- #1
n0_3sc
- 243
- 1
Well here's the question:
Find 'X' where 'AX = '
6
8
4
and 'A = '
1 2 4
3 1 2
0 2 4
Now i would go ('AX')*('A^-1') which would clearly give me 'X' BUT 'A' is not invertible! So how do i do this? (by the way the ' is not part of the question).
And then it says find the null vectors of A and hence the general solution. what do i do here?
Find 'X' where 'AX = '
6
8
4
and 'A = '
1 2 4
3 1 2
0 2 4
Now i would go ('AX')*('A^-1') which would clearly give me 'X' BUT 'A' is not invertible! So how do i do this? (by the way the ' is not part of the question).
And then it says find the null vectors of A and hence the general solution. what do i do here?