# Consider the matrix A = [[7,6,40,-8,1],[-2,-4,-16,0,-14],[-9,6,-24,-2,29],[2,3,14,7,24]] You may assume that its RREF is B= [[1,0,4,0,-1],[0,1,2,0,4],[0,0,0,1,2],[0,0,0,0,0]] We know that a basis...

Consider the matrix

A = [[7,6,40,-8,1],[-2,-4,-16,0,-14],[-9,6,-24,-2,29],[2,3,14,7,24]]

You may assume that its RREF is

B= [[1,0,4,0,-1],[0,1,2,0,4],[0,0,0,1,2],[0,0,0,0,0]]

We know that a basis for the row space of A is

X = <1,0,4,0,-1> , Y = <0,1,2,0,4> and Z = <0,0,0,1,2>

To build a basis for R^5 starting with X,Y,Z we need to add two vectors. Put them as columns in the following matrix.

rakesh05 | High School Teacher | (Level 1) Assistant Educator

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Because any basis of `R^5`  contain 5 linearly independent vectors in which the first element of every vector starts with 1.

We are givn the basis vectors `X=<1,0,4,0,-1>, Y=<0,1,2,0,4>, Z=<0,0,0,1,2>.`

Now we prepare the matrix corresponding to these vectors by considering each vector as a column vector.

`[[1,0,4,0,-1]^T,[0,1,2,0,4]^T,[0,0,0,1,2]^T]`

So we see that the third vector will have its first element 1 at third place and the fifth vector will have only the last nonzero element as 1.

So our final matrix will be

`[[1,0,0,0,0],[0,1,2,0,4],[0,0,1,0,0],[0,0,0,1,2],[0,0,0,0,1]]` .

pramodpandey | College Teacher | (Level 3) Valedictorian

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We have given matrix `A=[[7,6,40,-8,1],[-2,-4,-16,0,-14],[-9,6,-24,-2,-29],[2,3,14,7,24]]`

rref of A=B=`[[1,0,4,0,-1],[0,1,2,0,4],[0,0,0,1,2],[0,0,0,0,0]]`

we have given basis

x=<1,0,4,0,-1>, y=<0,1,2,0,4>,Z=<0,0,0,1,2>

We wish to prepare basis for `R^5` starting from  x,y,z .So we need matrix havs a five linearly independent columns .of which three columns are x,y,z. In x first component  ,in y second component  and in z fourth component  are 1. We need  to include a row three and five  whose respectively third and fifth component is 1.i.e.

u=<0,0,1,0,0> and v=<0,0,0,0,1>

So final matrix is

`B'=[[1,0,4,0,-1],[0,1,2,0,4],[0,0,1,0,0],[0,0,0,1,2],[0,0,0,0,1]]`

The columns of B' are linearly independent. So form basis for `R^5`