if A = [1 -2 -1] A = [2 -3 1] A = [-3 5 0] and B = [0 1 0] B = [-1 3 2] B = [1 -4 -2] find one matrix X such that AX = B. A and B are 3x3 matrix above
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$\displaystyle x=a^{-1} b.$
Originally Posted by Krizalid $\displaystyle x=a^{-1}b.$ But A and B both end up with a row of zeros, they don't have an inverse.
B doesn't need to be invertible, just A, and A is invertible.
Originally Posted by Krizalid B doesn't need to be invertible, just A, and A is invertible. How is A invertible? Won't one find a row of zeroes on the left-hand side of the augmented matrix, implying that A is singular? Incidentally, the same situation will arise with B.
what's its determinant?
Originally Posted by Krizalid what's its determinant? Such that it's not invertible.
ohh, i just realized that i got a typo in the third row when computing the determinant. anyway, there's no X, but at least the OP knows now how to solve the problem.
I have got the solution for the question, X = [-2 3 4] X = [-1 1 2] X = [0 0 0 ] thanks for the help
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