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describe the possible echelon forms of the matrix. Use the notation of Example 1 in Section 1.2.$A$ is a $2 \times 2$ matrix with linearly dependent columns.

$\left[ \begin{array} { l l } { \mathbf { 1 } } & { * } \\ { 0 } & { 0 } \end{array} \right] , \left[ \begin{array} { l l } { 0 } & { + } \\ { 0 } & { 0 } \end{array} \right]$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 7

Linear Independence

Introduction to Matrices

Oregon State University

McMaster University

Lectures

01:32

In mathematics, the absolu…

01:11

00:32

Describe the possible eche…

02:56

03:03

02:55

Find a row echelon form fo…

02:38

Write the matrix in row-ec…

02:25

In Exercises 29 and 30 , d…

01:26

CAPSTONE In your own words…

03:16

00:54

Complete each of the follo…

02:54

this video, we're gonna be solving a problem number. Um, 24 of section 1.7 was based on a linear independence. So here is asking us all the possible, uh, entries that are two bites you can have two by two matrix gonna have if it's linearly dependent for any arbitrary, too wet too. So ah, and it's long issues indication that the book used in section one point to where it had squares be, Ah, leading entry, which is a pivot and, um, position and ah star be any entry that is non zero. So here, for ah, set of vectors or column vectors to be linearly dependent that the X equals zero matrix equation has to have, uh, once at least one solution that is not the trivial solution, which is X equals a column vector of two zeros. In this case, since two by two can only has two columns. That means if you were to set this up as a linear equation, linear our system, it's a system of linear equations. Then we would solve for two variables, so that means it should have two entries, which is the triple solution. So the only way that a set of records can not only have the trivial solution of an answer is if, um, one of one of the columns is does not contain a pivot so that that column, or the variable associated with that column is a free variable where all the other variables in the in the system are written in terms of that variable. So if that variable changes, the value of the other variables also changes. So here, the first possibility here is, um is if the first entry is is a pivot, and the second is a non zero or a zero. Um, and if the bottom row is just zeros, because that is the only way, um, that one of the variables can be free as, um as the second row does not have, ah, pivot position where the second column doesn't have a position. And the second possibility is if, ah, the first orders have ah, pivot. But it has a non zero leading entry. So which would still results in the first column being free as um by Durham nine as its space that if a set of vectors contains zero becker than it is linearly dependent because 00 vector causes the value of that very well, corresponding to zero vector to be free variable.

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In mathematics, the absolute value or modulus |x| of a real number x is its …

Describe the possible echelon forms of the matrix. Use the notation of Examp…

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In Exercises 33 and $34, T$ is a linear transformation from $\mathbb{R}^{2}$…

02:13

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01:37

Show that if $A B$ is invertible, so is $B$

00:53

Unless otherwise specified, assume that all matrices in these exercises are …

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In Exercises $33-36,$ verify that det $E A=(\operatorname{det} E)(\operatorn…

02:58

Each statement in Exercises 33–38 is either true (in all cases) or false (fo…

07:38

With $A$ and $\mathbf{p}$ as in Exercise $7,$ determine if $\mathbf{p}$ is i…

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Exercises $23-26$ display a matrix $A$ and an echelon form of $A .$ Find a b…

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