One can create matrices using the "BasicMathInput" Pallette, like so :

A=(({{0, 0, 8, 16, 104/5}, {-9, -18, 1/5, 92/5, 193/25}, {2, 4, -4, -12, -172/15}, {5, 10, 0, -10, -4}, {-7, -14, 15, 44, 85/2}}))
MatrixForm[A]

Next piece of code produces a random 4-by-5 matrix M with integer entries between 1 and 2, displays it first as a list of rows, then displays it in the usual matrix form, then row-reduces M, and displays its RREF in matrix form. Semicolon surpresses the output of a given line.

M=Table[RandomInteger[{1,2}],{i,4},{j,5}]
"M is"MatrixForm[M]
R=RowReduce[M];
"RREF(M) is" MatrixForm[R]

d={{0},{0}, {0},{0}};
MatrixForm[d]

This is how Mathematica can be asked to find a solution to a linear system with the coefficient matrix R and right - hand - side vector d. Again, we ask for the result of LinearSolve operation to be displayed in a matrix form.

MatrixForm[LinearSolve[R, d]]
(*Not appropriate for homogeneous systems. Just gives one solution!*)

When we ask for the nullspace of M (i.e. for Ker (M)) we get a list of vectors (each written horizontally), such that the elements of the nullspace are exactly the linear combinations of these vectors. If you see just one vector, then the elements of the nullspace are exactly all of the multiples of this vector.

NullSpace[M]

Next we create a random element of (R^4) with integer entries between -10 and 10; we supress the output in the list form and only output the matrix form.

b=Table[RandomInteger[{-10,10}],{i,1,4},{j,1,1}];
MatrixForm[b]

MatrixForm[LinearSolve[R, b]]

NullSpace[R]

Here is a really fancy way to create 4 - by - 4 matrices :

Manipulate[L=List[
{a11,a12,a13,a14},{a21,a22,a23,a24},{a31,a32,a33,a34},{a41,a42,a43,a44}];
"L is" MatrixForm[L],
 {{a11,0}, -10, 10,1},
{{a21,0}, -10, 10,1},
{{a31,0}, -10, 10,1},
{{a41,0}, -10, 10,1},
{{a12,0}, -10, 10,1},
{{a22,0}, -10, 10,1},
{{a32,0}, -10, 10,1},
{{a42,0}, -10, 10,1},
{{a13,0}, -10, 10,1},
{{a23,0}, -10, 10,1},
{{a33,0}, -10, 10,1},
{{a43,0}, -10, 10,1},
{{a14,0}, -10, 10,1},
{{a24,0}, -10, 10,1/10},
{{a34,0}, -10, 10,1/20},
{{a44,0}, -10, 10}
]

... and elements of (R^4) too!

Manipulate[Y=Table[{{y1},{y2},{y3},{y4}}]; MatrixForm[Y],{{y1,0}, -10, 10,1},
{{y2,0}, -10, 10,1},
{{y3,0}, -10, 10,1},
{{y4,0}, -10, 10,1}]

LinearSolve[L,Y];
MatrixForm[LinearSolve[L,Y]]

Finally, a whole different way to solve linear systems :

Solve[{2 x+3 y-z==1,4 x+3z==2,-7x+4y+3z==6},{x,y,z}]

Solve[{2 x+3 y-z==1,-7x+4y+3z==6},{x,y,z}]

There is an alternative, especially when the right hand sides of some of the equations contain variable.

Reduce[2 x+3 y-z==1 &&4x+6y-2z==C,{x,y,z}]

Notice that Solve may "choke up" here :

Solve[{2 x+3 y-z==1,4x+6y-2z==C},{x,y,z}]