Homogenous Equation: Finding 'X' with Non-Invertible Matrix A | Solution Help

[n0_3sc]

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?

 

[outandbeyond2004]

I thought it went this way:

$$(A^{-1})AX = (A^{-1})\left(\begin{array}{c}6\\8\\4\end\right)$$

True, A is not invertible. But, if it were, that's how I would have written that.

 

[n0_3sc]

That is the way its meant to go, but their still has to be some way (apparently using row reduction) to solve for 'X'. because 'X = '
2
0
1
which works. ie
1 2 4 | 2 6
3 1 2 | 0 = 8
0 2 4 | 1 4
but how do they get 'X'?

 

[matt grime]

Go back and do your old fashioned solution by substitution, which is all you do with row operations. What you'll find is that you end up with an equation in two unknowns, say ax+by=c

let x=t some parameter, solve for y in terms of t, put thos back into one of the original equation to get z in terms of t - you've thus solved the equation for the family of solutions.

Other options might happen because for a singular matrix, their maybe a plane of, a line of, or no solutions.