if the matrix is squared, and it has an inverse, then the answer is YES
2006-11-28 16:52:32
·
answer #1
·
answered by Anonymous
·
1⤊
0⤋
Just as division is really multiplication by the reciprocal (or multiplicative inverse), think of the inverse of a matrix as its multiplicative inverse;
For matrix M, M^(-1) is the inverse
MM^(-1) = I , the identity.
So,
[M^(-1)}^2 = M^(-1)M^(-1)
So that MMM^(-1)M^(-1) = I
Just as you can have M^k you can have M(-k)..
Notice that if the determinant of a matrix is zero then it has no inverse. So, that's like being zero.
2006-11-28 17:02:05
·
answer #2
·
answered by modulo_function 7
·
0⤊
0⤋
imagine about what powers are. in simple terms repeated multiplication. In matrix multiplication, you may absolutely multiply matrices in the experience that they meet particular criteria about kind of columns and rows. attempting to multiply matrices that do not meet those criteria is incomprehensible, so i ought to wager that throughout maximum situations, absolutely sq. matrices must be raised to a plausible because the rows of one ought to equivalent the columns of the different, even although that i have not concept that with the help of totally, so i ought to nicely be incorrect. you may't have non-complete numbers because you may have fractional matrices.
2016-11-27 20:28:35
·
answer #3
·
answered by ? 4
·
0⤊
0⤋