# How do you solve the system -5 = -64a + 16b - 4c + d, -4 = -27a + 9b - 3c + d, -3 = -8a + 4b - 2c + d, 4 = -a + b - c + d?

Mar 5, 2017

$\left\{\begin{matrix}a = 1 \\ b = 9 \\ c = 27 \\ d = 23\end{matrix}\right.$

#### Explanation:

Given:

$\left\{\begin{matrix}- 5 = - 64 a + 16 b - 4 c + d \\ - 4 = - 27 a + 9 b - 3 c + d \\ - 3 = - 8 a + 4 b - 2 c + d \\ 4 = - a + b - c + d\end{matrix}\right.$

Consider the function:

$f \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d$

Note that the given system of equations is equivalent to:

$\left\{\begin{matrix}f \left(- 4\right) = - 5 \\ f \left(- 3\right) = - 4 \\ f \left(- 2\right) = - 3 \\ f \left(- 1\right) = 4\end{matrix}\right.$

Since the sampling points are at equal intervals of size $1$, we can conveniently find a formula for a cubic function satisfying this system by examining differences...

$\textcolor{b l u e}{- 5} , - 4 , - 3 , 4$

Write down the sequence of differences between successive terms:

$\textcolor{b l u e}{1} , 1 , 7$

Write down the sequence of differences between successive terms:

$\textcolor{b l u e}{0} , 6$

Write down the sequence of differences between successive terms:

$\textcolor{b l u e}{6}$

Then we can write down a formula for $f \left(x\right)$ using the initial term of each of these sequences (see footnote):

f(x) = color(blue)(-5)/(0!)+color(blue)(1)/(1!)(x+4)+color(blue)(0)/(2!)(x+4)(x+3)+color(blue)(6)/(3!)(x+4)(x+3)(x+2)

$\textcolor{w h i t e}{f \left(x\right)} = - 5 + x + 4 + {x}^{3} + 9 {x}^{2} + 26 x + 24$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} + 9 {x}^{2} + 27 x + 23$

So:

$\left\{\begin{matrix}a = 1 \\ b = 9 \\ c = 27 \\ d = 23\end{matrix}\right.$

$\textcolor{w h i t e}{}$
Footnote

Note that:

color(purple)(1/(0!))_(color(white)(1/1)) is a constant function taking the value $1$ when $x = - 4$

color(purple)(1/(1!)(x+4))_(color(white)(1/1)) is a linear function taking the value $0$ when $x = - 4$ and $1$ when $x = - 3$

color(purple)(1/(2!)(x+4)(x+3))_(color(white)(1/1)) is a quadratic function taking the value $0$ when $x = - 4$ or $x = - 3$ and the value $1$ when $x = - 2$

color(purple)(1/(3!)(x+4)(x+3)(x+2))_(color(white)(1/1)) is a cubic function taking the value $0$ when $x = - 4$, $x = - 3$ or $x = - 2$ and the value $1$ when $x = - 1$

So as we add suitable multiples of these functions in turn, we get a sequence of polynomials of increasing degree that match each of the sample points in turn. The suitable multiples are the differences we found.

Mar 5, 2017

$\left\{\begin{matrix}a = 1 \\ b = 9 \\ c = 27 \\ d = 23\end{matrix}\right.$

#### Explanation:

Given:

$\left\{\begin{matrix}- 5 = - 64 a + 16 b - 4 c + d \\ - 4 = - 27 a + 9 b - 3 c + d \\ - 3 = - 8 a + 4 b - 2 c + d \\ 4 = - a + b - c + d\end{matrix}\right.$

Write in matrix form:

$\left(\begin{matrix}- 64 & 16 & - 4 & 1 & - 5 \\ - 27 & 9 & - 3 & 1 & - 4 \\ - 8 & 4 & - 2 & 1 & - 3 \\ - 1 & 1 & - 1 & 1 & 4\end{matrix}\right)$

Perform some row operations to make the left hand side into a $4 \times 4$ identity matrix. These may not be optimal, but this is a sequence I found...

Subtract $3 \times \text{row} 2$ from $\text{row} 1$ to get:

$\left(\begin{matrix}17 & - 11 & 5 & - 2 & 7 \\ - 27 & 9 & - 3 & 1 & - 4 \\ - 8 & 4 & - 2 & 1 & - 3 \\ - 1 & 1 & - 1 & 1 & 4\end{matrix}\right)$

Add $2 \times \text{row} 1$ to $\text{row} 2$ to get:

$\left(\begin{matrix}17 & - 11 & 5 & - 2 & 7 \\ 7 & - 13 & 7 & - 3 & 10 \\ - 8 & 4 & - 2 & 1 & - 3 \\ - 1 & 1 & - 1 & 1 & 4\end{matrix}\right)$

Add $2 \times \text{row} 3$ to $\text{row} 1$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 7 & - 13 & 7 & - 3 & 10 \\ - 8 & 4 & - 2 & 1 & - 3 \\ - 1 & 1 & - 1 & 1 & 4\end{matrix}\right)$

Add $\text{row"1 + "row} 3$ to $\text{row} 2$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & - 12 & 6 & - 2 & 8 \\ - 8 & 4 & - 2 & 1 & - 3 \\ - 1 & 1 & - 1 & 1 & 4\end{matrix}\right)$

Add $8 \times \text{row} 1$ to $\text{row} 3$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & - 12 & 6 & - 2 & 8 \\ 0 & - 20 & 6 & 1 & 5 \\ - 1 & 1 & - 1 & 1 & 4\end{matrix}\right)$

Add $\text{row} 1$ to $\text{row} 4$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & - 12 & 6 & - 2 & 8 \\ 0 & - 20 & 6 & 1 & 5 \\ 0 & - 2 & 0 & 1 & 5\end{matrix}\right)$

Subtract $\text{row} 2$ from $\text{row} 3$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & - 12 & 6 & - 2 & 8 \\ 0 & - 8 & 0 & 3 & - 3 \\ 0 & - 2 & 0 & 1 & 5\end{matrix}\right)$

Subtract $\text{row"3+"row} 4$ from $\text{row} 2$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & - 2 & 6 & - 6 & 6 \\ 0 & - 8 & 0 & 3 & - 3 \\ 0 & - 2 & 0 & 1 & 5\end{matrix}\right)$

Divide $\text{row} 2$ by $- 2$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & - 3 & 3 & - 3 \\ 0 & - 8 & 0 & 3 & - 3 \\ 0 & - 2 & 0 & 1 & 5\end{matrix}\right)$

Subtract $4 \times \text{row} 4$ from $\text{row} 3$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & - 3 & 3 & - 3 \\ 0 & 0 & 0 & - 1 & - 23 \\ 0 & - 2 & 0 & 1 & 5\end{matrix}\right)$

Add $\text{row} 3$ to $\text{row} 4$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & - 3 & 3 & - 3 \\ 0 & 0 & 0 & - 1 & - 23 \\ 0 & - 2 & 0 & 0 & - 18\end{matrix}\right)$

Multiply $\text{row} 3$ by $- 1$ and divide $\text{row} 4$ by $- 2$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & - 3 & 3 & - 3 \\ 0 & 0 & 0 & 1 & 23 \\ 0 & 1 & 0 & 0 & 9\end{matrix}\right)$

Permute rows $2$, $3$ and $4$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 9 \\ 0 & 1 & - 3 & 3 & - 3 \\ 0 & 0 & 0 & 1 & 23\end{matrix}\right)$

Subtract $\text{row} 2$ from $\text{row} 3$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 9 \\ 0 & 0 & - 3 & 3 & - 12 \\ 0 & 0 & 0 & 1 & 23\end{matrix}\right)$

Divide $\text{row} 3$ by $- 3$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 9 \\ 0 & 0 & 1 & - 1 & 4 \\ 0 & 0 & 0 & 1 & 23\end{matrix}\right)$

Add $\text{row} 4$ to $\text{row} 3$ to get:

$\left(\begin{matrix}1 & - 3 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 9 \\ 0 & 0 & 1 & 0 & 27 \\ 0 & 0 & 0 & 1 & 23\end{matrix}\right)$

Add $3 \times \text{row"2 - "row} 3$ to $\text{row} 1$ to get:

$\left(\begin{matrix}1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 9 \\ 0 & 0 & 1 & 0 & 27 \\ 0 & 0 & 0 & 1 & 23\end{matrix}\right)$

We can now read off $a , b , c , d$ from the last column as:

$\left\{\begin{matrix}a = 1 \\ b = 9 \\ c = 27 \\ d = 23\end{matrix}\right.$