WEBVTT

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Here is the problem: [[2, negative 5, negative 5], [4, negative 8, negative 7], [8, 1, 8]].

2
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Find the determinant of a 3 times 3 matrix.

3
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Read the 3 times 3 matrix.

4
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Expand along the first row (cofactor expansion).

5
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Evaluate each 2 times 2 minor.

6
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Combine with the alternating signs plus minus +.

7
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So the answer is det(A) equals negative 14.

8
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The determinant is negative 14, which is non-zero, so the matrix is invertible and its rows are linearly independent.

9
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Computed exactly with fractions; row reduction gives the same value negative 14.