Answer to Question #206222 in Linear Algebra for Aiman Arif

Question #206222

determine whether or not the set of vectors {(1,2,-1),(0,3,1),(1,-5,3)} is a basis of R^3 also find dimension


1
Expert's answer
2021-06-14T15:36:29-0400

We can set up a matrix and use Gaussian elimination .


(121031153)\begin{pmatrix} 1 & 2 & -1 \\ 0 & 3 & 1 \\ 1 & -5 & 3 \\ \end{pmatrix}

R3=R3R1R_3=R_3-R_1

(121031074)\begin{pmatrix} 1 & 2 & -1 \\ 0 & 3 & 1 \\ 0 & -7 & 4 \\ \end{pmatrix}

R2=R2/3R_2=R_2/3

(121011/3074)\begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 1/3 \\ 0 & -7 & 4 \\ \end{pmatrix}

R1=R12R2R_1=R_1-2R_2

(105/3011/3074)\begin{pmatrix} 1 & 0 & -5/3 \\ 0 & 1 & 1/3 \\ 0 & -7 & 4 \\ \end{pmatrix}

R3=R3+7R2R_3=R_3+7R_2


(105/3011/30019/3)\begin{pmatrix} 1 & 0 & -5/3 \\ 0 & 1 & 1/3 \\ 0 & 0 & 19/3\\ \end{pmatrix}

R3=(3/19)R3R_3=(3/19)R_3


(105/3011/3001)\begin{pmatrix} 1 & 0 & -5/3 \\ 0 & 1 & 1/3 \\ 0 & 0 & 1\\ \end{pmatrix}

R1=R1+(5/3)R3R_1=R_1+(5/3)R_3


(100011/3001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1/3 \\ 0 & 0 & 1\\ \end{pmatrix}

R2=R2R3/3R_2=R_2-R_3/3


(100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}

Since the rank of the matrix is 3, then the set {(1,2,-1),(0,3,1),(1,-5,3)} is a basis of R3.R^3.

The dimension is 3.



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