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If an $n \times n$ matrix $K$ cannot be row reduced to $I_{n},$ what can you say about the columns of $K ?$ Why?

false.

Algebra

Chapter 2

Matrix Algebra

Section 3

Characterizations of Invertible Matrices

Introduction to Matrices

Missouri State University

McMaster University

Harvey Mudd College

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

03:33

Suppose $C A=I_{n}(\text {…

03:43

Explain why the columns of…

09:29

Let $A$ be a lower triangu…

03:03

Show that $A I_{n}=A$ when…

04:14

Let $\mathbf{A}$ be an $n …

02:13

Let $A$ be an $m \times n$…

01:08

Suppose that $D$ is an $n …

02:00

Let $A$ be an invertible $…

06:33

Suppose $A$ is an $n \time…

05:17

Suppose $A D=I_{m}$ (the $…

in this example, we're starting off with the Matrix K, which is of size and by end, meaning that this is a square matrix and Thea vertical metrics there, um then applies to the Matrix K. Let's add another assumption to this list. Let's suppose that K is not row equivalent to the end by N Identity matrix. Well, if this is the situation for this matrix K, then by then vertical matrix. Dirham, this implies that K in verse does not exist. Now, this is the beauty of the convertible matrix, dirham, all statements in that there are logically equivalent. That means if the first statement is that K inverse does not exist, negating that statement than every single statement in the vertical matrix theory is now false for this matrix K. So we can say that this implies immediately that the columns of a are linearly dependent, which is the negation of linear independence. Other things we can say about the columns of a are since all statements in the verdict, bold matrix there more faults are that the columns of a do not span are in and we could go on and on if the columns do not spend are in then saying que x equals B for a matrix equation. We can find vectors be such that the system is inconsistent. But for the major punchline, if case of size and by end not row equivalent to the identity matrix, then the columns of a are linearly dependent and they do not span all of our end.

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