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Divide by singular matrix

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1. If matrix A has determinate 0, let A = (10, 5, // 10, 5), and be a 2x2 matrix. let B = some random 2x2 matrix. Let C = AB. If C = (25, 40, // 25, 40) how can we work out B? Since the determinate is 0, we cannot use that method.
2. (Original post by Timizorzom)
If matrix A has determinate 0, let A = (10, 5, // 10, 5), and be a 2x2 matrix. let B = some random 2x2 matrix. Let C = AB. If C = (25, 40, // 25, 40) how can we work out B? Since the determinate is 0, we cannot use that method.
Try letting B = (a, b // c, d) So
(25, 40 // 25, 40) = (10, 5 // 10, 5)(a, b // c, d)
Solve for a, b, c, d
3. (Original post by Timizorzom)
If matrix A has determinate 0, let A = (10, 5, // 10, 5), and be a 2x2 matrix. let B = some random 2x2 matrix. Let C = AB. If C = (25, 40, // 25, 40) how can we work out B? Since the determinate is 0, we cannot use that method.
You can't find any unique solution because there exists an infinite amount of solutions.

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Updated: October 17, 2016
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