Find a basis for the row space of A, a basis for the column space of A, a basis

Question

Find a basis for the row space of A, a basis for the column space of A, a basis for the null space of A, the rank of A, and the nullity of A.

Please explain where I went wrong?

(1 point) Let A 0 0 0 2 2 10 Find a basis for the row space of A, a 0 0 0 -10 3 0 0 0 -1 0 3 basis for the column space of A, a basis for the null space of A, the rank of A, and the nullity of 1 0 0 0 0 0 1 0 -3 A. (Note that the reduced row echelon form of A is 0 0 0 0 1 -2 0 0 0 0 0 0 0 0 0 0 0 0 Row Space basis: 1,-3, -1,0,0,-1],[0,0,0,1,0,-3], [0,0,0,0,1,-2] EEE Column Space basis [1,2,0,0,0],[0,0,2,-1,-1],[0,0,2, 0,0] Null Space basis: 13,1,0,0,0],[1,0,1,0,0], [1,0,0,3,2,1] Rank 3 Nullity: 3 To enter a basis into WeBWork, place the entries of each vector inside of brackets, and enter a list of these vectors, separated by commas. For instance, if your basis is R 2 1 then you would enter [1,2,31, 1,1,11 in the answer blank Note: You can earn partial credit on this problem. Preview My Answers Submit Answers Your score was recorded You have attempted this problem 5 times. You received a score of 80% for this attempt. Your overall recorded score is 80%. You have unlimited attempts remaining.

Explanation / Answer

Solution :

Basis for the row space :

First, we must convert the matrix to reduced row echelon form:

Add (-2 * row1) to row2

Swapping row3 with row2

Divide row2 by 2

Add (1 * row2) to row4

Add (1 * row2) to row5

Swapping row4 with row3

Add (-1 * row3) to row5

Add (-1 * row3) to row2

Because we have only performed linear operations on rows, the non-zero rows in the reduced row echelon form of the matrix comprise a Basis for the Row Space of the matrix.

The rows highlighted below in BOLD comprise a Basis for the Row Space of our matrix :

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Basis for the column space :

Add (-2 * row1) to row2

Swapping row3 with row2

Add (1/2 * row2) to row4

Add (1/2 * row2) to row5

Swapping row4 with row3

Add (-1 * row3) to row5

First, we must reduce the matrix so we can calculate the pivots of the matrix (note that we are reducing to row echelon form, not reduced row echelon form) :

The matrix has 3 pivots (hilighted above in BOLD).

Because we have found pivots in columns 0, 3 and 4. We know that these columns in the original matrix define the Column Space of the matrix.

Therefore, the Column Space is given by the following equation:

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Basis for the null space :

First, let's put our matrix in Reduced Row Eschelon Form...

Add (-2 * row1) to row2

Swapping row3 with row2

Divide row2 by 2

Add (1 * row2) to row4

Add (1 * row2) to row5

Swapping row4 with row3

Add (-1 * row3) to row5

Add (-1 * row3) to row2

The matrix has 3 pivot columns (hilighted in BOLD) and 3 free columns; because the matrix has 3 pivots, the rank of the matrix is 3.

Let's take the 'free' part of the reduced row echelon form matrix (hilighted below in BOLD)

and turn it into its own matrix:

Let's multiply this matrix by -1:

Now, we add the Identity Matrix to the rows in our new matrix which correspond to the 'free' columns in the original matrix, making sure our number of rows equals the number of columns in the original matrix (otherwise, we couldn't multiply the original matrix against our new matrix):

Finally, the Null Space of our matrix is defined by scalar multiples of these column vectors:

-------------------------------------------------------------------------------------------------------------------------------------------------------------

The rank of the matrix is 3. It equals the number of leading entries.

The nullity of the matrix is 3. This is the dimension of the null space. It equals the number of columns without leading entries.

1 -3 -1 0 0 -1 0 0 0 0 0 0 0 0 0 2 2 -10 0 0 0 -1 0 3 0 0 0 -1 0 3

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