WEBVTT

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Here is the problem: [[0, 2, negative 4], [8, 1, negative 6], [8, 0, negative 8]].

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Find the determinant of a 3 times 3 matrix.

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Read the 3 times 3 matrix.

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Expand along the first row (cofactor expansion).

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Evaluate each 2 times 2 minor.

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Combine with the alternating signs plus minus +.

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So the answer is det(A) equals 64.

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The determinant is 64, which is non-zero, so the matrix is invertible and its rows are linearly independent.

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Computed exactly with fractions; row reduction gives the same value 64.