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…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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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.

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

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

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