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

$-28$

Algebra

Chapter 3

Determinants

Section 2

Properties of Determinants

Introduction to Matrices

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were given a matrix and rest to find the determinant of this matrix I wrote reducing it to echelon form. So this is the four by four Matrix with entries one negative 13 Sorry negative. 30 0154 Negative 1053 and three Negative. Three. Negative 23 And so first I'll add Row one to Row three and I'll subtract three of row one from Row four. So I get that The Matrix and one negative. One negative. 30 01 54 zero Negative one to three and then zero Negative. Three plus three zero negative, too. Plus nine seven and then three. Next I'm going to add wrote to to row three. So I get the Matrix. One negative, one negative. 30 0154 00 77 and 0073 Finally, I'll subtract Row three from RAV four and I'll get the Matrix. One negative, one negative. 30 0154 0077 and 00 zero. Negative four. Now that our matrix is an echelon form, we know that the determinant is the product of the diagonal entries which is negative 28. And this is our answer

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