Evaluate 80t - 20s + 60r - 50q + 70p

[anemone]

Given a system of simultaneous equations below:

$pqrs+pqrt+pqst+prst=-264$

$pqrs+pqrt+prst+qrst=-24$

$pqrs+pqrt+pqst+qrst=24$

$pqrs+pqst+prst+qrst=248$

$pqrt+pqst+prst+qrst=-16$

Evaluate $80t-20s+60r-50q+70p$.

 

[Opalg]

Re: Evaluate 80t-20s+60r-50q+70p

Spoiler

I think that the best way to do this is to use some linear algebra. Divide the equations by $pqrst$ and write them in matrix form: $$\begin{bmatrix}0&1&1&1&1 \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1 \\ 1&1&1&1&0 \end{bmatrix} \begin{bmatrix}p^{-1} \\q^{-1} \\r^{-1} \\s^{-1} \\t^{-1} \end{bmatrix} = \frac1{pqrst}\begin{bmatrix}-264 \\ -24 \\ 24 \\ 248 \\ -16 \end{bmatrix}.$$ That $5\times5$ matrix is equal to $5P-I$, where $P$ is the matrix of the projection onto the 1-dimensional subspace spanned by the unit vector $\frac1{\sqrt5}(1,1,1,1,1)$. The inverse of that matrix is $\tfrac54P-I = \dfrac14 \begin{bmatrix}-3&1&1&1&1 \\ 1&-3&1&1&1 \\ 1&1&-3&1&1 \\ 1&1&1&-3&1 \\ 1&1&1&1&-3 \end{bmatrix}.$ It follows that $$\begin{bmatrix}p^{-1} \\q^{-1} \\r^{-1} \\s^{-1} \\t^{-1} \end{bmatrix} = \frac1{4pqrst} \begin{bmatrix}-3&1&1&1&1 \\ 1&-3&1&1&1 \\ 1&1&-3&1&1 \\ 1&1&1&-3&1 \\ 1&1&1&1&-3 \end{bmatrix} \begin{bmatrix}-264 \\ -24 \\ 24 \\ 248 \\ -16 \end{bmatrix} = \frac1{pqrst}\begin{bmatrix}256 \\ 16 \\ -32 \\ -256 \\ 8 \end{bmatrix} = \frac1{pqrst}\begin{bmatrix}2^8 \\ 2^4 \\ -2^5 \\ -2^8 \\ 2^3 \end{bmatrix}.$$ If $\Pi = pqrst$ then $p^{-1} = \dfrac{2^8}{\Pi}$, $q^{-1} = \dfrac{2^4}{\Pi}$, $r^{-1} = -\dfrac{2^5}{\Pi}$, $s^{-1} = -\dfrac{2^8}{\Pi}$, $t^{-1} = \dfrac{2^3}{\Pi}.$ Multiply those five equations together to see that $\dfrac1{\Pi} = \dfrac{2^{28}}{\Pi^5}$, from which $\Pi^4 = 2^{28}$ and so $\Pi = 2^7$.

Thus $p = 1/2$, $q = 8$, $r = -4$, $s = -1/2$, $t = 16.$ If I have done the arithmetic correctly, then $80t-20s+60r-50q+70p = 685.$

Edit. The left sides of the original equations each contain four of the five products $pqrs$, $pqrt$, $pqst$, $prst$, $qrst$. When I first read the problem, I saw that the first equation uses the four products that include $p$, and the last equation uses the four products that include $t$. I assumed, without stopping to read everything carefully, that if the first equation was a "$p$" equation and the last equation was a "$t$" equation, then the intermediate equations would be the "$q$", "$r$", "$s$" equations, in that order. But anemone points out to me in a PM that the order of the "$q$" and "$r$" equations is reversed. This means that my values for $q$ and $r$ should be exchanged, and that obviously alters the result of the final calculation.

 

[anemone]

Re: Evaluate 80t-20s+60r-50q+70p

Thanks for participating, Opalg! Even though you have read the problem wrongly, I am glad to learn that there is another way (which is different than mine) to approach the problem.

My solution:

Spoiler

Divide all given equations by $pqrst$ we get:

$\dfrac{pqrs}{pqrst}+\dfrac{pqrt}{pqrst}+\dfrac{pqst}{pqrst}+\dfrac{prst}{pqrst}=-\dfrac{264}{pqrst}\rightarrow\;\dfrac{1}{t}+\dfrac{1}{s}+\dfrac{1}{r}+\dfrac{1}{q}=-\dfrac{264}{pqrst}-(1)$

$\dfrac{pqrs}{pqrst}+\dfrac{pqrt}{pqrst}+\dfrac{prst}{pqrst}+\dfrac{qrst}{pqrst}=-\dfrac{24}{pqrst} \rightarrow\;\dfrac{1}{t}+\dfrac{1}{s}+\dfrac{1}{q}+\dfrac{1}{p}=-\dfrac{24}{pqrst}-(2)$

$\dfrac{pqrs}{pqrst}+\dfrac{pqrt}{pqrst}+\dfrac{pqst}{pqrst}+\dfrac{qrst}{pqrst}=\dfrac{24}{pqrst} \rightarrow\;\dfrac{1}{t}+\dfrac{1}{s}+\dfrac{1}{r}+\dfrac{1}{p}=\dfrac{24}{pqrst}-(3)$

$\dfrac{pqrs}{pqrst}+\dfrac{pqst}{pqrst}+\dfrac{prst}{pqrst}+\dfrac{qrst}{pqrst}=\dfrac{248}{pqrst} \rightarrow\;\dfrac{1}{t}+\dfrac{1}{r}+\dfrac{1}{q}+\dfrac{1}{p}=\dfrac{248}{pqrst}-(4)$

$\dfrac{pqrt}{pqrst}+\dfrac{pqst}{pqrst}+\dfrac{prst}{pqrst}+\dfrac{qrst}{pqrst}=-\dfrac{16}{pqrst}\rightarrow\;\dfrac{1}{s}+\dfrac{1}{r}+\dfrac{1}{q}+\dfrac{1}{p}=-\dfrac{16}{pqrst}-(5)$

Equation (2) $-$ Equation (1):
$\dfrac{1}{p}-\dfrac{1}{r}=\dfrac{240}{pqrst}-(6)$

Equation (3) $-$ Equation (2):
$\dfrac{1}{r}-\dfrac{1}{q}=\dfrac{48}{pqrst}-(7)$

$\therefore \dfrac{1}{p}-\dfrac{1}{q}=\dfrac{288}{pqrst}--(*)$

Equation (4) $+$ Equation (5):
$\dfrac{1}{t}+\dfrac{1}{s}+2\left(\dfrac{1}{r}+ \dfrac{1}{q}+ \dfrac{1}{p} \right)=\dfrac{232}{pqrst}-(8)$

Equation (7) $-$ Equation (3):
$2\left(\dfrac{1}{q}\right)+ \dfrac{1}{r}+ \dfrac{1}{p}=\dfrac{208}{pqrst}-(9)$

Equation (8) $-$ Equation (6):
$3\left(\dfrac{1}{q}\right)+ \dfrac{1}{p}=\dfrac{160}{pqrst}-(**)$

Solving the equations (*) and (**) simultaneously we get:
$\dfrac{1}{q}=-\dfrac{32}{pqrst},\dfrac{1}{p}=\dfrac{256}{pqrst}, \dfrac{1}{r}=\dfrac{16}{pqrst}, \dfrac{1}{s}=-\dfrac{256}{pqrst}, \dfrac{1}{t}=\dfrac{8}{pqrst}$

Therefore, if we multiply these five equations together, we get
$\dfrac{1}{pqrst}=\dfrac{2^{28}}{(pqrst)^5}$

$\therefore pqrst=2^7$ or $\dfrac{1}{q}=-\dfrac{1}{4},\dfrac{1}{p}=2, \dfrac{1}{r}=\dfrac{1}{8}, \dfrac{1}{s}=-2, \dfrac{1}{t}=\dfrac{1}{16}$.

Hence $80t-20s+60r-50q+70p=80(16)-20(-\dfrac{1}{2})+60(8)-50(-4)+70(\dfrac{1}{2})=2005$.