let A= first row [3 -6] second row [-1 2]
construct a 2×2 matrix B such that AB is the zero matrix. Use two different nonzero columns for B.
when i get zero its only 2 and the other 2 numbers aren’t. I just can’t figure out how to get AB to be all 0.
choose any c, d
For all genuine numbers a and b, a*b = 0 if and provided that a = 0 or b = 0. regrettably, the analogous effect for matrices does no longer carry. here’s a trouble-free counterexample. permit A be first matrix and B the 2nd. [ a million 0 ][ 0 0 ]..=.[ 0 0 ] [ 0 0 ][ 0 a million ]…..[ 0 0 ] Then AB = 0 and A ? 0 yet B ? 0. notice: A is invertible if and provided that the determinant of A is nonzero. the undeniable fact that the Matrix A is nonzero does no longer mean that the Determinant is nonzero. so we will not end that A is invertible.