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = (– 2, 8, 0) + s( – 1, – 1, 1). O The general solution of Ax b is (x1,x2, x3) = s(- 2, 8, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x1, X2, x3) = s(- 1, - 1, 1).

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Find the vector form of the general solution of the given linear system Ax = b; then use that result to find the vector form of the general solution of Ax = 0. X1 + x2 + 2x3 6. + X3 -2 2x1 + x2 + 3x3 4 The general solution of Ax = b is (x1, x2, x3) = ( – 2, 8, 0) + s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = s( – 1, – 1, 1). O The general solution of Ax = b is (x1, X2 , X3) = (– 2, 8, 0) + s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2, X3) = (– 2, 8, 0). O The general solution of Ax = b is (x1, X2, x3) = s( – 2, 8, 0) + (- 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2 , X3) = s( – 2, 8, 0). O The general solution of Ax = b is(x1, x2, x3) = s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2, X3) = ( – 2, 8, 0) + s( – 1, – 1, 1). O The general solution of Ax = b is (x1, X2, X3) = s( – 2, 8, 0) + (- 1, – 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = s( – 1, – 1, 1).
Transcribed Image Text:Find the vector form of the general solution of the given linear system Ax = b; then use that result to find the vector form of the general solution of Ax = 0. X1 + x2 + 2x3 6. + X3 -2 2x1 + x2 + 3x3 4 The general solution of Ax = b is (x1, x2, x3) = ( – 2, 8, 0) + s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = s( – 1, – 1, 1). O The general solution of Ax = b is (x1, X2 , X3) = (– 2, 8, 0) + s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2, X3) = (– 2, 8, 0). O The general solution of Ax = b is (x1, X2, x3) = s( – 2, 8, 0) + (- 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2 , X3) = s( – 2, 8, 0). O The general solution of Ax = b is(x1, x2, x3) = s( – 1, – 1, 1); and the general solution of Ax = 0 is (x1, X2, X3) = ( – 2, 8, 0) + s( – 1, – 1, 1). O The general solution of Ax = b is (x1, X2, X3) = s( – 2, 8, 0) + (- 1, – 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = s( – 1, – 1, 1).
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