Linear Algebra 6. Given that a linear system Ax b has a solution if and only if b is in the column space1 1 11 2 4biof A, find all vectors b =b2that are in the column space of A =Hint: Put theb32 4 8augmented matix in REF.

Answered: 6. Given that a linear system Ax_b has…

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 60EQ

See similar textbooks

Related questions

Concept explainers

Question

Linear Algebra

6. Given that a linear system Ax b has a solution if and only if b is in the column space
1 1 1
1 2 4
bi
of A, find all vectors b =
b2
that are in the column space of A =
Hint: Put the
b3
2 4 8
augmented matix in REF.

Branch of mathematics concerned with mathematical structures that are closed under operations like addition and scalar multiplication. It is the study of linear combinations, vector spaces, lines and planes, and some mappings that are used to perform linear transformations. Linear algebra also includes vectors, matrices, and linear functions. It has many applications from mathematical physics to modern algebra and coding theory.

Expert Solution

Step 1

Given that a linear system $A x = b$ has a solution if and only if $b$ is in the column space of $A$

Where, $A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 2 & 4 & 8 \end{matrix}]$ and $b = [\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix}]$

We have to find the vector $b = [\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix}]$ that are the column space of $A$

Write the augmented matrix $(A | b)$

$(A | b) = (\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 2 & 4 & 8 \end{matrix} \begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix})$

Now, change the augmented matrix into row echelon form

$\begin{array}{l} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - 2 R_{1} \end{array}$

$(A | b) ~ (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 2 & 6 \end{matrix} \begin{matrix} b_{1} \\ b_{2} - b_{1} \\ b_{3} - 2 b_{1} \end{matrix})$

Now, $R_{3} \to R_{3} - 2 R_{2}$

$(A | b) ~ (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix} \begin{matrix} b_{1} \\ b_{2} - b_{1} \\ b_{3} - 2 b_{2} \end{matrix})$

Step by step

Solved in 2 steps

Knowledge Booster

$Linear Equations$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions