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[M] Use as many columns of $A$ as possible to construct a matrix $B$ with the property that the equation $B \mathbf{x}=\mathbf{0}$ has only the trivial solution. Solve $B \mathbf{x}=\mathbf{0}$ to verify your work.$A=\left[\begin{array}{rrrrrr}{12} & {10} & {-6} & {-3} & {7} & {10} \\ {-7} & {-6} & {4} & {7} & {-9} & {5} \\ {9} & {9} & {-9} & {-5} & {5} & {-1} \\ {-4} & {-3} & {1} & {6} & {-8} & {9} \\ {8} & {7} & {-5} & {-9} & {11} & {-8}\end{array}\right]$

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Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 7

Linear Independence

Introduction to Matrices

Campbell University

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:32

In mathematics, the absoluâ€¦

01:11

03:28

[M] Use as many columns ofâ€¦

02:31

Use matrix multiplication,â€¦

03:55

Use the matrices below to â€¦

03:44

01:38

Find matrix $A$ if$$B=â€¦

03:22

04:37

04:33

04:45

01:47

and this question, we're gonna be selling ball number 42 of, um, the literature textbook from section 1.7, which is based on a linear independence. So the problem here is asking us Thio use as many columns of a provided ah, that I wrote out here So constructive matrix equation be where such that B X equals zero has only the trivial solution, which in this case, um, but it's just a zero calling Becker with as many variables as possible from from a so that this means that be as linearly independent. So it's asking us to construct a set by using the columns of a thio constructive, linearly independent set of B using as many calls a ce possible which are linearly independent themselves. So to do this, you just have the solve the ah matrix equation X equals zero, which I set up right here. Azan augmented matrix where I just contaminated a with a zero vector and you wrote, is it? I recommend using a R e f calculator or just a regular echelon formed calculators because, um, if it's any bigger than like a four by four or four by three as it would take a long time to get this in echelon form. So I haven't written out the echelon form of Matrix. But basically, the rose would earn the columns without a pivot position, which is basically it, um, basically a leading entry in every column. So the columns without a leading entry was a pivot position are, um, for your variable. So in the in the final matrix that is reduced Columns three and five are free variables. So we cannot include this these columns and be as road using them would give us. Ah, would end up making them free variables. So making the making the entire set, which would make the entire set linearly dependent. So if you exclude these two columns, these other four columns 124 and six are linearly independent themselves because all of them contain, uh, leading entries that air pivots, uh, making it so that there are no free variables and be making the entire set of be linearly independent. So be could be written as dog went emitters that contains the columns 124 and six. So in this case it would be a 12 negative seven nine negative 48 10 Negative six nine Negative 37 We're just calling to and calling three are calling for It is negative. 37 Negative 56 Negative nine and column six is 10 if I make it a born mine negative idiot, which will be the Matrix for B, and it would literally independent.

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