Answer to Question #153455 in Linear Algebra for usman

Question

using the following table, find f(x) as a polynomial in x:

x: – 1 0 3 6 7

f(x): 3 – 6 39 822 1611.

Answer 1

Let $f(x)=ax^4+bx^3+cx^2+dx+e$, then we have a system of equations

$\begin{cases} f(-1)=a-b+c-d+e=3\\ f(3)=81a+27b+9c+3d+e=39\\ f(6)=1296a+216b+36c+6d+e=822\\ f(7)=2401a+343b+49c+7d+e=1611\\ f(0)=e=-6 \end{cases}$

Extended matrix of the system is

$\begin{pmatrix} 1&-1&1&-1&1&3\\ 81&27&9&3&1&39\\ 1296&216&36&6&1&822\\ 2401&343&49&7&1&1611\\ 0&0&0&0&1&-6 \end{pmatrix}$

Solve the system by Jordan method

$\begin{pmatrix} 1&-1&1&-1&1&3\\ 81&27&9&3&1&39\\ 1296&216&36&6&1&822\\ 2401&343&49&7&1&1611\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&-1&1&-1&0&9\\ 81&27&9&3&0&45\\ 1296&216&36&6&0&828\\ 2401&343&49&7&0&1617\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&-1&1&-1&0&9\\ 0&108&-72&84&0&-684\\ 0&1512&-1260&1302&0&-10836\\ 0&2744&-2352&2408&0&-19992\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&-1&1&-1&0&9\\ 0&9&-6&7&0&-57\\ 0&36&-30&31&0&-258\\ 0&49&-42&43&0&-357\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&0&\frac{1}{3}&-\frac{2}{9}&0&\frac{8}{3}\\ 0&9&-6&7&0&-57\\ 0&0&-6&3&0&-30\\ 0&0&-\frac{28}{3}&\frac{44}{9}&0&-\frac{140}{3}\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&0&\frac{1}{3}&-\frac{2}{9}&0&\frac{8}{3}\\ 0&9&-6&7&0&-57\\ 0&0&-2&1&0&-10\\ 0&0&-84&44&0&-420\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&0&0&-\frac{1}{18}&0&1\\ 0&9&0&4&0&-27\\ 0&0&-2&1&0&-10\\ 0&0&0&2&0&0\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&0&0&0&0&1\\ 0&9&0&0&0&-27\\ 0&0&-2&0&0&-10\\ 0&0&0&1&0&0\\ 0&0&0&0&1&-6 \end{pmatrix}\rightarrow$

$\rightarrow\begin{pmatrix} 1&0&0&0&0&1\\ 0&1&0&0&0&-3\\ 0&0&1&0&0&5\\ 0&0&0&1&0&0\\ 0&0&0&0&1&-6 \end{pmatrix}$

We obtain $a=1, b=-3, c=5, d=0, e=-6$, that is $f(x)=x^4-3x^3+5x^2-6$

Answer: $f(x)=x^4-3x^3+5x^2-6$