The polynomials $9−9x+12x^2$, $10+4x−2x^2$ and $−7−7x+6x^2$ linearly independent in $P_2$ so find $c_1$, $c_2$ and $c_3$ not all zero to prove that these polynomials are linearly independent when?
$c_1(9−9x+12x^2) + c_2(10+4x−2x^2) + c_3(−7−7x+6x^2) = 0$
You want to prove that the polynomials are linearly dependent.
Notice that $$9 -9x + 12x^2 = 3(10 + 4x -2x^2) + 3(-7 -7x + 6x^2)$$
So we can set $c_1 = 1$ and $c_2 = c_3 = -3$ and we get zero.
In general, with a set of linearly dependent vectors it will always be possible to write one of them as a linear combination of the others. My method here was to spot that $9 = 3 \cdot 10 + 3 \cdot -7$ and then check that this holds for the other corresponding terms, which it does.
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To do this using row reduction, observe that we can write our constraints as a set of linear simultaneous equations.
$$9c_1 + 10c_2 -7c_3 = 0$$ $$-9c_1 + 4c_2 + -7c_3 = 0$$ $$12c_1 -2c_2 + 6c_3 = 0$$
Then form an augmented matrix for this system.
$$ \begin{bmatrix} 9&10&-7&|&0\\ -9&4&-7&|&0\\ 12&-2&6&|&0 \end{bmatrix} $$
Row reduce this matrix to obtain
$$ \begin{bmatrix} 1&0&\frac{1}{3}&|&0\\ 0&1&-1&|&0\\ 0&0&0&|&0 \end{bmatrix} $$
which gives solutions $c_3 = t \in \mathbb{R}$ (a free parameter), $c_2 = c_3 = t$, and $c_1 = -\frac{1}{3}c_3 = -\frac{t}{3}$.
The solutions are therefore $(-\frac{t}{3}, t, t), t \in \mathbb{R}$, so in fact my solution is just one of an infinite number. Another solution would be $c_1 = -51698, c_2 = c_3 = 155094$, as a crazy example.
Note that it suffices to find a single nonzero solution to prove linear dependency.