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Let $A$ be a $3 \times 2$ matrix. Explain why the equation $A \mathbf{x}=\mathbf{b}$ cannot be consistent for all $\mathbf{b}$ in $\mathbb{R}^{3}$ . Generalize your argument to the case of an arbitrary $A$ with more rows than columns.

In general, when matrix is $m \times n$ matrix and $m>n$ (more rows than columns) it can span $\mathbb{R}^{n}$ and not $\mathbb{R}^{m}$ because it can have at most $n$ pivot columns.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 4

The Matrix Equation Ax D b

Introduction to Matrices

Oregon State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

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in this video, we're starting off with a matrix. A. It's going to be of size three by two. So what we want to do here is to consider the Matrix equation eight times X equals a Vector B. Now he has chosen arbitrarily on the right hand side. So a big question about this equation is whether it is consistent or not. That's hard to do when we don't know what The Matrix A is specifically but will use the size of the matrix toe. Analyze the pivots. Let's start by imagining what a could look like. It's going to be of size three by two. So I want to write something that has three columns and or excuse me three rows and two columns, so they'll be entry here, here and here. I'm just going to put little dots here to give them blink arbitrary entries as well as entries here, here and here. Now, the best case for this Matrix is that we have a pivot in the first column, then in the second column. But to get any more pivots, we would require 1/3 column. So there is always going to be a row if we're dealing with three by two where there is no pivots. So this matrix a has cannot have a pivot in every row, so a cannot have a pivot in every row. What this tells us, then, is that a X equals B is not consistent for all vectors. Be in our be sure to use the three here. So for all vectors, be in our three. It doesn't tell us that a X equals B is always inconsistent. What it does tell us is that we can find a vector B for which it will not be consistent. No, we can ask ourselves why did this happen? Suppose a is and m by N matrix. Our matrix was a size three by two. And if m is bigger than n so that we have more roses than columns just like in our case, then we will know immediately that there cannot be a pivot in every row and we can generalize even further to say that for such a situation, a X equals B will not be consistent. These air s is sorry for every vector Be in are now We have to be careful. It would be always this dimension First, just like we picked this dimension. So in our M and that generalizes the three by two situation to an end by N one where specifically their arm or rose than columns.

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