Find a basis for the row space and a basis for the column sp

Find a basis for the row space and a basis for the column space of the matrix

What can you say about the dimension of the solution space of the homogenous linear system with this matrix of coefficients?

37611 108 36 11 8 156 1( 2575 1231

Solution

Let the given matrix be denoted by A. To answer the given questions, we will reduce A to its RREF as under:

Add -2 times the 1st row to the 2nd row

Add -3 times the 1st row to the 3rd row

Add -1 times the 1st row to the 4th row

Add -1 times the 2nd row to the 3rd row

Add -3 times the 2nd row to the 4th row

Multiply the 3rd row by 1/2

Add -5 times the 3rd row to the 4th row

Add -3 times the 3rd row to the 1st row

Add -2 times the 2nd row to the 1st row

Then the RREF of A is

1

0

-5

0

0

1

3

0

0

0

0

1

0

0

0

0

Now, it is apparent that c₃ = -5c₁+3c₂, where c_is (1 i 4)denote the columns of A. Also, only c₁,c₂,c₄ are linearly independent. Hence, a basis for Col(A) is {(1,2,3,1)^T,(2,5,7,5)^T, (3,6,11,8)^T} or, {(1,0,0,0)^T,(0,1,0,0)^T,(0,0,0,1)^T} .

In the row operations stated above, there have not been any row interchanges so that a basis for Row(A) is { (1,2,1,3),(2,5,5,6),(3,7,6,11)} or,{ (1,0,-5,0),(0,1,3,0),(0,0,01)}.

If X = (x,y,z,w)^T, then the given homogeneous linear system AX = 0 is equivalent to x -5z = 0 or, x = 5z, y +3z = 0 ory = -3z and w = 0 so that X = ( 5z,-3z,z,0)^T= z(5,-3,1,0)^T. Here, z is an arbitrary real number so that the given homogeneous linear system has infinite solutions. A basis for the solution space is {(5,-3,1,0)^T}.

1	0	-5	0
0	1	3	0
0	0	0	1
0	0	0	0