How to Find the General Solution for a Matrix with a Free Variable?

[Ad123q]

I was just wondering, if I had a matrix in reduced-row echelon form, say,

1 0 0 2
0 1 3 1
0 0 0 0

then I could write the general solution as a1= -2a4 , a2= -3a3-a4, with a3 and a4 defined in terms of these. (I obtained this solution by putting a1, a2, a3 and a4 under the first, second, third and fourth columns, equating the matrix to zero, and solving for the unknowns under the leading 1's).

But how would I find the general solution of, say

0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0 ?

Any help much appreciated.

 

[moo5003]

If you are having a hard time deciphering the matrix representation you can always put them back in equations

In the latter case:

0a + b + 0c + 0d = 0
Thus b = 0

0a + 0b + c + 0d = 0
Thus c = 0

0a + 0b + 0c + d = 0

and 0 = 0 for the last equation

Thus the only vectors to satisfy the matrix are:

[x, 0, 0, 0] transpose. Where x is any real number (I assume these are matrices over R)

 

[Ad123q]

Thanks, so if I was to write x1, x2, x3, x4 under the matrix in question, would the solution space for the general solution just be x = {x1 x2 x3 x4} = {x1 0 0 0} ?

 

[Defennder]

But how would I find the general solution of, say

0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0 ?

I assume the last column is the RHS of a system of linear equations. Note that the last entry of the third row of the matrix is non-0 whereas all the coefficients of the unknowns you are solving for are 0. What does that tell you?

 

[HallsofIvy]

It would have helped a lot if you had asked the question correctly! There is no such thing as a "general solution of a matrix" any more than there is a general solution of a number.

What you were asking for is a general solution to the equation Ax= 0, or, equivalently, the null space of A.
The matrix equation
\left[\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 &1\\ 0 & 0 & 0 & 0\end{array}\right]\left[\begin{array}{c}x_1 \\ x_2\\ x_3\\ x_4\end{array}\right]= \left]\begin{array}{c}0 \\ 0 \\ 0 \\ 0\end{array}\right]
reduces, as you say, to x₁= 0, x₂= 0, x₃= 0 and 0= 0. Obviously, x₁, x₂, and x₃ must all be 0 but there is no restriction at all on x₄. The "general solution" is [0 0 0 x] where x can be any number. The null space is the subspace of R⁴ spanned by [0 0 0 1].

 

[adriank]

Almost right, except I think you got your matrix multiplication backwards; as said before, x₂, x₃, and x₄ must be zero, and there is no restriction on x₁.

:)

 

[CalculusSandwich]

Correct you write your matrix in terms of the free variable which in this case would be x1.

So [1,0,0,0] would be the solution set. Also if this is an augmented matrix the general solution would be inconsistent since, 0x1 + 0x2 + 0x3 = 1, cannot be true.