Question #324395

For what values of α are vectors (1,1,2,1), (2,1,2,3), (1,4,2,1) (-1,3,5,α) are linearly inde-


pendent

1
Expert's answer
2022-04-07T14:39:08-0400

Let v1=(1,1,2,1),  v2=(2,1,2,3),  v3=(1,4,2,1),  v4=(1,3,5,α).v_1=(1, 1, 2, 1),\ \ v_2=(2, 1, 2, 3),\ \ v_3=(1,4,2,1),\ \ v_4=(-1,3,5,\alpha).

These vectors are linearly independent if and only if 112121231421135α0\begin{vmatrix} 1&1&2&1\\2&1&2&3\\1&4&2&1\\-1&3&5&\alpha \end{vmatrix}\neq0 .


112121231421135α=112101210300047α+1=112101310063001α+5=11210131001α+50063=11210131001α+50006α27=1(1)(1)(6α27)=6α+270\begin{vmatrix} 1&1&2&1\\2&1&2&3\\1&4&2&1\\-1&3&5&\alpha \end{vmatrix}=\begin{vmatrix}1&1&2&1\\0&-1&-2&1\\0&3&0&0\\0&4&7&\alpha+1 \end{vmatrix}= \begin{vmatrix} 1&1&2&1\\0&-1&-3&1\\0&0&-6&3\\0&0&-1&\alpha+5 \end{vmatrix}=-\begin{vmatrix} 1&1&2&1\\0&-1&-3&1\\0&0&-1&\alpha+5 \\0&0&-6&3 \end{vmatrix} =-\begin{vmatrix} 1&1&2&1\\0&-1&-3&1\\0&0&-1&\alpha+5 \\0&0&0&-6\alpha-27 \end{vmatrix}=-1\cdot(-1)\cdot(-1)\cdot(-6\alpha-27)=6\alpha+27\neq 0



Answer: for all values α4.5\alpha\neq -4.5



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