Let 4 be the matrix given by A 2 1 0 1 3 1 3 4 1 Find Carte

Let .4 be the matrix given by A = (2 1 0 1 3 1 3 4 1). Find Cartesian equations for Col(A) and Null(A). Are the columns of A linearly independent? Give full reasons for your answer (based on the definition of linear independence).

Solution

(a) We will reduce A to its RREF as under:

multiply the 1st row by ½

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

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

Multiply the 2nd row by 2/5

Add -5/2 times the 2nd row to the 3rd row

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

              Then the RREf of A is

1

0

-1/5

0

1

2/5

0

0

0

This implies that the 3^rd column of A is a linear combination of the first two columns of A. Also, Col(A) = span{ (2,1,3)^T, (1,3,4)^T }

Further, Null(A) is the set of solutions of the equation AX = 0. If X = (x,y,z)^T , then it is equivalent to x-z/5 = 0 and y+2z/5 = 0. Now, if z = 5t, then x = t and y= -2t so that X = t(1,-2,5)^T Null(A) = span {( 1,-2,5)^T}.

Let the cartesian equation for Col(A) be ax+by+cz = 0. Then we have 2a+b+3c = 0 and a+3b+4c = 0 so that a = b = -c. Then the cartesian equation for Col(A) is x+y-z = 0.

As regards, Null(A), it may be observed that it being the span of a single vector, is a line. Its Cartesian equation is (x,y,z) = c(1,-2,5) where c is an arbitrary real number.

(b) As may be observed from the RREF of A, the columns of A are linearly dependent as the 3^rd column of A is a linear combination of the first two columns of A.

1	0	-1/5
0	1	2/5
0	0	0