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Unless otherwise specified, assume that all matrices in these exercises are $n \times n .$ Determine which of the matrices in Exercises $1-10$ are invertible. Use as few calculations as possible. Justify your answers.$$\left[\begin{array}{rrrr}{-1} & {-3} & {0} & {1} \\ {3} & {5} & {8} & {-3} \\ {-2} & {-6} & {3} & {2} \\ {0} & {-1} & {2} & {1}\end{array}\right]$$

invertible

Algebra

Chapter 2

Matrix Algebra

Section 3

Characterizations of Invertible Matrices

Introduction to Matrices

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01:32

In mathematics, the absolu…

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Unless otherwise specified…

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For the following exercise…

in this example, we have a four by four matrix a provided, and our job here is to determine if this is in verbal. Well, in order to get to that, we're going to use row reduction here, and we're going to reduce until we are in echelon form. So to that end, we'll start with this pivot position here and eliminate the entries in Rows two and Row three that air shaded. So let's start off by copying in row one. It's negative. One negative three is your on one, and the operations will be to first multiply row one by three at the result to row two will have zero, then negative four, eight and zero for the next operation. Let's multiply row one by negative too. At that result to row three will have zero. It looks like zero here as well. A three and another zero. Now we already have a zero here, so we can just copy in this row. Now it's analyzed the pivots. So far, we have a negative four here, and if we like, we could eliminate this entry by doing the following rule operation divide Row two by four At the results to Row four. First, let me copy The Matrix rose 12 and three. So there's our row one. Row two is zero negative. 48 and zero. We're copping and row three since no changes necessary there either. And now we're ready to begin. So we're dividing by negative for here. Adding the result to the fourth row will have 00 divided by a negative four will have a negative two plus two. So another zero and then copying in a one. Well, that was pretty quick, since this was a nice matrix toe work with were already it already initial on form and along the maid diagonal. We see that every column is now a pivot column. This implies immediately that a must be raw equivalent to the four by four identity matrix, but for using the vertebral matrix there, um, notice that we don't even need to use that step. If a is of size four by four. And we found four pivots, then the vertical matrix theory. Um, on its own already says that a is convertible, so this example is complete

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