Question

If vectors i+j+2k, i+pj+5k and 5i+3j+4k are linearly dependent, the value of p is what?​

Answer 1

Answer:

$p = 2$ 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 $\vec u = (1,1,2)$, $\vec v = (1,p,5)$ and $\vec w = (5,3,4)$, the linear combination is:

$\alpha_{1}\cdot (1,1,2)+\alpha_{2}\cdot (1,p,5)+\alpha_{3}\cdot (5,3,4) =(0,0,0)$

In other words, the following system of equations must be satisfied:

$\alpha_{1}+\alpha_{2}+5\cdot \alpha_{3}=0$ (Eq. 1)

$\alpha_{1}+p\cdot \alpha_{2}+3\cdot \alpha_{3}=0$ (Eq. 2)

$2\cdot \alpha_{1}+5\cdot \alpha_{2}+4\cdot \alpha_{3}=0$ (Eq. 3)

By Eq. 1:

$\alpha_{1} = -\alpha_{2}-5\cdot \alpha_{3}$

Eq. 1 in Eqs. 2-3:

$-\alpha_{2}-5\cdot \alpha_{3}+p\cdot \alpha_{2}+3\cdot \alpha_{3}=0$

$-2\cdot \alpha_{2}-10\cdot \alpha_{3}+5\cdot \alpha_{2}+4\cdot \alpha_{3}=0$

$(p-1)\cdot \alpha_{2}-2\cdot \alpha_{3}=0$ (Eq. 2b)

$3\cdot \alpha_{2}-6\cdot \alpha_{3} = 0$ (Eq. 3b)

By Eq. 3b:

$\alpha_{3} = \frac{1}{2}\cdot \alpha_{2}$

Eq. 3b in Eq. 2b:

$(p-2)\cdot \alpha_{2} = 0$

If $p = 2$ if given vectors must be linearly independent.