Question

5. Determine, with proof, whether each of the following subsets S of a vector space V is linearly dependent or independent: a) V = R. S = {(2, 8.-1.4), (3.2. 4.0), (-1,-5, 2, 3), (0.0.7, 2)} 1112×2

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Answer #1

a) Given S=\{(2,8,-1,4),(3,2,4,0),(-1,-5,2,3),(0,0,7,2)\}

Now clearly vectors (3,2,4,0) & (0,0,7,2) are Linearily Independent as one vector has Non zero component corresponds to same place Another vector has zero component

Now suppose

a(3,2,4,0)+b(0,0,7,2)=(2,8,-1,4)\\ ==>3a=2,2b=4,2a=8,4a+7b=-1

These four equation give different value of constant a&b which is not possible

==>\{(3,2,4,0), (0,0,7,2), (2,8,-1,4)\} is Linearily Independent

Next suppose

a(3,2,4,0)+ b(0,0,7,2)+ c(2,8,-1,4)=(-1,-5,2,3) \\==> 3a+2c=-1---(1) \\2a+8c=-5--(2)\\ 4a+7b-c=2--(3)\\ 2b+4c=3---(4)

Solving equation (1)&(2) we obtain

a=1/10,c=-13/20 Substitute in equation (4) we get b=14/5

But values obtained in equation (1),(2)&(4) doesn't satisfies equation (3)

\{(3,2,4,0), (0,0,7,2), (2,8,-1,4), (-1,-5,2,3) \} is Linearily Independent

b) Let \ v_1=\begin{bmatrix} 1 &0 \\ 0 & -1 \end{bmatrix},v_2=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}\&v_3=\begin {bmatrix} 1&1\\-1&1\end {bmatrix}

Clealy v_1\&v_2 \ are \ Linearly \ Independent {As v_1\ has \ Non-zero \ value \ whereas \ v_2 \ has \ zero \ at \ that \ places }

Now

If av_1+bv_2=v_3 \\ a\begin {bmatrix} 1&0\\0&-1\end {bmatrix} +b\begin{bmatrix} 0&1\\1&0\end {bmatrix} =\begin {bmatrix} 1&1\\-1&1\end {bmatrix} \\==>\begin{bmatrix} a&b\\b&-a\end{bmatrix} =\begin {bmatrix} 1&1\\-1&1\end {bmatrix} \\ this \ give \ following \ equations \\ a=1,a=-1,b=1,b=-1

Which make no sense

===>S=\{v_1,v_2,v_3\} \ is \ Linearly \ Independent

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