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Should be solved without performing row operations. [Hint: Write $A \mathbf{x}=\mathbf{0}$ as a vector equation.Given $A=\left[\begin{array}{rrr}{4} & {1} & {6} \\ {-7} & {5} & {3} \\ {9} & {-3} & {3}\end{array}\right],$ observe that the first column.plus twice the second column equals the third column. Find a nontrivial solution of $A \mathbf{x}=\mathbf{0} .$

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Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 7

Linear Independence

Introduction to Matrices

Missouri State University

Campbell University

Harvey Mudd College

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

02:55

Should be solved without p…

05:08

Determine the solution set…

04:11

Use Cramer's rule to …

02:25

01:14

Write the system of equati…

02:58

04:44

02:27

Construct a $3 \times 3$ n…

03:16

06:29

uh, this video Where you calling Paul? Number 32 of section 1.7 of the book, which is on linear Independence. So in this problem, we're getting given a Matrix with Rose 416 negative. 7539 Negative three and three And we're also given the ah, the first call them. Let's call it a one plus a second goal in this Call it a two plus two times a second column equals the third column. 83 So here I wrote the equation, though, Um, one sec. Let me change my tip over Penn. Um, so a one plus 282 equals a three. And it's asking us to find a nontrivial solution for the matrix equation. X equals zero. So X equals zero is just another way of writing a constant. So ah, a solution to this inspector would be a nontrivial solution Will be C one sea to N. C. Three, where it's three scaler numbers because we have three columns in our original matrix. Therefore, we're gonna be solving for three variables, and this solution to this equation should give us three variables. So if you multiply that by a one we should get C one times One are if you multiply X by the matrix. Avery She got C one times a one was C two times 82 waas. See, three times eight that we equals zero. So when we look at this, this is indicative of a bad equation. So if we get the first equation that is given to us into this form, we know that a one plus two a two was our, um minus 83 equals zero. And since these two questions are in the same form, one trivial solution to this equation without solving this setting this matrix to zero and I reducing it as it states in the prompted the problem, we can just assume that there are constants constants in front of, um, each of the each of a one a two and a three. So we know that one trivial solution would be X equals 12 and negative one. That is one possible trivial solution out of

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