How to build a matrix out of one equation so that it can be solved with Gaussian elimination?

Question

Find such $ a, b, c, d ∈ ℝ $ that

$a(x^3 − x^2 + x − 1) + b(x^3 + x^2 + 3x−2) + c(x^2 + 3x + 1) +d(x^3 + 2x^2 − 2) + 7 = 0$

$∀x ∈ ℝ$ using Gauss' elimination.

How to get from the one equation to such matrix that Gaussian elimination can be used?

Answer 1

Hint:

Calculating the coefficients in each degree, you have to solve the system of linear equations: [\begin{cases}\begin{aligned} a +b+d&=0\\ -a+b+2d&=0\\ a+3b+2c&=0 \\ a+2b -c+2d&=7 \end{aligned} \end{cases} Can you take it from there?

Answer 2

We have the following constraint

• The coefficient of $x^3$ must be zero. ($a+b+d=0$)
• The coefficient of $x^2$ must be zero. ($-a+b+c+2d=0$)
• The coefficient of $x$ must be zero. ($a+3b+3c=0$)
• The constant value must be zero. ($-a-2b+c-2d+7=0$)

In matrix form:$$\begin{bmatrix}1&1&0&1\\-1&1&1&2\\1&3&3&0\\-1&-2&1&-2\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\0\\0\\-7\end{bmatrix}$$Therefore we need to perform operations on the following matrix$$\begin{bmatrix}1&1&0&1&0\\-1&1&1&2&0\\1&3&3&0&0\\-1&-2&1&-2&-7\end{bmatrix}$$as below$$\begin{bmatrix}1&1&0&1&0\\0&2&1&3&0\\0&2&3&-1&0\\0&-1&1&-1&-7\end{bmatrix}$$

$$\begin{bmatrix}2&0&-1&-1&0\\0&2&1&3&0\\0&0&2&-4&0\\0&0&3&1&-14\end{bmatrix}$$

$$\begin{bmatrix}4&0&0&-6&0\\0&4&0&10&0\\0&0&2&-4&0\\0&0&0&14&-28\end{bmatrix}$$or simply

$$\begin{bmatrix}2&0&0&-3&0\\0&2&0&5&0\\0&0&1&-2&0\\0&0&0&1&-2\end{bmatrix}$$

Let's continue what we were doing

$$\begin{bmatrix}2&0&0&0&-6\\0&2&0&0&10\\0&0&1&0&-4\\0&0&0&1&-2\end{bmatrix}$$which finally yields to

$$\begin{bmatrix}1&0&0&0&-3\\0&1&0&0&5\\0&0&1&0&-4\\0&0&0&1&-2\end{bmatrix}$$and at last$$\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}-3\\5\\-4\\-2\end{bmatrix}$$