WEBVTT

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Here is the problem: [[1, 2, 3], [0, 1, 4], [5, 6, 0]].

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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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Clear column 1 using the pivot in R1.

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Clear column 2 using the pivot in R2.

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Clear column 3 using the pivot in R3.

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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 [[-24, 18, 5], [20, negative 15, negative 4], [-5, 4, 1]].

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The determinant is 1, 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.