Question: Do the polynomials $\displaystyle t^{3} + 2t + 1$, $\displaystyle t^{2} - t + 2$, $\displaystyle t^{3} +2$, $\displaystyle -t^{3} + t^{2} - 5t + 2$ span $\displaystyle P_{3}$?

My attempt: Let $\displaystyle at^{3} + bt^{2} + ct + d$ be an arbitrary vector in $\displaystyle P_{3}$, then:

$\displaystyle c_{1}(t^{3} + 2t + 1) + c_{2}(t^{2} - t + 2) + c_{3}(t^{3} +2) + c_{4}(-t^{3} + t^{2} - 5t + 2) = at^{3} + bt^{2} + ct + d$

Combining the terms on the left side and equating corresponding coefficients gives the linear system:

$\displaystyle c_{1} + c_{3} - c_{4} = a$

$\displaystyle c_{2} + c_{4} = b$

$\displaystyle 2c_{1} - c_{2} - 5c_{4} = c$

$\displaystyle c_{1} + 2c_{2} + 2c_{3} + 2c_{4} = d$

Which becomes the augmented matrix:

$\displaystyle \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\2&-1&0&-5&c\\1&2&2&2&d\end{array}\right]$

Row reducing it, I come to the matrix:

$\displaystyle \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&0&0&-\frac{1}{2}(c-2a+b)(d-a-2b)\end{array}\right]$

Since there are a row of zeros, then $\displaystyle -\frac{1}{2}(c-2a+b)(d-a-2b)$ must be equal to zero for the system to be consistent, and hence, the polynomials do not span $\displaystyle P_{3}$. However, my book concluded that the reason that the polynomials do not span $\displaystyle P_{3}$ is because it is an inconsistent system, that no value for a, b, c, and d would yield 0 for the 4th row? How did they come to this conclusion? Thanks in advance.