Suppose A 2 4 6 1 1 1 2 3 1 5 3 6 9 2 12 a Find the rank an

Suppose A = [-2 4 -6 1 -1 1 -2 3 1 5 3 -6 9 2 12]. (a) Find the rank and nullity of A. (b) Find a basis for the column space of A that consists of columns of A. (c) Find a basis for the null space of A. (d) Find a basis for the row space of A. (e) Find a basis for the row space of A that consists of rows of A. Does the set {1 + x. 3 = x^2, 1 + 4x + x^2) span P_2? If not, then write one vector as a linear combination of the other two vectors, then add one of the standard basis vectors of P_2 to the set to form a bask for P_2. Consider the transformation T: R^3 rightarrow R^4. T(x, y, z) = (x + 2y + 3x, y + z, x + 3y + 4z, x + z). (a) Find the image of v = (2, 3, -5). (b) Find the preimage of w = (0, -1.-1, 2). (c) Find a basis for the range of T. (d) Find a basis for the kernel of T.

Solution

11. In order to determine the answers to all the given questions, we will reduce to its RREF,as under, the given matrix A =

-2

4

-6

1

-1

1

-2

3

1

5

3

-6

9

2

12

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 -1/4

Add 12 times the 2nd row to the 3rd row;Multiply the 3rd row by -1

Add 1/8 times the 3rd row to the 2nd row;Add -1/2 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

0

0

0

0

1

-3/2

0

-1

0

0

0

1

3

(a). The rank of A, being the number of non-zero rows in its RREF is 3

(b). The 1^st,2^nd and 4 The 1^st,2^nd and 4^th columns of A are linearly independent. The 3^rd column is (-3/2)1^st column and 5^th column is (-1)1^st column + 3*4^th column. Hence, a basis for Col(A), which contains columns of A, is { (2,-1,3)^T, (4,-2,-6)^T,(1,1,2)^T}.

(c ). Null(A) is the setof solutions to the equation AX = 0. In view of the RREF of A, if X = (x₁,x₂,x₃,x₄,x₅)^T, then this equation is equivalent to the linear system x₁= 0, x₂ -3x₃/2 –x₅ = 0 or, x₂ = 3x₃/2+ x₅ and x₄+3x₅ = 0 or, x₄ = -3x₅.Hence, X = (0, 3x₃/2+ x₅, x₃, -3x₅,x₅ )^T= x₃(0, 3/2,1,0,0)^T+x₅(0,1,0,-3,1)^T. Hence a basis for Null(A) is {(0, 3/2,1,0,0)^T ,(0,1,0,-3,1)^T}.

(d). A basis for the row space of A is { (1,0,0,0,0),(0,1,-3/2,0,-1),(0,0,0,1,3)}.

(e ). A basis for the row space of A, that contains the rows of A is { (-2,4,-6,1,-1),(1,-2,3,1,5),                           (3,-6,9,2,12)}.

12. Let A be the matrix whose columns are coefficients of the vectors in the given set S (say). Then A =

1

3

1

1

0

4

0

-1

1

We will reduce A to its RREF as under:

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

Multiply the 2nd row by -1/3

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

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

Then the RREF of A is

1

0

4

0

1

-1

0

0

0

Now, it is apparent that 1+4x+x² = 4(1+x)-(3-x²). Hence the given vectors in S are not linearly independent and also do not span P₂.

13. We have T(x,y,z) = (x+2y+3z, y+z,x+3y+4z,x+z).

(a). The image of (2,3,-5) under T is ( 2+6-15, 3-5,1+9-20,2-5), i.e. ( -7,-2,-10,-3).

(b).Let the pre-image of w = (0,-1,-1,2) be X =(x,y,z). Then, we have x+2y+3z = 0, y+z = -1,x+3y+4z =-1, and x+z=2. Let A be the augmented matrix of this linear system. Then A=

1

2

3

0

0

1

1

-1

1

3

4

-1

The RREF of A is

1

0

1

2

0

1

1

-1

0

0

0

0

Then the above linear system is equivalent to x+z = 2, or, x = 2-z, andy +z = -1 or y = -1-z. Then X =                  ( 2-z,-1-z,z) where z is arbitrary. Thus, w has several pre-images under T, which are ( 2-z,-1-z,z) where z is arbitrary.

(c ). The standard matrix A of T has T(e₁), T(e₂ ),andT(e₃ ) as columns, where {e₁,e₂,e₃} is the standard basis of R³. Here, T(e₁) = (1,0,1,1), T(e₂)= (2,1,3,0) and T(e₃)= ( 3,1,4,1). Then A =

1

2

3

0

1

1

1

3

4

1

0

1

The RREF of A is

1

0

1

0

1

1

0

0

0

0

0

0

Hence the Range of T , which is same as Col(A) is { (1,0,0,0)^T,(0,1,0,0)^T}.

(d). Ker(T) is same as Null(A). If X = (x,y,z)^T, then Null (A) is the set of solutions to the equation AX = 0 or, (x+z) = 0, i.e. x = -z, and y +z = 0, i.e. y = -z. Then X = (-z,-z,z)^T = z(-1,-1,1)^T. Hence, a basis for Ker (T) is {{ (-1,-1,1)^T}.

-2	4	-6	1	-1
1	-2	3	1	5
3	-6	9	2	12