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In Exercises $1-4,$ determine if the vectors are linearly independent. Justify each answer.$\left[\begin{array}{l}{0} \\ {0} \\ {2}\end{array}\right],\left[\begin{array}{r}{0} \\ {5} \\ {-8}\end{array}\right],\left[\begin{array}{r}{-3} \\ {4} \\ {1}\end{array}\right]$

Linearly Independent

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 7

Linear Independence

Introduction to Matrices

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Harvey Mudd College

Lectures

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In Exercises $1-4,$ determ…

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in this video, we're gonna be solving problem number two of section seven off. Um, chapter one, which is on dinner. Independence. Um, so we're given three column vector 00205 native eight negative 34 and one which I, um which I combine into an argument in matrix for an X equals zero equation. What we're doing here is basically, just rewriting, um, a system of equations into an augmented matrix where we get zero x plus zero y plus native three equals zero and zero x plus five wipe lists for Z equals zero. And so on, Um, and that system of creations history wrote into this argument matrix can only have the trivial solution for it to be a linearly independent. Um, which means that all the variables, uh, can have to be zero. So in this case, X has the equal 000 since, um, there's three columns, which means there's gonna be we're solving for three variables, um, and each each entry in this column represents a solution to one of these variables. So to get that to see if, um, this solution has a has on Lee the trivial solution um, we have toe reduce it. So to do that, I will flip a road three under 01 because it makes it easier. Um, because we already are. Get a staircase pattern, which we want and which is an integral part of the echelon form. Um, and then we want ones at this diagonal. So I just divide the first row by two. Can I get a negative one or one negative? Four and one, huh? The second row by five. So I get zero one and 4/5 in the third row by negative three sawyer 001 And on the right side. Hey, um, So since this matrix is now in basic Ashland form, weaken. And since they're ones at every leading entry, which is also the position, we can conclude that there there can be no free variables possible for this for this augmented matrix. So are for this system. So they're no free variables in the system. That means, um that means this right. Call them the right most home must be the solution to the system, which is X equals 000 And we don't have to solve this any further because we started with zeros on the right and ended with zeros on the right. Um, and we didn't performing your operations where we were. The values of these zeros changed. So we can conclude that the final solution of this side of system of this system would be X equals 000 in a column vector. Ah, And since this set of vectors on Lee has, ah, the trivial solution, that means this set of vectors is linearly independent on, and that is the answer to this.

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