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Suppose $A$ is a $3 \times n$ matrix whose columns span $\mathbb{R}^{3}$ . Explain how to construct an $n \times 3$ matrix $D$ such that $A D=I_{3}$ .

The matrix $D$ can be obtained from the matrix $A$ of order $3 \times n$ which spans $\mathbb{R}^{3}$ .

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

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in this video, we have a three by N Matrix. So that's three rows and columns, and we're going to assume that the columns of a are going to span all of the space are three. Next, let's addict something extra to this picture. Let's say assume de is a three or N by three matrix, for which eight times D equals I three. So what we want to determine here is what does this matrix de equal in particular? Well, let's start with I three. We know that I three is the three by three identity matrix, so it is formed by columns E one E two e three, where e one is 100 E two is 010 anti three is 001 so altogether were then looking for a matrix D, such that when we multiply a times d, we obtain this three by three identity matrix we see displayed here. First, let's also note that since the columns of a span or three each time we write a matrix equation eight times X equals B, where be is in are three. This matrix equation is always consistent. This is because knowing that the columns of a spin or three is logically equivalent to the statements that a X equals B is always a consistent system of equations. Let's use that fact next to construct our matrix de here, we can say that the systems eight times X equals e one eight times X equals e to eight times X equals e three were e one e two e three are the columns of our identity matrix here. Each one of these systems are consistent, since we could have placed anything in these positions on the right hand sides and they're still consistent. Let's suppose then that let's call this d one de to d three our solutions. So in other words, let's say what we mean exactly were saying that since eight times X equals e one e two or three here is always consistent. It follows that a Times D one must be equal to the one. Since this is our first solution. Eight times D, too, is E to and likewise, eight times D three is E three. Now let's then define the matrix D By columns D one do to d three, where these came from. The solutions that we had constructed. Then when we take a times D, this is what will produce by the definition of a matrix times another matrix. We take a Times first D one, which came from here. Then the second column of the product is a times D to and lastly, we have eight times d three for the third column of this product. So this is what eight times D looks like. But we also know that eight times d produces e one e two, any three so we could make a substitution here, here and here to get the following. We have the one e two and e three, but this is now the three by three identity matrix. So we've just shown how can to construct the particular matrix de such that eight times d equals I three.

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