# forall u, v, ((2,3,5,7), (13,17,19,23)) * ((64,28,-18), (-64,-27,18), (15,5,-5), (0,0,1)) * ((1), (u), (v)) = ((11), (29)) ?

Aug 12, 2018

Please refer to a Proof in the Explanation.

#### Explanation:

Let, ${\left[A\right]}_{2 \times 4} = \left[\begin{matrix}2 & 3 & 5 & 7 \\ 13 & 17 & 19 & 23\end{matrix}\right]$,

${\left[B\right]}_{4 \times 3} = \left[\begin{matrix}64 & 28 & - 18 \\ - 64 & - 27 & 18 \\ 15 & 5 & - 5 \\ 0 & 0 & 1\end{matrix}\right]$,

${\left[C\right]}_{3 \times 1} = \left[\begin{matrix}1 \\ u \\ v\end{matrix}\right]$.

Then, ${\left[A\right]}_{2 \times 4} \cdot {\left[B\right]}_{4 \times 3} \cdot {\left[C\right]}_{3 \times 1}$ is defined, and, is a

$\left(2 \times 1\right)$ Matrix. .

Now, ${\left[B\right]}_{4 \times 3} \cdot {\left[C\right]}_{3 \times 1}$ is a $4 \times 1$ Matrix, given by,,

$= \left[\begin{matrix}64 + 28 u - 18 v \\ - 64 - 27 u + 18 v \\ 15 + 5 u - 5 v \\ v\end{matrix}\right]$.

Next, ${\left[A\right]}_{2 \times 4} \cdot \left\{{\left[B\right]}_{4 \times 3} \cdot {\left[C\right]}_{3 \times 1}\right\}$ is a $2 \times 1$ Matrix

and, ${\left[A\right]}_{2 \times 4} \cdot \left\{{\left[B\right]}_{4 \times 3} \cdot {\left[C\right]}_{3 \times 1}\right\}$,

$= \left[\begin{matrix}2 & 3 & 5 & 7 \\ 13 & 17 & 19 & 23\end{matrix}\right] \cdot \left[\begin{matrix}64 + 28 u - 18 v \\ - 64 - 27 u + 18 v \\ 15 + 5 u - 5 v \\ v\end{matrix}\right]$.

Now,

$2 \left(64 + 28 u - 18 v\right) + 3 \left(- 64 - 27 u + 18 v\right) + 5 \left(15 + 5 u - 5 v\right) + 7 \left(v\right) ,$

$= \left(128 - 192 + 75\right) + u \left(56 - 81 + 25\right) + v \left(- 36 + 54 - 25 + 7\right)$,

$= 11$, and,

$13 \left(64 + 28 u - 18 v\right) + 17 \left(- 64 - 27 u + 18 v\right) + 19 \left(15 + 5 u - 5 v\right) + 23 \left(v\right) ,$

$= \left(832 - 1088 + 285\right) + u \left(364 - 459 + 95\right) + v \left(- 234 + 306 - 95 + 23\right)$,

$= 29$.

$\therefore \left[\begin{matrix}2 & 3 & 5 & 7 \\ 13 & 17 & 19 & 23\end{matrix}\right] \cdot \left[\begin{matrix}64 + 28 u - 18 v \\ - 64 - 27 u + 18 v \\ 15 + 5 u - 5 v \\ v\end{matrix}\right] = \left[\begin{matrix}11 \\ 29\end{matrix}\right]$.

From here on, consider the variables exchange

$\alpha = 3 + u - v$

$\beta = u$

$A B C = K$

$A D E = K$

$D = \left(\begin{matrix}10 & 18 & 10 \\ - 10 & - 18 & - 9 \\ 0 & 5 & 0 \\ 3 & - 1 & 1\end{matrix}\right) , E = \left(\begin{matrix}1 \\ \alpha \\ \beta\end{matrix}\right)$

$D E = B C$

$E = M C$

$D M C = B C \implies D M = B \implies M = \left(\begin{matrix}1 & 0 & 0 \\ 3 & 1 & - 1 \\ 0 & 1 & 0\end{matrix}\right)$