WEBVTT

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Here is the problem: [[2, 0, 0], [0, 3, 0], [0, 0, 4]].

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Find the inverse of a 3 times 3 matrix.

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Check that the determinant is non-zero.

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Write the augmented matrix [A | I].

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Divide R1 by 2.

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Divide R2 by 3.

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Divide R3 by 4.

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The right half is now the inverse.

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Check: A times A to the power minus 1 equals I.

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So the answer is A to the power minus 1 equals [[1 over 2, 0, 0], [0, 1 over 3, 0], [0, 0, 1 over 4]].

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The determinant is 24, so A is invertible. Every entry of the inverse is exact.

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Multiplying A by the result gives the identity matrix exactly.