According to the textbook, the basis of the null space for the following matrix: A = (1 3 − 2 0 2 0 2 6 − 5 − 2 4 − 3 0 0 5 10 0 15 2 6 0 8 4 18) is found by first finding the reduced row echelon
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The dimension of the null space comes up in the rank theorem, which posits that the rank of a matrix is the difference between the dimension