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Let $A=\left[\begin{array}{rr}{2} & {-1} \\ {-6} & {3}\end{array}\right]$ and $\mathbf{b}=\left[\begin{array}{l}{b_{1}} \\ {b_{2}}\end{array}\right] .$ Show that the equation $A \mathbf{x}=\mathbf{b}$ does not have a solution for all possible $\mathbf{b},$ and describe the set of all $\mathbf{b}$ for which $A \mathbf{x}=\mathbf{b}$ does have a solution.

$-3 b_{1}$ only for these set of values the system is consistent

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 4

The Matrix Equation Ax D b

Introduction to Matrices

Campbell University

Oregon State University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

01:11

01:28

If a system has an infinit…

01:19

Solve using the eliminatio…

06:44

Determine all values of th…

01:07

Solve each system graphica…

01:09

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01:43

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01:20

01:03

Determine whether each equ…

in this video, we have a two by two matrix a a Vector B, and we put them together so that we can consider the Matrix equation A X equals B. Let's see what the solutions might be for such your equation. We can put the entire system in the form of two native six negative 13 and augment with B one and B two. If we do, this will be able to reduce to echelon form to determine whether or not this system is consistent. So, first, to get this in echelon form, let's eliminate the negative six indicated here I'll copy Row one, which is to negative one B one, and our operation will be to multiply row one by three at the result to Row three and record the result here. So we'll obtain altogether 00 and be too plus three B one. So that's our system. So far, let's analyze where the pivots might be. Well, there's definitely a pivot here, but I say might be the A pivot here because that just depends on the value of B two and three B one. So now we can say if B two plus three B one is not equal to zero, then the right most column has a pivot when we know that when there's a pivot in the right, most column, then a X equals B R system in matrix form is inconsistent, meaning there will not be solutions. Now, if be to plus three B one is equal to zero, then the system will be consistent. And that's because if we have this quantity equal zero, that means there is a zero here, and this highlighted position is not a pivot position, forcing the system to be consistent. Notice that this equation could also be ascribed at described as be too equals negative three B one. So this is a line in the variables B two and B sub one, and this line would go through the origin. So as long as you choose values for B two and B one that's on that line described here will have a consistent system for this matrix equation.

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