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Let $\mathbf{v}_{1}=\left[\begin{array}{r}{1} \\ {0} \\ {-2}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{-3} \\ {1} \\ {8}\end{array}\right],$ and $\mathbf{y}=\left[\begin{array}{r}{h} \\ {-5} \\ {-3}\end{array}\right] .$ For whatvalue(s) of $h$ is $\mathbf{y}$ in the plane generated by $\mathbf{v}_{1}$ and $\mathbf{v}_{2} ?$

$h = - 3.5$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 3

Vector Equations

Introduction to Matrices

Missouri State University

Campbell University

Baylor University

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Okay, so for a problem 18 we wouldn't want to find the value of age. So that, uh, Victor why he's saying that he's on the plane generated to buy of You want to be too. So, by the giving information we know first that next next to Sorry, I'll just first right down. Do you want me to? So you wanna be too? Will be 10 connected to and neck of 318 And Victor, why will be age next five and collective three. So that is actually to consider this Al Commended Matrix age Negative five. Elective three. We want this mission this matrix to be consistent. That is the say, a matrix v spend by spend by V one and V two. So we're considering the system. He actually was. Why? Okay, So we need to do what we need to do here is to performed the cash indignation. So it's not a complicated matri. So let's do this by hand. So way first used just 1/3 row plus twice up the first rule. So row three, us twice off row one. So we can cancel the first entry on the third row. So He's a general and connective six here and plus eight. So we'll be inactive. Started positive, too. And for the constant term, we have connective three plus two age. Okay. And I'll keep the kitty out of bettors. One negative three and a JJ. Okay, so the next we want to cancel the third the second entry off on 1/3 row. So to do that, we we use Ah, the role minus twice off the second roll. So, bro. Three minus twice off road, too. So it will be one first. Right down. Ian changed those, and we have 01 and negative five. So for the last roll, there are zero to minus two times one. So that zero and negative three plus two age miners two times negative. Five. So? So it's ah, um, right down. So elective three lasts to age minus two times. Row two will be negative. Five. So here it will be naked. Three plus two age plus 10. And this will be seven thus to age. Okay, so too to make they're all commended me may trace. Um, consistent. We need to make the last entry on the right hand side to be zero. So we require, in this case, age to be inactive. Seven have so that day sentry will be zero than the system can be solved. Otherwise the system cannot be solved.

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