WEBVTT

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Here is the problem: [[1, 0, 2], [2, 1, 0], [0, 3, 1]].

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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 13.

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The determinant is 13, 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 13.