Kernel is basis of domain space.
The mapping A(x,y,z,w) =(x+2y+3z+1w,x+3y+5z-2w,3x+8y+13z-3w).
/*Each row of the matrix */
Thus after reducing equations of A:
x + 0y -z + 7w = 0
0x + y +2z -3w= 0
By these 2 equations we get:-
z = (-y+3w)/2
x = (-y-11w)/2
Maximum basis of A is 5. But, by solving we get:
S = ( (-y-11w)/2,y,(-y+3w)/2,w)
Now we know that, S is a subset of A. Conversely, to show A is subset of S.
As S = {y(-1/2,1,-1/2,0) + w(-11/2,0,3/2,1)} is space of A
with elts={(-1/2,1,-1/2,0), (-11/2,0,3/2,1)} /*As calculated above*/
Thus, S = A So, basis of kernel is {(-1/2,1,-1/2,0), (-11/2,0,3/2,1)}.
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Now, basis of image of A:- Standard basis of R4.......
A(1,0,0,0) = (1,1,3)
A(0,1,0,0) = (2,3,8)
A(0,0,1,0) = (3,5,13)
A(0,0,0,1) = (1,-2,-3)
Reduce RHS from a 4*3 matrix, to get 4*2 matrix /*row echelon form*/
As maximum non empty rows are 2. Thus, basis is 2.
So, {(1,1,3),(2,3,8)} is basis of image of A.