Ytukyg

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?

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?

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

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

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

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

Let's continue what we were doing

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

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