Chat with our AI personalities
I bet it can be done, but I'll be darned if I can!
The identity matrix is a square one with ones (1s) down its main diagonal and zeroes (0s) elsewhere. That is, it must have the same number of rows as columns, and where the row number is the same as the column number, the entry must be 1, elsewhere, it must be 0.
It will either be a 1*23 row matrix or a 23*1 column matrix.
No.
Using the method derived from the usual definition using the minors, the complexity is O(n!). But it seems that one could just do the Gaussian elimination on the matrix, turning the matrix into a triangular one while keeping track of any neccessary row swaps, and then just multiply the values of the diagona. This method would get the complexity of O(n*n).