You will be able to
You plug your measurement into an equation used to estimate the overall height of an adult male based on femur length:
H=1.88cmcm(L)+81.3 cmH = 1.88 \frac{\text{cm}}{\text{cm}}(L) + 81.3 \text{ cm}H=1.88cmcm(L)+81.3 cm
H=(1.88×47.5) cm+81.3 cmH = (1.88 × 47.5) \text{ cm} + 81.3 \text{ cm}H=(1.88×47.5) cm+81.3 cm
H=170.6 cmH = 170.6 \text{ cm}H=170.6 cm
f^(x1)=β0+β1⋅x1=81.3 cm+1.88cmcm⋅x1=y\hat{f}(x_1)= \beta_0 + \beta_1 \cdot x_1 = 81.3 \text{ cm}+ 1.88 \frac{\text{cm}}{\text{cm}} \cdot x_1 = y f^(x1)=β0+β1⋅x1=81.3 cm+1.88cmcm⋅x1=y
someone put knowledge about the world in a formula fff (model)
We will use matrix notation most of the time
f(x1)=β⃗×x⃗=[β0β1]×[1x1]=[81.31.88]×[1x1]f(x_1) = \vec{\beta} \times \vec{x} = \begin{bmatrix} \beta_0 & \beta_1 \end{bmatrix} \times \begin{bmatrix} 1 \\ x_1 \end{bmatrix} = \begin{bmatrix} 81.3 & 1.88 \end{bmatrix} \times \begin{bmatrix} 1 \\ x_1 \end{bmatrix} f(x1)=β×x=[β0β1]×[1x1]=[81.31.88]×[1x1]
=81.3+1.88⋅x1= 81.3 + 1.88 \cdot x_1 =81.3+1.88⋅x1
10 min
p(s)=65+s⋅20 bpm5 kphp(s)=65 + s \cdot \frac{20 \text{ bpm}}{5 \text{ kph}}p(s)=65+s⋅5 kph20 bpm
parameters of the model
145 bpm145 \text{ bpm}145 bpm, but we do not know whether this prediction is valid
p(s)=β⃗⋅x⃗p(s) = \vec{\beta} \cdot \vec{x}p(s)=β⋅x =[β0β1]×[1s]=[654]×[1s]= \begin{bmatrix}\beta_0 & \beta_1 \end{bmatrix} \times \begin{bmatrix} 1 \\ s \end{bmatrix} = \begin{bmatrix} 65 & 4 \end{bmatrix} \times \begin{bmatrix} 1 \\ s \end{bmatrix}=[β0β1]×[1s]=[654]×[1s]
A=[651547318268167]A = \begin{bmatrix} 65 & 154 \\ 73 & 182 \\ 68 & 167 \\ \end{bmatrix}A=⎣⎡657368154182167⎦⎤
A=(ai,j)=[a1,1a1,2a2,1a2,2a3,1a3,2]A = (a_{i,j}) = \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ a_{3,1} & a_{3,2} \\ \end{bmatrix}A=(ai,j)=⎣⎡a1,1a2,1a3,1a1,2a2,2a3,2⎦⎤
has only one row or column
x⃗=[x1x2x3]\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}x=⎣⎡x1x2x3⎦⎤
transposed vector
x⃗′=[x1x2x3]\vec{x}' = \begin{bmatrix} x_1 & x_2 & x_3 \\ \end{bmatrix}x′=[x1x2x3]
(x⃗′)′=x(\vec{x}')' =x (x′)′=x
A+B=[a1,1a1,2a2,1a2,2a3,1a3,2]+[b1,1b1,2b2,1b2,2b3,1b3,2]=[a1,1+b1,1a1,1+b1,2a2,1+b2,1a2,1+b2,2a3,1+b3,1a3,1+b3,2]A + B = \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ a_{3,1} & a_{3,2} \\ \end{bmatrix} + \begin{bmatrix} b_{1,1} & b_{1,2} \\ b_{2,1} & b_{2,2} \\ b_{3,1} & b_{3,2} \\ \end{bmatrix} = \begin{bmatrix} a_{1,1} + b_{1,1} & a_{1,1} + b_{1,2} \\ a_{2,1} + b_{2,1} & a_{2,1} + b_{2,2} \\ a_{3,1} + b_{3,1} & a_{3,1} + b_{3,2} \\ \end{bmatrix} A+B=⎣⎡a1,1a2,1a3,1a1,2a2,2a3,2⎦⎤+⎣⎡b1,1b2,1b3,1b1,2b2,2b3,2⎦⎤=⎣⎡a1,1+b1,1a2,1+b2,1a3,1+b3,1a1,1+b1,2a2,1+b2,2a3,1+b3,2⎦⎤
c⋅A=[c⋅a1,1c⋅a1,2c⋅a2,1c⋅a2,2c⋅a3,1c⋅a3,2]c \cdot A = \begin{bmatrix} c \cdot a_{1,1} & c \cdot a_{1,2} \\ c \cdot a_{2,1} & c \cdot a_{2,2} \\ c \cdot a_{3,1} & c \cdot a_{3,2} \\ \end{bmatrix} c⋅A=⎣⎡c⋅a1,1c⋅a2,1c⋅a3,1c⋅a1,2c⋅a2,2c⋅a3,2⎦⎤
A×B=[a1,1a1,2a2,1a2,2]⋅[b1,1b1,2b2,1b2,2]A \times B = \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{bmatrix} \cdot \begin{bmatrix} b_{1,1} & b_{1,2} \\ b_{2,1} & b_{2,2} \\ \end{bmatrix} A×B=[a1,1a2,1a1,2a2,2]⋅[b1,1b2,1b1,2b2,2]
=[a1,1b1,1+a1,2b2,1a1,1b1,2+a1,2b2,2a2,1b1,1+a2,2b2,1a2,1b1,2+a2,2b2,2]= \begin{bmatrix} a_{1,1} b_{1,1} + a_{1,2} b_{2,1} & a_{1,1} b_{1,2} + a_{1,2} b_{2,2} \\ a_{2,1} b_{1,1} + a_{2,2} b_{2,1} & a_{2,1} b_{1,2} + a_{2,2} b_{2,2} \\ \end{bmatrix} =[a1,1b1,1+a1,2b2,1a2,1b1,1+a2,2b2,1a1,1b1,2+a1,2b2,2a2,1b1,2+a2,2b2,2]
5 minutes
[223451]+[100001]\begin{bmatrix} 2 & 2 \\ 3 & 4 \\ 5 & 1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{bmatrix}⎣⎡235241⎦⎤+⎣⎡100001⎦⎤
[2234]×[1000]\begin{bmatrix} 2 & 2 \\ 3 & 4 \\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} [2324]×[1000]
[223451]+[100001]=[323452]\begin{bmatrix} 2 & 2 \\ 3 & 4 \\ 5 & 1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 3 & 4 \\ 5 & 2 \\ \end{bmatrix}⎣⎡235241⎦⎤+⎣⎡100001⎦⎤=⎣⎡335242⎦⎤
[2234]⋅[1000]=[2⋅1+2⋅02⋅0+2⋅03⋅1+4⋅03⋅0+4⋅0]=[2030]\begin{bmatrix} 2 & 2 \\ 3 & 4 \\ \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 + 2 \cdot 0 & 2 \cdot 0 + 2 \cdot 0 \\ 3 \cdot 1 + 4 \cdot 0 & 3 \cdot 0 + 4 \cdot 0 \\ \end{bmatrix}= \begin{bmatrix} 2 & 0 \\ 3 & 0 \\ \end{bmatrix} [2324]⋅[1000]=[2⋅1+2⋅03⋅1+4⋅02⋅0+2⋅03⋅0+4⋅0]=[2300]
Caloric energy of Sample 1
We could write a for-loop!
for
Caloric energy
[100g10g1g70g20g2g90g10g1g]⋅[0kcal/g4kcal/g0kcal/g]=[40kcal80kcal40kcal]\begin{bmatrix} 100 g & 10 g& 1g \\ 70 g & 20 g& 2g \\ 90 g & 10 g& 1g \\ \end{bmatrix} \cdot \begin{bmatrix} 0 kcal/ g \\ 4 kcal/ g \\ 0 kcal/ g \\ \end{bmatrix} = \begin{bmatrix} 40 kcal \\ 80 kcal \\ 40 kcal \\ \end{bmatrix} ⎣⎡100g70g90g10g20g10g1g2g1g⎦⎤⋅⎣⎡0kcal/g4kcal/g0kcal/g⎦⎤=⎣⎡40kcal80kcal40kcal⎦⎤
Task
Price
[100g10g1g70g20g2g90g10g1g]⋅[0.00€/g0.02€/g0.10€/g]=[.........]\begin{bmatrix} 100 g & 10 g& 1g \\ 70 g & 20 g& 2g \\ 90 g & 10 g& 1g \\ \end{bmatrix} \cdot \begin{bmatrix} 0.00 €/ g \\ 0.02 €/ g \\ 0.10 €/ g \\ \end{bmatrix} = \begin{bmatrix} ... \\ ... \\ ... \\ \end{bmatrix} ⎣⎡100g70g90g10g20g10g1g2g1g⎦⎤⋅⎣⎡0.00€/g0.02€/g0.10€/g⎦⎤=⎣⎡.........⎦⎤
[100g10g1g70g20g2g90g10g1g]⋅[0.00€/g0.02€/g0.10€/g]=[100⋅0+10⋅0.02+1⋅0.170⋅0+20⋅0.02+2⋅0.190⋅0+10⋅0.02+1⋅0.1]€=\begin{bmatrix} 100 g & 10 g& 1g \\ 70 g & 20 g& 2g \\ 90 g & 10 g& 1g \\ \end{bmatrix} \cdot \begin{bmatrix} 0.00 €/ g \\ 0.02 €/ g \\ 0.10 €/ g \\ \end{bmatrix} = \begin{bmatrix} 100 \cdot 0 + 10 \cdot 0.02 + 1 \cdot 0.1 \\ 70 \cdot 0 + 20 \cdot 0.02 + 2 \cdot 0.1 \\ 90 \cdot 0 + 10 \cdot 0.02 + 1 \cdot 0.1 \\ \end{bmatrix} \text{€} = ⎣⎡100g70g90g10g20g10g1g2g1g⎦⎤⋅⎣⎡0.00€/g0.02€/g0.10€/g⎦⎤=⎣⎡100⋅0+10⋅0.02+1⋅0.170⋅0+20⋅0.02+2⋅0.190⋅0+10⋅0.02+1⋅0.1⎦⎤€=
[0.30€0.60€0.30€]\begin{bmatrix} 0.30 \text{€} \\ 0.60 \text{€} \\ 0.30 \text{€} \\ \end{bmatrix} ⎣⎡0.30€0.60€0.30€⎦⎤
even more convenient:
[100g10g1g70g20g2g90g10g1g]⋅[0kcal0.00€/g4kcal0.02€/g0kcal0.10€/g]=[40kcal0.30€80kcal0.60€40kcal0.30€]\begin{bmatrix} 100 g & 10 g& 1g \\ 70 g & 20 g& 2g \\ 90 g & 10 g& 1g \\ \end{bmatrix} \cdot \begin{bmatrix} 0 kcal & 0.00 \text{€} / g \\ 4 kcal & 0.02 \text{€} / g \\ 0 kcal & 0.10 \text{€} / g \\ \end{bmatrix}= \begin{bmatrix} & 40 kcal & 0.30 \text{€} \\ & 80 kcal & 0.60 \text{€} \\ & 40 kcal & 0.30 \text{€} \\ \end{bmatrix} ⎣⎡100g70g90g10g20g10g1g2g1g⎦⎤⋅⎣⎡0kcal4kcal0kcal0.00€/g0.02€/g0.10€/g⎦⎤=⎣⎡40kcal80kcal40kcal0.30€0.60€0.30€⎦⎤
numpy
1 Matrix Data and numpy
45 minutes
Equation I 2y+x=62y + x = 62y+x=6 can be written as y=3−0.5xy = 3 - 0.5 xy=3−0.5x
Equation II: 2y+2=3x2y + 2 = 3x2y+2=3x can be written as y=−1+1.5xy = -1 + 1.5xy=−1+1.5x
We can rewrite this system in matrix form
A×b⃗=c⃗A \times \vec{b} = \vec{c} A×b=c
[12−6−322]×[xy1]=[00] \begin{bmatrix} 1 & 2 & -6 \\ -3 & 2 & 2 \\ \end{bmatrix} \times \begin{bmatrix} x \\ y \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}[1−322−62]×⎣⎡xy1⎦⎤=[00]
Equation I: 3y−3x=33y - 3x = 33y−3x=3
Equation II: 2y+2=2x2y + 2 = 2x2y+2=2x
there is no solution
[−333−222]×[xy1]=[00] \begin{bmatrix} -3 & 3 & 3 \\ -2 & 2 & 2 \\ \end{bmatrix} \times \begin{bmatrix} x \\ y \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}[−3−23232]×⎣⎡xy1⎦⎤=[00]
You want to create a new growth medium on industrial scale.
You base the the new medium on two existing products (A and B).
You want to create 400 kg of the new mixture.
Component A costs 18 €. Component B costs 22 €.
How much (kg) of A and B do You need, if the new mixture should cost 19,50 €?
First, create two formulas by what You know. Then reformulate them as a matrix multiplication and solve them using numpy.
Product of Two Matrices or Two Vectors