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Suppose $A$ is $n \times n$ and the equation $A \mathbf{x}=\mathbf{0}$ has only the trivial solution. Explain why $A$ has $n$ pivot columns and $A$ is row equivalent to $I_{n} .$ By Theorem $7,$ this shows that $A$ must be invertible. (This exercise and Exercise 24 will be cited in Section $2.3 . )$

the matrix $A$ must convert to row equivalent identity matrix.

Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

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in this example, we start out with this assumption that the Matrix A is of size and by end from there we're going to start adding another condition. Let's assume that the A matrix equation eight times X equals a zero Vector has Onley the trivial solution, So we have not said much about this matrix A. It's a square matrix, and we know that X equals zero is the only solution to that matrix equation. From here we have some surprising consequences. Notice that when we say a X equals zero has only the trivial solution. That means there's a unique solution, and this only happens if a X equals a zero Vector has no free variable. So now that we know that there is no free variable, we can say something further about this particular matrix. A. So this implies when there is no free variable that a has a pivot in every column. Let's pause here for a moment with that statement and use the fact that a is a square matrix. So if a was of size three by three for Justin example to consider on the margin here and if we had a pivot in every column. Then there would be a pivot here, which we could force into a one here and here. But as soon as we find pivots in every column, it means that the pivot is along the main diagonal. And if there was any number here, say a four a row operation would eliminate it. Since we can place zero here and here, then there'll be zeros above and below all the pivot positions. So this would be role equivalent to that to buy or three by three identity matrix. If a row equivalents to this matrix here. So this tells us then that's a ISRO equivalent to the n by N identity matrix. But then, as soon as we know that it's Roy equivalent to that matrix that immediately implies this matrix A is in vertebral that deserves an exclamation mark. Always started with was the assumption that the matrix is square of size in by in. And this equation has on lee the trivial solution. That's all it takes to determine that the Matrix A is in vertebral

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