WEBVTT

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

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The determinant is 0, so the matrix is singular: it has no inverse and its rows are linearly dependent.

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Computed exactly with fractions; row reduction gives the same value 0.