Homepage › Solution manuals › Gilbert Strang › Introduction to Linear Algebra › Exercise 3.3.6

Introduction to Linear Algebra

Exercise 3.3.6

Answers

(a): Solvable if $b_{2} = 2 b_{1}$ and $3 b_{1} - 3 b_{3} + b_{4} = 0 .$ Then $x = [\begin{matrix} 5 b_{1} - 2 b_{3} \\ b_{3} - 2 b_{1} \end{matrix}] = x_{p} .$
(b): Solvable if $b_{2} = 2 b_{1}$ and $3 b_{1} - 3 b_{3} + b_{4} = 0 .$ Then $x = [\begin{matrix} 5 b_{1} - 2 b_{3} \\ b_{3} - 2 b_{1} \\ 0 \end{matrix}] + x_{3} [\begin{matrix} - 1 \\ 1 \end{matrix}] .$

mathchakra

2022-01-23 14:55

Comments