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Find the determinants in Exercises 5–10 by row reduction to echelon form.$$\left|\begin{array}{rrrr}{1} & {3} & {2} & {-4} \\ {0} & {1} & {2} & {-5} \\ {2} & {7} & {6} & {-3} \\ {-3} & {-10} & {-7} & {2}\end{array}\right|$$

$-10$

Algebra

Chapter 3

Determinants

Section 2

Properties of Determinants

Introduction to Matrices

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okay. In this question, we want to find a determinant by road reduction to the festive steps that we do is we'll go right. Three minus two times. I wrote one, and we'll put that into three. And I wrote full plus three tons of growing one. And put that integrate for that Use 32 months full. 01 too modest by now. Zero. So seven months. Six gives you 16 Minus forgives you, too. Native. Three plus eight gives you five no los road. So there's zero here, minus 10. Plus my gives you minus one negative. 76 gives you modest one as well on Dhe too. Minus 12. Gives you negative. 10. Yeah. Now will perform two more operations, so we'll get right. Three and one minus one. Right here too, sir. Thanks. Electable three when you get work for or possibly two thanks to that fool. So 132 minus 412 My spy. So this is there a two months to kit is your five minus negative by gives you 10. So 01 plus minus 102 minus one. Did you one negative? 10 minus five. Minus 50. Okay, Now we will swap. Right? Three. Well, slow, right? Three. No, Stop that with rope. For so that gives you 132 months for their one to spy. 001 minus 15 on there. Now, be careful because we we swap Rose. We had two times this by minus one. So this is an upper triangle matrix. So determined is just simply the product of the diagonal. So we should just 10 times one time, one times 1 10 and then we have two times it. Why must one this negative 10?

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