i)
( 1 0 0 1 2 0 1 2 3 ) \begin{pmatrix}
1 & 0 & 0 \\
1 & 2 & 0 \\
1 & 2 & 3 \\
\end{pmatrix} ⎝ ⎛ 1 1 1 0 2 2 0 0 3 ⎠ ⎞ R 2 = R 2 − R 1 R_2=R_2-R_1 R 2 = R 2 − R 1
( 1 0 0 0 2 0 1 2 3 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
1 & 2 & 3 \\
\end{pmatrix} ⎝ ⎛ 1 0 1 0 2 2 0 0 3 ⎠ ⎞ R 3 = R 3 − R 1 R_3=R_3-R_1 R 3 = R 3 − R 1
( 1 0 0 0 2 0 0 2 3 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 2 & 3 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 2 2 0 0 3 ⎠ ⎞ R 2 = R 2 / 2 R_2=R_2/2 R 2 = R 2 /2
( 1 0 0 0 1 0 0 2 3 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 2 & 3 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 2 0 0 3 ⎠ ⎞ R 3 = R 3 − 2 R 2 R_3=R_3-2R_2 R 3 = R 3 − 2 R 2
( 1 0 0 0 1 0 0 0 3 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 3 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 0 3 ⎠ ⎞ R 3 = R 3 / 3 R_3=R_3/3 R 3 = R 3 /3
( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ The rank of the matrix is 3, so the given vectors span a subspace of dimension 3, hence they span R3 .
ii)
a ( 1 1 1 ) + b ( 0 2 2 ) + c ( 0 0 3 ) = ( 0 0 0 ) a\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}+b\begin{pmatrix}
0 \\
2 \\
2
\end{pmatrix}+c\begin{pmatrix}
0 \\
0 \\
3
\end{pmatrix}=\begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix} a ⎝ ⎛ 1 1 1 ⎠ ⎞ + b ⎝ ⎛ 0 2 2 ⎠ ⎞ + c ⎝ ⎛ 0 0 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞
( 1 0 0 0 1 2 0 0 1 2 3 0 ) → ( 1 0 0 0 0 1 0 0 0 0 1 0 ) \begin{pmatrix}
1 & 0 & 0 & & 0 \\
1 & 2 & 0 & & 0 \\
1 & 2 & 3 & & 0 \\
\end{pmatrix}\to\begin{pmatrix}
1 & 0 & 0 & & 0 \\
0 & 1 & 0 & & 0 \\
0 & 0 & 1 & & 0 \\
\end{pmatrix} ⎝ ⎛ 1 1 1 0 2 2 0 0 3 0 0 0 ⎠ ⎞ → ⎝ ⎛ 1 0 0 0 1 0 0 0 1 0 0 0 ⎠ ⎞ a = b = c = 0 a=b=c=0 a = b = c = 0
The given vectors are linearly independent.
iii) A subset S S S of a vector space V V V is called a basis if
1. S S S is a spanning set
2. S S S is linearly independent.
Therefore the set B = { ( 1 , 1 , 1 ) , ( 0 , 2 , 2 ) , ( 0 , 0 , 3 ) } B=\{{(1,1,1),(0,2,2),(0,0,3)\}} B = { ( 1 , 1 , 1 ) , ( 0 , 2 , 2 ) , ( 0 , 0 , 3 )} is a basis for R3 .
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