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In Exercises $5-8,$ determine if the columns of the matrix form a linearly independent set. Justify each answer.$\left[\begin{array}{rrrr}{1} & {-3} & {3} & {-2} \\ {-3} & {7} & {-1} & {2} \\ {0} & {1} & {-4} & {3}\end{array}\right]$

Linearly Dependent

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 7

Linear Independence

Introduction to Matrices

Missouri State University

Campbell University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

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In Exercises $5-8,$ determ…

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In Exercises 13 and $14,$ …

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In Exercises $1-4,$ determ…

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[M] In Exercises $37-40$ ,…

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In Exercises 24–26, use de…

in this video, we're gonna be solving question number eight from section 1.7, which is based on linear independence here. Were given a matrix that is three by four matrix initially and were asked, uh, were asked if it is new. If the columns of this matrix are literally independent, Um, to find that out, we need to learn about Leonard. Defendants. Linear independence is basically a set of vectors is linearly independent? If, um, if if the matrix equation X equals zero only has the triple solution, which, in this case is this right here a column of four zeros. Because here, uh, we know that if you split this matrix into individual column vectors, you have four column vectors, which means that that's how many variables we're gonna be solving for if we were to set this question up as a system of equations Ah, and setting them equal to it like, uh, zero. Hence X equals zero. So this is how you set up in augmented matrix of X equals zero right here. Um, and to solve this ah problem. In an easy step, you just consider the roommate from the book and I'd recommend reading that chapters of the book or a section of the book as they're very concise. And the book does a really good job of explaining how to do every proper, like the concept of how to do every problem so they're made of. The book states that less. Let's say we have n representing the number of row vectors and appear representing the number of column vectors. So here N is three as their three row vectors and, um, so n equals three and P equals four as there's four, um, Colin vectors as we're selling for four variables. Ah, so four is greater than three. Therefore, this system is literally independent, because solving this system would result in a free variable where one of the variables, uh, is called a free variable, where you can choose any value so there on the number line. So there's infinitely possible values on, and that free variable causes the values of the rest of the variables to change, as the rest of the variables will be in terms of that variable. So once you perform elementary row operations on this, um, this augmented matrix you'll end up with this matrix where um there's leading entries in three of the four columns, but the four ah, one of the columns does not have a leading entry, which means not call them. If he labeled this x y, uh x one x two x three and x four x three would be a free variable. This means that, um, this means that this system has more solutions than just the trivial solution. Therefore, um, the system is linearly dependent. That's what Jeremy it is really saying.

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In Exercises $5-8,$ determine if the columns of the matrix form a linearly i…

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