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# Divide by singular matrix Watch

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.

Updated: October 17, 2016
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