[6, 12, 16]
The dimension of the above matrix is 1 X 3
True
False

Step-by-step explanation:

1. Find a basis for the row space, column space, and null space of the matrix given

below:

A =









3 4 0 7

1 −5 2 −2

−1 4 0 3

1 −1 2 2









Solution. reef(A) =





1 0 0 1

0 1 0 1

0 0 1 1



. Thus a basis for the row space of A is

{[1, 0, 0, 1], [0, 1, 0, 1], [0, 0, 1, 1]}. Since the first, second, and third columns of

rref(A) contain a pivot, a basis for the column space of A is {









3

1

−1

1









,









4

−5

4

−1









,









0

2

0

2









}.

If we solve Ax = 0, we find that x4 is a free variable, so we set x4 = r. We

obtain









x1

x2

x3

x4









= r









−1

−1

−1

1









, so {









−1

−1

−1

1









} is a basis for the nullspace of A.

2. What is the maximum number of linearly independent vectors that can be found

in the nullspace of

A =









1 2 0 3 1

2 4 −1 5 4

3 6 −1 8 5

4 8 −1 12 8









Solution. rref(A) has three columns with pivots and two columns without

pivots. Thus the dimension of the nullspace of A is 2, so at most 2 linearly

independent vectors can be found in the nullspace of A.

3. Let T : R3 → R3 be the linear transformation defined by

T([x1, x2, x3]) = [2x1 + 3x2, x3, 4x1 − 2x2].

Find the standard matrix representation of T. Is T invertible? If so, find a

formula for T

−1

.

Mathematics Department 1

tthe dimension of the above matrix is 1 x 3 is true

4400

step-by-step explanation:

a

step-by-step explanation:

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