January 13, 2022 – Physicspupil

Let $x=2$ , $y=3$ , and $z=5$ . Then,

$\begin{aligned} &\quad 2x+3y+5z \\ &= 2(2)+3(3)+5(5) \\ &= 38 \\ & \\ &\quad 3x+5y+2z \\ & = 3(2)+5(3)+2(5) \\ & = 31 \\ & \\ & \quad 5x+2y+3z \\ & = 5(2)+2(3)+3(5) \\ & = 31 \\ \end{aligned}$

That is to say, I have made a set of three equations:

$\begin{cases} 2x+3y+5z =38 \\ 3x+5y+2z =31 \\ 5x+2y+3z =31 \\ \end{cases}$

in matrix form,

$\begin{bmatrix} 2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix} = \begin{bmatrix} 38 \\ 31 \\ 31 \\ \end{bmatrix}$

complicating it,

$\begin{bmatrix} 2 & 0 & 3 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & 0 & 5 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 5 & 0 & 2 & 0 & 3 \\ \end{bmatrix}\begin{bmatrix} x \\ 0 \\ y \\ 0 \\ z\\ \end{bmatrix} = \begin{bmatrix} 38 \\ 0 \\ 31 \\ 0 \\ 31 \\ \end{bmatrix}$

there should be no loss of information; instead, twenty $0$ ‘s are gained .

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31

Day: January 13, 2022

202201131704 Practice in the Making