I have a question in my work book that I think I understand, but I'm somehow struggling to arrive at the answer as it's given in the book. Here it is: For which values of $\lambda$ are the following vectors $(\lambda,-\frac{1}{2},-\frac{1}{2}), (-\frac{1}{2},\lambda,-\frac{1}{2}), (-\frac{1}{2},-\frac{1}{2},\lambda)$ linearly dependent in $\mathbb{R}^3$.
Well I know that I can use the determinant of the matrix whose columns are given by the above vectors to determine for which values of $\lambda$ the determinant is $0$.
Working out the determinant I get $\lambda^3-\frac{1}{4}\lambda$.
Then $\lambda^3-\frac{1}{4}\lambda = 0$, works out to $\lambda = 0, \lambda = -\frac{1}{2}, \lambda = \frac{1}{2}$
The solution in my work book gives a solution of $\lambda = 1, \lambda = -\frac{1}{2}$.
Where am I going wrong?
Best Answer
A more explicit way to do this would be by considering
\begin{align} v_1 = (\lambda,a,a),\\ v_2 = (a,\lambda,a), \\ v_3 = (a,a,\lambda),\end{align} such that $a\in\mathbb{R}$ is fixed. (In this case, $a=-1/2$.)
We want to find $\lambda$ such that $v_1,v_2,v_3$ are LD. If these vectors are LD, then $\exists c,d \in\mathbb{R}: cv_1+dv_2=v_3$. That is:
\begin{align}\begin{cases}c\lambda+da=a \\ ca+d\lambda=a \\ (c+d)a=\lambda\end{cases}\implies \begin{cases}(c+d)(a+\lambda) = 2a \\ (c+d)a=\lambda\end{cases}\implies \begin{cases}(c+d+1)\lambda = 2a \\ (c+d)a=\lambda\end{cases}\end{align}
And we can deduce for $a,\lambda\neq 0$:
\begin{align}&(c+d)(c+d+1)\lambda a = 2\lambda a \\ \implies &(c+d)(c+d+1)= 2 \\ \implies &(c+d) = 1 \text{ or } -2.\end{align}
So, from $(c+d)a=\lambda$, we have $\lambda = -2a,\lambda = a$. In the case $a=-1/2$ we have, indeed, $\lambda=1,\lambda=-1/2$.