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\n

Nullity of $A=\\;$[[1]]

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Row 1

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Row 2

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Row 3

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Row 4

", "

Row 5

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Row 6

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Determine a basis of $\\operatorname{row}(A)$ consisting of rows of $R$ by clicking on the appropriate choices of rows of $R$.

\n

[[0]]

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Column 1

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Column 2

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Column 3

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Column 4

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Column 5

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\n

Click on the choices of columns of $A$ which form a basis for the column space.

\n\n

Note that there are more than one possible choice of columns and if the columns you choose form a basis then this will be marked as correct. If you Reveal the answer you will see one possible such choice.

\n

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Express the following columns in terms of the columns $c_\\var{echform[0]}$, $c_\\var{echform[1]}$, $c_\\var{echform[2]}$

\n

$c_{\\var{othercols[0]}}$=[[0]]$c_\\var{echform[0]}$+[[1]]$c_{\\var{echform[1]}}$+[[2]]$c_{\\var{echform[2]}}$

\n

$c_{\\var{othercols[1]}}$=[[3]]$c_\\var{echform[0]}$+[[4]]$c_{\\var{echform[1]}}$+[[5]]$c_{\\var{echform[2]}}$

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Determine a basis for $\\operatorname{Null}(A)$.

\n

Enter the basis elements as the columns of a matrix:

\n

Note that any such basis will be marked as correct. If you Reveal the answer you will see one possible such basis.

\n

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The row-reduced echelon form $R$ of the following matrix $A$

\n

$A=\\var{testmatrix}$ is given by $R=\\var{echmatrix}$.

\n

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Given a matrix in row reduced form use this to find bases for the null, column and row spaces of the matrix.

"}, "advice": "

a) The following shows how $A$ is reduced to row-echelon form.

\n

\n

{solution(record_ops_matrix,record_ops_message)}

\n

The rank of A is the number of columns of R with pivots, i.e. 3.

\n

The nullity of A is the number of columns of R without pivots, i.e. 2.

\n

b) Given the row-echelon form we see that the first three rows of $R$ form a basis for the row space of $A$.

\n

c) One basis for the column-space of $A$ is given by the original columns which in the row-reduced echelon form have a pivot.

\n

For example, $\\{c_i:i \\in \\var{set(echform)}\\}$

\n

$\\var{column_basis}$

\n

There are other bases which you could have chosen and would also have been marked as correct.

\n

d) If you look at the columns $c_{\\var{othercols[0]}}$ and $c_{\\var{othercols[1]}}$ we can simply read off these columns as combinations of the other columns by looking at the numbers above the pivot element in each of the columns $c_{\\var{othercols[0]}}$ and $c_{\\var{othercols[1]}}$.

\n

e) Basis for the null-space.

\n

The dimension of the null-space is $2$ and so we should have $2$ vectors in the basis.

\n

The null space is given by $\\operatorname{Null}(A)=\\operatorname{Null}(R)= $  \\[\\left\\{(x_1,x_2,x_3,x_4,x_5):\\simplify{{ec}*vector(x_1,x_2,x_3,x_4,x_5)=vector(0,0,0,0,0,0)}\\right\\}\\]

\n

This gives three equations:

\n

\\begin{align}\\simplify{{ec[0][0]}x_1+{ec[0][1]}x_2+{ec[0][2]}x_3+{ec[0][3]}x_4+{ec[0][4]}x_5}&=0\\\\\\simplify{{ec[1][0]}x_1+{ec[1][1]}x_2+{ec[1][2]}x_3+{ec[1][3]}x_4+{ec[1][4]}x_5}&=0\\\\\\simplify{{ec[2][0]}x_1+{ec[2][1]}x_2+{ec[2][2]}x_3+{ec[2][3]}x_4+{ec[2][4]}x_5}&=0 \\end{align}

\n

The free columns (without a pivot element) are $c_{\\var{nbasis_column[0]}},c_{\\var{nbasis_column[1]}}$

\n

We obtain $2$ vectors which are linearly independent by:

\n

1) Setting: $x_{\\var{nbasis_column[0]}}=1,x_{\\var{nbasis_column[1]}}=0$ gives $\\var{rowvector(null_basis[0])}$ as a basis element for the null-space.

\n

2) Setting: $x_{\\var{nbasis_column[0]}}=0,x_{\\var{nbasis_column[1]}}=1$ gives $\\var{rowvector(null_basis[1])}$ as another basis element for the null-space.

\n

Hence $\\{\\var{rowvector(null_basis[0])},\\var{rowvector(null_basis[1])}\\}$

\n

is a basis for the null-space.

\n

Note that this is just one possible basis, and if you input another correct basis for the null-space then it will be marked as correct.

\n

Alternatively.

\n

Note that on using the equations above, the variables $x_{\\var{echform[0]}},\\;x_{\\var{echform[1]}}$ and  $x_{\\var{echform[2]}}$ can be expressed in terms of the variables  $x_{\\var{nbasis_column[0]}},x_{\\var{nbasis_column[1]}}$.

\n

If we do so we find that:

\n

$ (x_1,x_2,x_3,x_4,x_5) \\in \\operatorname{Null}(A) \\iff$ \\[(x_1,x_2,x_3,x_4,x_5)= x_{\\var{nbasis_column[0]}}\\var{rowvector(null_basis[0])}+x_{\\var{nbasis_column[1]}}\\var{rowvector(null_basis[1])}.\\]

\n

As $ x_{\\var{nbasis_column[0]}},\\;x_{\\var{nbasis_column[1]}}$ can take any values we see that the vectors

\n

$\\{\\var{rowvector(null_basis[0])},\\var{rowvector(null_basis[1])}\\}$ form a basis as they span  $\\operatorname{Null}(A)$ and there are $2$ vectors which agrees with the nullity.

\n

\n

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'+mess+'
'+m+'