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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 X1 + X3 -4 2x1 + x2 + 3x3 1 O The general solution of Ax = b is (x1, x2, x3) = (- 4,9, 0) + s(- 1, - 1, 1); and the general solution of Ax = 0 is (x), x2, x3) = (- 4, 9, 0). O The general solution of Ax = b is (x1, x2, x3) = (- 4, 9, 0) + s( – 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, X3) = s(- 1, 1, 1). %3D O The general solution of Ax = b is (x1, x2, x3) = s(- 4, 9, 0) + (- 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) = s(- 4, 9, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, X3) = s( - 4,9, 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) = (- 4,9, 0) + 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. X1 + x2 + 2x3 X1 + X3 -4 2x1 + x2 + 3x3 1 O The general solution of Ax = b is (x1, x2, x3) = (- 4,9, 0) + s(- 1, - 1, 1); and the general solution of Ax = 0 is (x), x2, x3) = (- 4, 9, 0). O The general solution of Ax = b is (x1, x2, x3) = (- 4, 9, 0) + s( – 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, X3) = s(- 1, 1, 1). %3D O The general solution of Ax = b is (x1, x2, x3) = s(- 4, 9, 0) + (- 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) = s(- 4, 9, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, X3) = s( - 4,9, 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) = (- 4,9, 0) + s( – 1, – 1, 1).

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 2BEXP
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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 X1 + X3 -4 2x1 + x2 + 3x3 1 O The general solution of Ax = b is (x1, x2, x3) = (- 4,9, 0) + s(- 1, - 1, 1); and the general solution of Ax = 0 is (x), x2, x3) = (- 4, 9, 0). O The general solution of Ax = b is (x1, x2, x3) = (- 4, 9, 0) + s( - 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, X3) = s(- 1, 1, 1). %3D O The general solution of Ax = b is (x1, x2, x3) = s( - 4, 9, 0) + (- 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) = s(- 4, 9, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = s( - 4, 9, 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) = (- 4,9, 0) + 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 X1 + X3 -4 2x1 + x2 + 3x3 1 O The general solution of Ax = b is (x1, x2, x3) = (- 4,9, 0) + s(- 1, - 1, 1); and the general solution of Ax = 0 is (x), x2, x3) = (- 4, 9, 0). O The general solution of Ax = b is (x1, x2, x3) = (- 4, 9, 0) + s( - 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, X3) = s(- 1, 1, 1). %3D O The general solution of Ax = b is (x1, x2, x3) = s( - 4, 9, 0) + (- 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) = s(- 4, 9, 0) + (- 1, - 1, 1); and the general solution of Ax = 0 is (x1, x2, x3) = s( - 4, 9, 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) = (- 4,9, 0) + s( – 1, – 1, 1).
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