Linear algebra help with general solutions

[amy21]

. write the system of linear equ x1 +2x2 +2x4 = 3 (a) as AX=b
2x1 -3x3 +2x4 =6
x1 -x2 -x3 = 0

b find rank and nullity of matrix A

c find general solution of the system .Do the solution form a subspace in R4..indicate the particular solution of the inhomogeneous system and the general solution of the corresponding homogenous system

d give the bases for the row space, column space and null space of A



i have done parts a and b but i am lost on part c and d
can some one explain

 

[DryRun]

For part (c), to find the general solution, i think you're being asked to find the nullspace, which consists of all the special solutions for Ax=0.
To find the particular solution, solve Ax=b; convert to row echelon form and express the pivot variables in terms of the free variables.
Now, for the part where it asks you "Do the solution form a subspace in R4?" -- I'm not so sure about this, as i think "4" should refer to the number of rows in the basis vector(s) of the nullspace.

For part (d), the bases refer to the set of linearly independent vectors. For the row space, just take the rows corresponding to the pivots, directly from the ref of the matrix A. For the column space, take the corresponding pivots columns from the original matrix A. And you can find the basis for the nullspace if you solved the general solution.

 

[HallsofIvy]

So you have AX= b as
$$\begin{bmatrix}1 & 2 & 0 & 2 \\ 2 & 0 & -3 & 2 \\ 1 & -1 & -1 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix}= \begin{bmatrix}3 \\ 6 \\ 0 \end{bmatrix}$$

If you subtract the second equation from the first you get -x1+ 2x2+ 3x3= -3 eliminating x4. That, together with the third equation, x1- x2- x3= 0, gives two equations in three unknowns. Adding those two equations, x2+ 2x3= -3 so that x2= -2x3- 3. Putting that into the previous equation, x1- (-2x3- 3)- x3= x1+x3+3= 0 so x1= -x3- 3. Finally, putting those into x1+ 2x2+ 2x4= (-x3- 3)+ 2(-2x3- 3)+ 2x4= -5x3- 9+ 2x4= 3, x4= -3+ (5/2)x3.

Any solution can be written in the form <x1, x2, x3, x4>= <-x3- 3, -2x3- 3, x3, -3+ (5/2)x3>= <3, -3, 1, -3>x3+ <-3, -3, 0, -5/8>. That should tell you everything you need.

It might be simpler to "row-reduce" the augmented matrix but you should never lose track of the basic concept- to solve the system of equations.