User:SigmaJargon/Sandbox

$PI={\frac {Score_{Final}}{Score_{Raw}}}$ --- $PI\approx {\frac {(Score_{Lake}+{Average\ Value}_{Junk}*Fish_{Junk})^{2}}{Fish_{Lake}+Fish_{Junk}}}$ --- $Score_{Final}\approx {\frac {Score_{Lake}+{Average\ Value}_{Junk}*Fish_{Junk}}{Fish_{Lake}+Fish_{Junk}}}$ --- $Fish_{Junk}\approx {\frac {10*Score_{Final}-Score_{Lake}*Average\ Value_{Junk}-{\sqrt {20*Score_{Final}*(5*Score_{Final}+Average\ Value_{Junk}*(Fish_{Lake}*Average\ Value_{Junk}-Score_{Lake}))}}}{Average\ Value_{Junk}^{2}}}$

$Score_{Final}-5*Score_{Lake}*Average\ Value_{Junk}\pm {\sqrt {20*Score_{Final}*(5*Score_{Final}+4*Average\ Value_{Junk}*(Fish_{Lake}*Average\ Value_{Junk}-Score_{Lake}))}}$

$Fish_{Junk}=Fish_{Total}-Fish_{Lake}\$ --- $Score_{Raw}=Score_{Bassken}+Score_{Gambino}+Score_{Durem}+Score_{Junk}\$

--- $Score_{Lake}=Score_{Bassken}+Score_{Gambino}+Score_{Durem}\$ --- $Fish_{Total}=Fish_{Bassken}+Fish_{Gambino}+Fish_{Durem}+Fish_{Junk}\$

--- $Fish_{Lake}=Fish_{Bassken}+Fish_{Gambino}+Fish_{Durem}\$

9.) More on block matrices

a.) False! Consider the following submatrices:

$A_{1,1}={\begin{bmatrix}1&0\\0&0\\\end{bmatrix}}A_{2,1}={\begin{bmatrix}0&0\\0&1\\\end{bmatrix}}A_{1,2}={\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}A_{2,2}={\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}$

Obviously, $det(A_{1,1})=det(A_{2,1})=0\,\!$, so $det(A_{1,1})*det(A_{2,2})-det(A_{1,2})*det(A_{2,1})=0\,\!$. However, if you consider the whole matrix $A={\begin{bmatrix}A_{1,1}&A_{1,2}\\A_{2,1}&A_{2,2}\\\end{bmatrix}}={\begin{bmatrix}1&0&1&0\\0&0&0&1\\0&0&1&0\\0&1&0&1\\\end{bmatrix}}$

Then you see that $det(A)=-1$ by expansion along the 3rd row and then the 2nd row.

b.) Also false! Consider the following matrices:

$B={\begin{bmatrix}1&0&0\\0&1&0\\\end{bmatrix}}C={\begin{bmatrix}0&0&1\\0&0&0\\\end{bmatrix}}D={\begin{bmatrix}0&0&0\\1&0&0\\\end{bmatrix}}$

Then $A={\begin{bmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\\end{bmatrix}}$

Now, $rank(B)=2\,\!$, $rank(D)=1\,\!$, but $rank(A)=4\,\!$