Answer to Question #249498 in Linear Algebra for parth

Question #249498

Which one of the following sets of vectors is linearly independent?


{(1,1,0),(1,0,1),(−5,−8,4)}

{(1,1,0),(1,0,1),(4,0,4)}

{(1,1,0),(1,0,1),(2,1,1)}

{(1,1,0),(1,0,1),(−5,−9,4)}


1
Expert's answer
2021-11-01T09:18:12-0400

Solution;

A set of vectors,v1,v2,v3 are linearly independent if the only scalars that satisfy;

k1v1+k2v2+k3v3=0.....(1)

Are;

k1=k2=k3=0.

The equivalent homogeneous solution of (1) is;

"\\begin{bmatrix}\n | & |&| \\\\\n v_1& v_2&v_3\\\\\n|&|&|\n\\end{bmatrix}" "\\begin{bmatrix}\n k_1 \\\\\n k_2\\\\\nk_3\n\\end{bmatrix}=0"

For set 1;

The vector matrix is;

"\\begin{bmatrix}\n 1 & 1&-5\\\\\n 1 & 0&-8\\\\\n0&1&4\n\\end{bmatrix}"

Reduced row echelon matrix form is;

"\\begin{bmatrix}\n 1 & 0&0 \\\\\n 0 & 1&0\\\\\n0&0&1\n\\end{bmatrix}"

From the echelon form,it seen that;

k3=0and it follows that k2=k1=0

Hence,the set of vectors are linearly independent.

For set 2;

The vector matrix is;

"\\begin{bmatrix}\n 1 & 1&4 \\\\\n 1& 0&0\\\\\n0&1&4\n\\end{bmatrix}"

The row reduced echelon form is;

"\\begin{bmatrix}\n 1 & 0&0\\\\\n 0 & 1&4\\\\\n0&0&0\n\n\\end{bmatrix}"

This shows that there exists a nontrivial linear combination of the vectors. Hence they are linearly dependent.

Fro set 3;

The vector matrix is;

"\\begin{bmatrix}\n 1 & 1&2\\\\\n 1 & 0&1\\\\\n0&1&1\n\\end{bmatrix}"

The reduced row echelon form is;

"\\begin{bmatrix}\n 1 & 0&1\\\\\n 0& 1&1\\\\\n0&0&0\n\\end{bmatrix}"

Hence the vectors are linearly dependent.

For set 3;

The vector matrix is;

"\\begin{bmatrix}\n 1 & 1&-5 \\\\\n 1&0& -9\\\\\n0&1&4\n\\end{bmatrix}"

The row reduced echelon form is;

"\\begin{bmatrix}\n 1 & 0&-9 \\\\\n 0& 1&4\\\\\n0&0&0\n\\end{bmatrix}"

Hence the vectors are linearly dependent.

Answer;

{(1,1,0),(1,0,1),(-5,-8,4)} is linearly independent.




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