Linear Algebra - determinant and linear independence I have a question that asks for values of alpha so that the polynomials $\alpha t^2+t+1$, $ t^2+\alpha t+1$ and $ t^2+t+\alpha$ are linearly independent.
For that to be true, the answers for $c_1(\alpha t^2+t+1) + c_2(t^2+\alpha t+1) +c_3(t^2+t+\alpha) $ must be trivial, with $c_1=c_2=c_3 = 0$
$t^2(\alpha c_1+c_2+c_3)+t(c_1+\alpha c_2 +c_3)+1(c_1+c_2+\alpha c_3)=0 $
So the following system has non-trivial solution if the determinant is equal to $0$, which would be a linearly dependent system:
$$
\left\{ 
\begin{array}{c}
\alpha c_1+c_2+c_3=0\\
c_1+\alpha c_2 +c_3=0\\
c_1+c_2+\alpha c_3=0
\end{array}
\right. 
$$
$$
\det \begin{vmatrix}
\alpha & 1 & 1\\
1 & \alpha & 1\\
1 & 1 & \alpha
\end{vmatrix} =0
$$
$Row_1 - \alpha Row_3$ and $ Row_2 - Row_3$ : 
$$
=\det \begin{vmatrix}
0 & 1-\alpha & 1-\alpha^2\\
0 & \alpha-1 & 1-\alpha\\
1 & 1 & \alpha
\end{vmatrix} =0
$$
In the $3\times 3$ matrix above I use the rule of Sarrus to get
$ ( 1-\alpha^2) - (1-\alpha^2)(\alpha-1)=0$
$= ( 1-\alpha^2)(2-\alpha)=0$
$\alpha =1$ or $\alpha=2$
This would mean that for all $\alpha \neq 1$ and $\alpha \neq 2 $, the polynomials are linearly independent. However the ansers are $\alpha \neq 1$ and $\alpha \neq -2 $. Where am I going wrong? Thanks everyone in advance.
 A: It might be easiest to just multiply it out without the row operations and find the roots.
$(a^3 -3a + 2) = 0\\
(a-1)(a^2+a-2)\\
(a-1)(a + 2)(a-1)$
But, there are some ways you might guess your way to an answer.
If $a = 1$ then all of the rows are identical and hence linearly independent.
If $a = -2$ all of the row sum to $0.$  And that means that the vector $(1,1,1)$ is in the kernel, and the matrix is singular.
If you know about eigenvalues....
What are the eignevalues of 
$\begin{bmatrix} 0&1&1\\1&0&1\\1&1&0\end{bmatrix}$
The sum of the eigenvalues equal the trace of the matrix, and the product of the eigenvalues equals the determinant of the matrix.
We know that $-1, 2$ are eigenvalues, the trace is $0$ and the determinant is $2.$
The last eigenvalue must be $-1$
A: Set $f_1(t) = \alpha t^2 + t +1$, $f_2(t) = t^2+\alpha t + 1$, $f_3(t) = t^2 + t+1$.
If $\{f_1, f_2, f_3\}$ is linearly dependent, then $\exists \alpha_1, \alpha_2,\alpha_3$ scalars not all zero such that $\sum_{i=1}^3 \alpha_if_i = 0$. Taking the derivative we also get $\sum_{i=1}^3 \alpha_if_i' = 0$ and $\sum_{i=1}^3 \alpha_if_i'' = 0$ so the matrix
$$\begin{bmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t) \\ f_1''(t) & f_2''(t) & f_3''(t) \\\end{bmatrix}$$
has linearly dependent columns.
In particular the determinant (known as the Wronskian) is equal to $0$:
$$0= \begin{vmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t) \\ f_1''(t) & f_2''(t) & f_3''(t) \\\end{vmatrix} = \begin{vmatrix} \alpha t^2 + t +1 & t^2+\alpha t + 1 & t^2+t+\alpha \\ 2\alpha t + 1 & 2t+\alpha & 2t+1 \\ 2\alpha & 2 & 2 \\\end{vmatrix} = -2(\alpha-1)^2(\alpha+2)$$
Hence $\alpha = 1$ or $\alpha = -2$.
Therefore if $\alpha \notin \{1,-2\}$, the set $\{f_1, f_2, f_3\}$ is surely linearly independent.
If $\alpha = 1$ then $f_1 = f_2 = f_3$, and if $\alpha = -2$ then $\sum_{i=1}^3 f_i = 0$ so in those cases we clearly have linear dependence.
