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In Exercises 11 and $12,$ determine if $\mathbf{b}$ is a linear combination of $\mathbf{a}_{1}, \mathbf{a}_{2},$ and $\mathbf{a}_{3} .$$$\mathbf{a}_{1}=\left[\begin{array}{r}{1} \\ {-2} \\ {2}\end{array}\right], \mathbf{a}_{2}=\left[\begin{array}{l}{0} \\ {5} \\ {5}\end{array}\right], \mathbf{a}_{3}=\left[\begin{array}{l}{2} \\ {0} \\ {8}\end{array}\right], \mathbf{b}=\left[\begin{array}{c}{-5} \\ {11} \\ {-7}\end{array}\right]$$

$\mathrm { B }$ is not a linear combination of $a _ { 1 } , a _ { 2 } ,$ and $a _ { 3 }$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 3

Vector Equations

Introduction to Matrices

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we'll get so far a problem. Tell work, even three vectors. They want you to a three, and we're also keeping the victor be. So we want to determine if bees alina combination off. You wanted to be three. So that is actually to find out the Matrix a spent by the white to a three. So that is one negative, too. Two and 05 by and 208 and Victor be that is negative. Five 11 7 Elected seven. So here we only need two. The question that want to answer instead whether whether this system is consistent. So to find out what the system is consistent, we need to consider the augmented matrix. So he's 102 negative five and negative. 250 11 and 258 Active seven. So by apply gushing elimination, we'll find, uh, re reduced rate. You should inform off this album it'd reach matrix. I'll just read it out here. It will be one. There are two zero and 01 for fifth and zero last role will be 0001 so you can find that the last rule it's not possible because zero cannot be won. If we have a vector times zero vectors, then you know how it has to be there around 11 side. But on the right hand side, we have the we have the, uh that's the return. So that means that system is not consistent. So one more do better answer our question. That is to say, he's not a linear combination all the way to a three not linear combination.

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