执行行操作R2=R2-2R1使2,1处的项为0。 [127272-2⋅12-2(27)7-2(27)] 化简R2。 [127270107457] [127270107457][127270107457] 将R2R2的每个元素乘以710710,使2,22,2的项为11。 点击获取更多步骤... 将R2的每个元素乘以710,使2,2的项为1。
The documentation pages (numpy.linalg.svd, andnumpy.compress) are opaque to me. I learned to do this by creating the matrixC = [A|0], finding the reduced row echelon form and solving for variables by row. I can't seem to follow how it's being done in these examples. ...
M. K. Ebrahimpour, M. Zare, M. Eftekhari, G. Aghamolaei, Occam's razor in dimension reduction: Using reduced row Echelon form for finding linear independent features in high dimensional microarray datasets, Engineering Applications of Artificial Intelligence, 62 (2017), 214-221....
The method is based on combining existing software to compute the minimal generating set for a “pointed cone” together with standard software to compute the Reduced Row Echelon Form.Dimitrije Jevremović and Daniel BoleyBiosystems