Answer:
[tex]p = 2[/tex] if given vectors must be linearly independent.
Step-by-step explanation:
A linear combination is linearly dependent if and only if there is at least one coefficient equal to zero. If [tex]\vec u = (1,1,2)[/tex], [tex]\vec v = (1,p,5)[/tex] and [tex]\vec w = (5,3,4)[/tex], the linear combination is:
[tex]\alpha_{1}\cdot (1,1,2)+\alpha_{2}\cdot (1,p,5)+\alpha_{3}\cdot (5,3,4) =(0,0,0)[/tex]
In other words, the following system of equations must be satisfied:
[tex]\alpha_{1}+\alpha_{2}+5\cdot \alpha_{3}=0[/tex] (Eq. 1)
[tex]\alpha_{1}+p\cdot \alpha_{2}+3\cdot \alpha_{3}=0[/tex] (Eq. 2)
[tex]2\cdot \alpha_{1}+5\cdot \alpha_{2}+4\cdot \alpha_{3}=0[/tex] (Eq. 3)
By Eq. 1:
[tex]\alpha_{1} = -\alpha_{2}-5\cdot \alpha_{3}[/tex]
Eq. 1 in Eqs. 2-3:
[tex]-\alpha_{2}-5\cdot \alpha_{3}+p\cdot \alpha_{2}+3\cdot \alpha_{3}=0[/tex]
[tex]-2\cdot \alpha_{2}-10\cdot \alpha_{3}+5\cdot \alpha_{2}+4\cdot \alpha_{3}=0[/tex]
[tex](p-1)\cdot \alpha_{2}-2\cdot \alpha_{3}=0[/tex] (Eq. 2b)
[tex]3\cdot \alpha_{2}-6\cdot \alpha_{3} = 0[/tex] (Eq. 3b)
By Eq. 3b:
[tex]\alpha_{3} = \frac{1}{2}\cdot \alpha_{2}[/tex]
Eq. 3b in Eq. 2b:
[tex](p-2)\cdot \alpha_{2} = 0[/tex]
If [tex]p = 2[/tex] if given vectors must be linearly independent.
