1
sriraghav
·2009-03-28 11:02:22
In stmt 2: shd it not be " B is of order n x m"??
1
sriraghav
·2009-03-28 11:22:21
some one answer this one..
1
The Scorpion
·2009-03-28 11:27:24
not necessarily... only required condition is dat "no. of rows in second matrix must b equal to no. of coloumns in first matrix... to multiply two matrices..."
well, u can also conclude lyk dis... AB=A => B is an identity matrix, and identity matrices are always square matrices... :)
1
sriraghav
·2009-03-28 11:30:16
so in stmt 2 : there can b any value in n x __ ....And it will bcome stmt 2 as wrong one....
1
sriraghav
·2009-03-28 11:30:46
coz stmt2 is not always true
1
The Scorpion
·2009-03-28 11:33:42
nope... see let me tell u clearly now...
dis is d basic step of multiplication...
(m x n).(n x p) = (m x p) ...... rite...!!!
now since RHS is A... and first matrix in LHS is also A, so both of them should have same order... i.e., (m x n)... hence p=n... which implies dat order of B must be (n x n)... implies it has to b a square matrix...
got it now... :)
11
Subash
·2009-03-28 19:27:03
The second statement is wrong i too answered it the same way
@mak
it is not true because for only the product to be defined the order of B can be nxp (p ≥1)
1
Optimus Prime
·2009-03-28 20:14:56
sriraghav how much did u get in the test?
1
The Scorpion
·2009-03-29 10:07:47
oh yeah... i'm sorry... i read statement 2 in ref. wid statement 1... so made dat blunder...
well, considering statement 2 individually, it's false... i agree... :)