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

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Algebra

Chapter 3

Determinants

Section 2

Properties of Determinants

Introduction to Matrices

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University of Michigan - Ann Arbor

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were given a matrix and were asked to find the determinant of this matrix by reducing it to echelon form. This is the four by four matrix. 1302 He's with Rose negative to negative. 574 35 to 1 and one negative one to negative three. And so first I'm going to add twice the first road with second row. Subtract three times the first row from the third row and I'll subtract the first road from the fourth row. So I get determinant of the Matrix. 13 02 zero one seven eight zero five minus nine is negative. Four zero. Sorry to, I mean and one minus six is negative. Five and then zero negative one minus three is negative. Four to and then negative three minus two is negative. Five. And then row reducing further, although it really isn't necessary at this point, we're now going to add four times, wrote to to row three and four times wrote to to row four. So I get the matrix. 1302 zero one 78 00 Then we have seven times four is 28 plus two is 30 and eight times four is 30 to minus five is 27 and the last row is also 00 30 27. And now I will subtract Row three from RAV four and doing so. I obtained the Matrix 1302 0178 00 30 27 and 0000 So this row of zeros and it's clear, since this is an upper triangular matrix determined is Thebe, product of the diagonal entries, and so this is zero.

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