MPZI_vj05_s_izradom -- Sage

Vektori i matrice

Vektori

1. Zadatak

Definirajte vektore

$\mathbf{a} = (2, 3, 1)$ i

$\mathbf{b} = (-1, 5, 4)$ ,

a zatim izračunajte:

$2\mathbf{a} + \mathbf{b}$ ,
$\mathbf{a}- 2\mathbf{b}$ ,
skalarni umnožak $\mathbf{a}\cdot\mathbf{b}$ vektora $\mathbf{a}$ i $\mathbf{b}$ ,
vektorski umnožak $\mathbf{a}\times\mathbf{b}$ vektora $\mathbf{a}$ i $\mathbf{b}$ i provjerite da je rezultat okomit na $\mathbf{a}$ i $\mathbf{b}$ .

a = vector ([2,3,1])
b = vector ([-1,5,4])
print "a =", a
print "b =", b

a = (2, 3, 1)
b = (-1, 5, 4)

a = (2, 3, 1)
b = (-1, 5, 4)

2*a + b

(3, 11, 6)

(3, 11, 6)

a - 2*b

(4, -7, -7)

(4, -7, -7)

Skalarni umnožak:

a.dot_product (b)

dot_product (a, b)                    # opća funkcija dot_product ne postoji

Traceback (click to the left of this block for traceback)
...
NameError: name 'dot_product' is not defined

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "_sage_input_7.py", line 10, in <module>
    exec compile(u'open("___code___.py","w").write("# -*- coding: utf-8 -*-\\n" + _support_.preparse_worksheet_cell(base64.b64decode("ZG90X3Byb2R1Y3QgKGEsIGIpICAgICAgICAgICAgICAgICAgICAjIG9wxIdhIGZ1bmtjaWphIGRvdF9wcm9kdWN0IG5lIHBvc3Rvamk="),globals())+"\\n"); execfile(os.path.abspath("___code___.py"))
  File "", line 1, in <module>
    
  File "/tmp/tmpRGFvyL/___code___.py", line 2, in <module>
    exec compile(u'dot_product (a, b)                    # op\u0107a funkcija dot_product ne postoji
  File "", line 1, in <module>
    
NameError: name 'dot_product' is not defined

a*b

Vektorski umnožak:

c = a.cross_product (b) 
c

(7, -9, 13)

(7, -9, 13)

cross_product (a, b)                    # opća funkcija cross_product ne postoji

Traceback (click to the left of this block for traceback)
...
NameError: name 'cross_product' is not defined

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "_sage_input_10.py", line 10, in <module>
    exec compile(u'open("___code___.py","w").write("# -*- coding: utf-8 -*-\\n" + _support_.preparse_worksheet_cell(base64.b64decode("Y3Jvc3NfcHJvZHVjdChhLCBiKSAgICAgICAgICAgICAgICAgICAgIyBvcMSHYSBmdW5rY2lqYSBjcm9zc19wcm9kdWN0IG5lIHBvc3Rvamk="),globals())+"\\n"); execfile(os.path.abspath("___code___.py"))
  File "", line 1, in <module>
    
  File "/tmp/tmpsenxvP/___code___.py", line 2, in <module>
    exec compile(u'cross_product(a, b)                    # op\u0107a funkcija cross_product ne postoji
  File "", line 1, in <module>
    
NameError: name 'cross_product' is not defined

Provjera da je vektorski umnožak okomit na faktore produkta (ako su vektori $\mathbf{a}$ i $\mathbf{b}$ međusobno okomiti, onda je $\mathbf{a}\cdot\mathbf{b} = 0$ ):

a*c

b*c

2. zadatak

Za vektore $\mathbf{a}$ i $\mathbf{b}$ , definirane u 1. zadatku, izračunajte:

duljinu vektora $\mathbf{a}$ ,
kut (u stupnjevima) između $\mathbf{a}$ i $\mathbf{b}$ .

Na kraju uklonite varijable a i b.

Duljina ili norma vektora:

a.norm()

sqrt(14)

sqrt(14)

a.norm().n()

3.74165738677394

3.74165738677394

norm (a)

sqrt(14)

sqrt(14)

N (norm (a))

3.74165738677394

3.74165738677394

Kut između vektora:

$c = \cos\, (\mathbf{a},\,\mathbf{b}) = \dfrac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\cdot \|\mathbf{b}\|} \quad\Rightarrow\quad \angle\,(\mathbf{a},\,\mathbf{b}) = \mathrm{arccos}\: c$

coskuta = (a*b)/(norm (a) * norm (b))
N (coskuta)

0.701068184015974

0.701068184015974

arccos (coskuta)

arccos(17/588*sqrt(42)*sqrt(14))

arccos(17/588*sqrt(42)*sqrt(14))

N (arccos (coskuta))

0.793901974719713

0.793901974719713

N (arccos(coskuta) * 180./pi)

45.4872324985413

45.4872324985413

Uklanjanje varijabli:

reset('a b')

print a, b

Traceback (click to the left of this block for traceback)
...
NameError: name 'a' is not defined

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "_sage_input_17.py", line 10, in <module>
    exec compile(u'open("___code___.py","w").write("# -*- coding: utf-8 -*-\\n" + _support_.preparse_worksheet_cell(base64.b64decode("cHJpbnQgYSwgYg=="),globals())+"\\n"); execfile(os.path.abspath("___code___.py"))
  File "", line 1, in <module>
    
  File "/tmp/tmpmlEqGM/___code___.py", line 2, in <module>
    exec compile(u'print a, b
  File "", line 1, in <module>
    
NameError: name 'a' is not defined

3. zadatak

Na materijalnu točku $\mathbf{A}$ djeluju dvije sile prikazane vektorima $\vec{a}_1=(2, -1, 2)$ i $\vec{a}_2=(4, 1, 1)$ , a na točku $\mathbf{B}$ sile prikazane vektorima $\vec{b}_1=(-1, 3, -1)$ i $\vec{b}_2=(-3, -2, 5)$ . Na koju od točaka djeluje rezultanta silâ većega intenziteta?

Uputa: rezultanta silâ dobiva se zbrajanjem sila, a intenzitet je sile jednak duljini vektora kojim je prikazana.

Prvo ćemo definirati sile i prikazati ih pomoću naredbe print:

a1 = vector ([2, -1, 2])
a2 = vector ([4, 1, 1])
b1 = vector ([-1, 3, -1])
b2 = vector ([-3, -2, 5])
print a1, a2, b1, b2

(2, -1, 2) (4, 1, 1) (-1, 3, -1) (-3, -2, 5)

(2, -1, 2) (4, 1, 1) (-1, 3, -1) (-3, -2, 5)

Zatim ćemo izračunati rezultante $\vec{r}_A$ i $\vec{r}_B$ :

ra = a1 + a2
rb = b1 + b2
print ra, rb

(6, 0, 3) (-4, 1, 4)

(6, 0, 3) (-4, 1, 4)

Sada možemo izračunati njihove intenzitete:

norm (ra); norm (rb)

3*sqrt(5)
sqrt(33)

3*sqrt(5)
sqrt(33)

N (norm (ra)); N (norm (rb))

6.70820393249937
5.74456264653803

6.70820393249937
5.74456264653803

Zaključujemo da na točku $\mathbf{A}$ djeluje sila većega intenziteta. Možemo to i provjeriti:

norm (ra) > norm (rb)

3*sqrt(5) > sqrt(33)

3*sqrt(5) > sqrt(33)

bool (norm (ra) > norm (rb))

True

True

norm (ra) < norm (rb)

3*sqrt(5) < sqrt(33)

3*sqrt(5) < sqrt(33)

bool (norm (ra) < norm (rb))

False

False

4. zadatak

Definirajte dva vektora. Komponente prvoga neka su svi parni brojevi između 1 i 30 (uključivo), a komponente drugoga svi neparni brojevi u istom intervalu. Izračunajte kut između tih vektora u radijanima i stupnjevima.

v1 = vector (srange (2, 31, 2))    # ili: v1 = vector (srange (2, 32, 2))
v2 = vector (srange (1, 30, 2))    # ili: v2 = vector (srange (1, 31, 2))
print v1 
print v2

(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30)
(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29)

(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30)
(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29)

Provjera jednakosti dimenzija (broja komponenata):

print len (v1); len (v2); len (v1) == len (v2)

15
15
True

15
15
True

Izračunavanje kuta:

coskuta = (v1*v2) / (norm (v1) * norm (v2))
coskuta

118/139345*sqrt(4495)*sqrt(310)

118/139345*sqrt(4495)*sqrt(310)

N (coskuta)

0.999623166286920

0.999623166286920

kut_radijani = arccos (coskuta)
N (kut_radijani)

0.0274538661296681

0.0274538661296681

kut_stupnjevi = kut_radijani * 180/pi 
N (kut_stupnjevi)

1.57299066054714

1.57299066054714

1. zadaci za vježbu

Definirajte vektore $\mathbf{a}=(1;\; 3;\; 2,\!1)$ i $\mathbf{b}=(-1;\; 5,\!3;\; 0)$ pa zatim izračunajte:

vektor $\mathbf{c} = 3\mathbf{a} - 4\mathbf{b}$ ,
duljine vektora $\mathbf{a}$ i $\mathbf{b}$ ,
skalarni i vektorski umnožak vektora $\mathbf{a}$ i $\mathbf{b}$ ,
kut između vektora $\mathbf{a}$ i $\mathbf{c}$ ; rezultat neka bude u stupnjevima,
ploštinu paralelograma određenoga vektorima $\mathbf{a}$ i $\mathbf{b}$ ; uputa: ploština je paralelograma $\|\mathbf{a}\times\mathbf{b}\|$ .

Rješenja se nalaze na kraju datoteke.

Matrični račun

5. zadatak

Zadajte matrice

$\mathbf{a}$ kojoj su reci $(0, 2, 1)$ , $( 4, 1, 0)$ i $(-2, 5, 1)$ i

$\mathbf{b}$ kojoj su reci $(0, 2, 1)$ , $(1; 1,\!5; 2)$ i $(0, 4, 2)$

pa izračunajte:

$3\mathbf{a}- 2\mathbf{b}$
umnožak $\mathbf{a}\,\mathbf{b}$ i
inverznu matricu one matrice kojoj je determinanta različita od nule.

Na kraju uklonite varijable a i b.

a = matrix ([[0, 2, 1], [4, 1, 0], [-2, 5, 1]])
b = matrix ([[0, 2, 1], [1, 1.5, 2], [0, 4, 2]])
print a
print     # prazan redak 
print b

[ 0  2  1]
[ 4  1  0]
[-2  5  1]

[0.000000000000000  2.00000000000000  1.00000000000000]
[ 1.00000000000000  1.50000000000000  2.00000000000000]
[0.000000000000000  4.00000000000000  2.00000000000000]

[ 0  2  1]
[ 4  1  0]
[-2  5  1]

[0.000000000000000  2.00000000000000  1.00000000000000]
[ 1.00000000000000  1.50000000000000  2.00000000000000]
[0.000000000000000  4.00000000000000  2.00000000000000]

3*a - 2*b

[0.000000000000000  2.00000000000000  1.00000000000000]
[ 10.0000000000000 0.000000000000000 -4.00000000000000]
[-6.00000000000000  7.00000000000000 -1.00000000000000]

[0.000000000000000  2.00000000000000  1.00000000000000]
[ 10.0000000000000 0.000000000000000 -4.00000000000000]
[-6.00000000000000  7.00000000000000 -1.00000000000000]

a * b

[2.00000000000000 7.00000000000000 6.00000000000000]
[1.00000000000000 9.50000000000000 6.00000000000000]
[5.00000000000000 7.50000000000000 10.0000000000000]

[2.00000000000000 7.00000000000000 6.00000000000000]
[1.00000000000000 9.50000000000000 6.00000000000000]
[5.00000000000000 7.50000000000000 10.0000000000000]

det (a)

det (b)

0.000000000000000

0.000000000000000

a^-1

[ 1/14  3/14 -1/14]
[ -2/7   1/7   2/7]
[ 11/7  -2/7  -4/7]

[ 1/14  3/14 -1/14]
[ -2/7   1/7   2/7]
[ 11/7  -2/7  -4/7]

(a^-1).n()

[ 0.0714285714285714   0.214285714285714 -0.0714285714285714]
[ -0.285714285714286   0.142857142857143   0.285714285714286]
[   1.57142857142857  -0.285714285714286  -0.571428571428571]

[ 0.0714285714285714   0.214285714285714 -0.0714285714285714]
[ -0.285714285714286   0.142857142857143   0.285714285714286]
[   1.57142857142857  -0.285714285714286  -0.571428571428571]

reset ('a b')

6. zadatak

Neka su

1. $A$ točka u ravnini i $\vec{r}_{\!A}$ njezin radijus–vektor i

2. $\mathbf{R}_\alpha$ matrica oblika $\left[ \begin{array}{cc} \cos\,\alpha & -\sin\,\alpha \\ \sin\,\alpha & \cos\,\alpha \end{array} \right]$ .

Umnožak $\mathbf{R}_\alpha \, \vec{r}_{\!A}$ je radijus–vektor $\vec{r}_B$ točke $B$ u koju prelazi točka $A$ pri rotaciji ravnine za kut $\alpha$ , pri čemu je ishodište središte rotacije, a rotacija je za $\alpha > 0$ u smislu suprotnom od smisla vrtnje kazaljke na satu, dok je za $\alpha < 0$ u smislu vrtnje kazaljke na satu.

Pronađite točku $B$ u koju prelazi točka $A=(2, 3)$ pri rotaciji ravnine za kut $\displaystyle \alpha = \frac{\pi}{3}$ . Provjerite da je radijus–vektor novodobivene točke jednake duljine kao i radijus–vektor polazne točke.

Dodatak: Rotacija za kut $-\alpha$ točku $B$ „vraća” u početni položaj, u točku $A$ ; ta je rotacija, prema tome, operacija koja je inverzna rotaciji za kut $\alpha$ , pa je $\mathbf{R}_{-\alpha} = \mathbf{R}_\alpha^{-1}$ .

Kako su $\sin\,(-\alpha) = -\sin\,\alpha$ i $\cos\,(-\alpha) = \cos\,\alpha$ , bit će i $\mathbf{R}_{-\alpha} = \mathbf{R}_\alpha^{\mathrm{T}}$ .

„Vratite” točku $B$ u početni položaj množenjem vektora $\vec{r}_B$ slijeva matricom $\mathbf{R}_\alpha^{\mathrm{T}}$ te množenjem vektora $\vec{r}_B$ slijeva matricom $\mathbf{R}_\alpha^{-1}$ .

Prvi način — sa simboličkim i cjelobrojnim veličinama:

matrica $\mathbf{R}_\alpha$ i vektor $\vec{r}_{\!A}$ :

R_alfa = matrix ([[cos(pi/3), -sin(pi/3)], 
                  [sin(pi/3), cos(pi/3)]])
show (R_alfa)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2} \, \sqrt{3} \\ \frac{1}{2} \, \sqrt{3} & \frac{1}{2} \end{array}\right)$

r_A = vector ([2, 3])
show (r_A)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2,\,3\right)$

rotacija točke $A$ :

r_B = R_alfa * r_A
show (r_B)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{3}{2} \, \sqrt{3} + 1,\,\sqrt{3} + \frac{3}{2}\right)$

provjera jednakosti duljina:

show (norm (r_A)); show (norm (r_B))

$\newcommand{\Bold}[1]{\mathbf{#1}}\sqrt{13}$

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{2} \, \sqrt{{\left(3 \, \sqrt{3} - 2\right)}^{2} + {\left(2 \, \sqrt{3} + 3\right)}^{2}}$

show (norm (r_B) . simplify_full())

$\newcommand{\Bold}[1]{\mathbf{#1}}\sqrt{13}$

norm (r_A) == norm (r_B)

sqrt(13) == 1/2*sqrt((3*sqrt(3) - 2)^2 + (2*sqrt(3) + 3)^2)

sqrt(13) == 1/2*sqrt((3*sqrt(3) - 2)^2 + (2*sqrt(3) + 3)^2)

bool (norm (r_A) == norm (r_B))

True

True

N (norm (r_A)) == N (norm (r_B))

True

True

proračun je proveden s točnim cjelobrojnim i simboličkim veličinama, a „realne” su aproksimacije izračunane tek na kraju, pa su međusobno jednake.

Dodatak: vraćanje u početni položaj pomoću transponirane matrice:

R_alfa_T = R_alfa.T
show (R_alfa_T)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} \frac{1}{2} & \frac{1}{2} \, \sqrt{3} \\ -\frac{1}{2} \, \sqrt{3} & \frac{1}{2} \end{array}\right)$

r_C = R_alfa_T * r_B
show (r_C)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{1}{4} \, \sqrt{3} {\left(2 \, \sqrt{3} + 3\right)} - \frac{3}{4} \, \sqrt{3} + \frac{1}{2},\,\frac{1}{4} \, \sqrt{3} {\left(3 \, \sqrt{3} - 2\right)} + \frac{1}{2} \, \sqrt{3} + \frac{3}{4}\right)$

show (r_C.simplify_full())

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2,\,3\right)$

vraćanje u početni položaj pomoću inverzne matrice:

R_alfa_inv = R_alfa^(-1)
show (R_alfa_inv)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} \frac{1}{2} & \frac{1}{2} \, \sqrt{3} \\ -\frac{1}{2} \, \sqrt{3} & \frac{1}{2} \end{array}\right)$

r_D = R_alfa_inv * r_B
show (r_D.simplify_full())

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2,\,3\right)$

Drugi način — s „realnim” brojevima:

matrica $\mathbf{R}_\alpha$ i vektor $\vec{r}_{\!A}$ :

R_alfa = matrix ([[cos (N(pi)/3), -sin (N(pi)/3)], 
                  [sin (N(pi)/3), cos (N(pi)/3)]])
show (R_alfa)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} 0.500000000000000 & -0.866025403784439 \\ 0.866025403784439 & 0.500000000000000 \end{array}\right)$

r_A = vector ([2., 3.])                  
r_A

(2.00000000000000, 3.00000000000000)

(2.00000000000000, 3.00000000000000)

rotacija točke $A$ :

r_B = R_alfa * r_A;
r_B

(-1.59807621135332, 3.23205080756888)

(-1.59807621135332, 3.23205080756888)

provjera jednakosti duljina:

print norm (r_A)
print norm (r_B)
norm (r_A) == norm (r_B)

3.60555127546399
3.60555127546399
False

3.60555127546399
3.60555127546399
False

- kako se cijeli proračun provodio s približnim „realnim” brojevima, stvarna se jednakost izgubila: iako se brojevi u ispisanomu dijelu znamenaka ne razlikuju, „interni” se zapis brojeva razlikuje:

norm (r_A).exact_rational(); norm (r_B).exact_rational()

8118979690322419/2251799813685248
2029744922580605/562949953421312

8118979690322419/2251799813685248
2029744922580605/562949953421312

Dodatak: vraćanje u početni položaj pomoću transponirane matrice:

R_alfa_T = R_alfa.T
show (R_alfa_T)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} 0.500000000000000 & 0.866025403784439 \\ -0.866025403784439 & 0.500000000000000 \end{array}\right)$

r_C = R_alfa_T * r_B
r_C

(2.00000000000000, 3.00000000000000)

(2.00000000000000, 3.00000000000000)

r_C == r_A

False

False

- proračun je proveden s „realnim” brojevima, pa ne možemo očekivati jednakost; interni je zapis koordinata točaka:

r_A[0].exact_rational(), r_A[1].exact_rational();  r_C[0].exact_rational(), r_C[1].exact_rational()

(2, 3)
(2, 6755399441055745/2251799813685248)

(2, 3)
(2, 6755399441055745/2251799813685248)

vraćanje u početni položaj pomoću inverzne matrice:

R_alfa_inv = R_alfa^(-1)
show (R_alfa_inv)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} 0.500000000000000 & 0.866025403784439 \\ -0.866025403784439 & 0.500000000000000 \end{array}\right)$

r_D = R_alfa_inv * r_B
r_D

(2.00000000000000, 3.00000000000000)

(2.00000000000000, 3.00000000000000)

r_D == r_A

False

False

r_D == r_C

False

False

- interni je zapis koordinate točke $D$ :

r_D[0].exact_rational(), r_D[1].exact_rational()

(9007199254740991/4503599627370496, 6755399441055745/2251799813685248)

(9007199254740991/4503599627370496, 6755399441055745/2251799813685248)

7. zadatak

Definirajte matricu

$\mathbf{D} = \left[ \begin{array}{rrr} 5,\!0 & -2,\!0 & -1,\!0 \\ -2,\!0 & 5,\!0 & -2,\!0 \\ -1,\!0 & -2,\!0 & 5,\!0 \end{array} \right]$ .

Provjerite je li matrica $\mathbf{D}$ regularna i ako jest, izračunajte njezin inverz. Provjerite svojstvo $\mathbf{D}\,\mathbf{D}^{-1} = \mathbf{D}^{-1}\mathbf{D} = \mathbf{I}$ .

Definicija matrice:

D = matrix ([[5., -2., -1.], 
             [-2., 5., -2.], 
             [-1., -2., 5.]])

show (D.n (digits = 2))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 5.0 & -2.0 & -1.0 \\ -2.0 & 5.0 & -2.0 \\ -1.0 & -2.0 & 5.0 \end{array}\right)$

Provjera regularnosti:

D.is_singular()

False

False

ili (matrica je regularna ako je $\mathrm{det}(\mathbf{D}) \neq 0$ ):

D.det()

72.0000000000000

72.0000000000000

Izračunavanje inverzne matrice:

Dinv = D.inverse()
show (Dinv)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.291666666666667 & 0.166666666666667 & 0.125000000000000 \\ 0.166666666666667 & 0.333333333333333 & 0.166666666666667 \\ 0.125000000000000 & 0.166666666666667 & 0.291666666666667 \end{array}\right)$

ili:

Dinv = D^-1

„Provjera” svojstva inverzne matrice:

D * Dinv == identity_matrix (RR, 3)

False

False

Jednakost nije zadovoljena, jer „realna” aritmetika na računalu nije matematička realna aritmetika beskonačne točnosti, pa neke komponente, koje su u jediničnoj matrici nule:

show (identity_matrix (RR, 3))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 1.00000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 \end{array}\right)$

u umnošku D * Dinv nisu nule:

show (D * Dinv)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1.00000000000000 & 1.11022302462516 \times 10^{-16} & 1.11022302462516 \times 10^{-16} \\ -1.66533453693773 \times 10^{-16} & 1.00000000000000 & -1.11022302462516 \times 10^{-16} \\ -1.11022302462516 \times 10^{-16} & 0.000000000000000 & 1.00000000000000 \end{array}\right)$

Štoviše, iako su ispisane kao jedinice, ni „jedinice” nisu jedinice, kao što „interni” zapisi pokazuju:

print (D * Dinv)[0,0].exact_rational()
print (D * Dinv)[1,1].exact_rational()
print (D * Dinv)[2,2].exact_rational()

4503599627370497/4503599627370496
4503599627370495/4503599627370496
4503599627370495/4503599627370496

4503599627370497/4503599627370496
4503599627370495/4503599627370496
4503599627370495/4503599627370496

Isto tako, nije zadovoljeno ni svojstvo

D * Dinv == Dinv * D

False

False

2. zadaci za vježbu

Definirajte matrice

$\mathbf{A} = \left[ \begin{array}{rrrr} 2 & 3 & -1 & 0 \\ 4 & 1 & 0 & 2 \\ -2 & 5 & 1 & 4 \\ -1 & 5 & 2 & 4 \end{array} \right] \qquad\text{i}\qquad \mathbf{B} = \left[ \begin{array}{llll} 0 & 2 & 1 & 1,\!5 \\ 2 & 0,\!5 & 1 & 1,\!5 \\ 1 & 1,\!5 & 2 & 3 \\ 3 & 0 & 4 & 2 \end{array} \right]$ .

Izračunajte

$\mathbf{A}\,\mathbf{B}$ ,
determinantu matrice $\mathbf{A}$ ,
inverznu matricu matrice $\mathbf{A}$ i množenjem sa $\mathbf{A}$ provjerite da je rezultat jedinična matrica,
transponiranu matricu matrice $\mathbf{B}$ .

Sve matrice prikažite pomoću funkcije show().

Rješenja 1. zadataka za vježbu

Definirajte vektore $\mathbf{a}=(1;\; 3;\; 2,\!1)$ i $\mathbf{b}=(-1;\; 5,\!3;\; 0)$

a = vector ([1, 3, 2.1]); b = vector ([-1, 5.3, 0]) 
print(a) 
print(b)

(1.00000000000000, 3.00000000000000, 2.10000000000000)
(-1.00000000000000, 5.30000000000000, 0.000000000000000)

(1.00000000000000, 3.00000000000000, 2.10000000000000)
(-1.00000000000000, 5.30000000000000, 0.000000000000000)

pa zatim izračunajte:

$\mathbf{c} = 3\mathbf{a} - 4\mathbf{b}$

c = 3*a - 4*b                                     
c

(7.00000000000000, -12.2000000000000, 6.30000000000000)

(7.00000000000000, -12.2000000000000, 6.30000000000000)

duljine (ili: euklidske norme) vektora $\mathbf{a}$ i $\mathbf{b}$

print (a.norm())                                 
print (b.norm())

3.79605057922046
5.39351462406472

3.79605057922046
5.39351462406472

skalarni umnožak vektora $\mathbf{a}$ i $\mathbf{b}$

a*b

14.9000000000000

14.9000000000000

a.dot_product (b)

14.9000000000000

14.9000000000000

vektorski umnožak vektora $\mathbf{a}$ i $\mathbf{b}$

a.cross_product (b)

(-11.1300000000000, -2.10000000000000, 8.30000000000000)

(-11.1300000000000, -2.10000000000000, 8.30000000000000)

kut između vektora $\mathbf{a}$ i $\mathbf{c}$

coskuta = (a*c)/(norm (a) * norm (c))

kut_radijan = arccos (coskuta)
kut_radijan

1.85438864827183

1.85438864827183

# u stupnjevima:
N (kut_radijan*180/pi)

106.248643122946

106.248643122946

ploštinu paralelograma određenoga vektorima $\mathbf{a}$ i $\mathbf{b}$ (ploština je $\|\mathbf{a}\times\mathbf{b}\|$ )

norm (a.cross_product (b))

14.0419692351180

14.0419692351180

Rješenja 2. zadataka za vježbu

Definirajte matrice

A = matrix ([[2, 3, -1, 0], 
             [4, 1, 0, 2], 
             [-2, 5, 1, 4], 
             [-1, 5, 2, 4]] );
show(A)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 2 & 3 & -1 & 0 \\ 4 & 1 & 0 & 2 \\ -2 & 5 & 1 & 4 \\ -1 & 5 & 2 & 4 \end{array}\right)$

B = matrix ([[0, 2, 1, 1.5], 
             [2, 0.5, 1, 1.5], 
             [1, 1.5, 2, 3], 
             [3, 0, 4, 2]] )
show(B)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 0.000000000000000 & 2.00000000000000 & 1.00000000000000 & 1.50000000000000 \\ 2.00000000000000 & 0.500000000000000 & 1.00000000000000 & 1.50000000000000 \\ 1.00000000000000 & 1.50000000000000 & 2.00000000000000 & 3.00000000000000 \\ 3.00000000000000 & 0.000000000000000 & 4.00000000000000 & 2.00000000000000 \end{array}\right)$

Izračunajte:

$\mathbf{A}\,\mathbf{B}$

show (A*B)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 5.00000000000000 & 4.00000000000000 & 3.00000000000000 & 4.50000000000000 \\ 8.00000000000000 & 8.50000000000000 & 13.0000000000000 & 11.5000000000000 \\ 23.0000000000000 & 0.000000000000000 & 21.0000000000000 & 15.5000000000000 \\ 24.0000000000000 & 3.50000000000000 & 24.0000000000000 & 20.0000000000000 \end{array}\right)$

determinantu matrice $\mathbf{A}$

det (A)

ili:

A.det()

inverznu matricu matrice $\mathbf{A}$ ( $\det(\mathbf{A}) \ne 0$ , pa matrica ima inverznu)

show (A.inverse())

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} \frac{1}{14} & \frac{1}{7} & -\frac{3}{14} & \frac{1}{7} \\ \frac{11}{42} & -\frac{1}{7} & -\frac{5}{42} & \frac{4}{21} \\ -\frac{1}{14} & -\frac{1}{7} & -\frac{11}{14} & \frac{6}{7} \\ -\frac{23}{84} & \frac{2}{7} & \frac{41}{84} & -\frac{8}{21} \end{array}\right)$

ili:

show (A^-1)

show (N(A^-1))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 0.0714285714285714 & 0.142857142857143 & -0.214285714285714 & 0.142857142857143 \\ 0.261904761904762 & -0.142857142857143 & -0.119047619047619 & 0.190476190476190 \\ -0.0714285714285714 & -0.142857142857143 & -0.785714285714286 & 0.857142857142857 \\ -0.273809523809524 & 0.285714285714286 & 0.488095238095238 & -0.380952380952381 \end{array}\right)$

provjera da je $\mathbf{A}\,\mathbf{A}^{\!-1} = \mathbf{I}$

show (A * A^-1)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$

A * A^-1 == identity_matrix (ZZ, 4)

True

True

A^-1 * A  ==  A * A^-1

True

True

- budući da je u izračunavanju inverzne matrice upotrijebljena racionalna (a ne „realna”) aritmetika, rezultat je točan, a ne približan, pa je i rezultat množenja točan

show (A.n (digits = 3))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 2.00 & 3.00 & -1.00 & 0.000 \\ 4.00 & 1.00 & 0.000 & 2.00 \\ -2.00 & 5.00 & 1.00 & 4.00 \\ -1.00 & 5.00 & 2.00 & 4.00 \end{array}\right)$

show (N(A)^-1)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 0.0714285714285713 & 0.142857142857143 & -0.214285714285714 & 0.142857142857143 \\ 0.261904761904762 & -0.142857142857143 & -0.119047619047619 & 0.190476190476190 \\ -0.0714285714285718 & -0.142857142857143 & -0.785714285714286 & 0.857142857142857 \\ -0.273809523809524 & 0.285714285714286 & 0.488095238095238 & -0.380952380952381 \end{array}\right)$

N(A) * N(A)^-1  ==  identity_matrix (RR, 4)

False

False

N(A) * N(A)^-1  ==  N(A)^-1 * N(A)

False

False

- za razliku od cjelobrojne, „realna” je aritmetika samo približna

transponiranu matricu matrice $\mathbf{B}$

show ( B.transpose() )

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 0.000000000000000 & 2.00000000000000 & 1.00000000000000 & 3.00000000000000 \\ 2.00000000000000 & 0.500000000000000 & 1.50000000000000 & 0.000000000000000 \\ 1.00000000000000 & 1.00000000000000 & 2.00000000000000 & 4.00000000000000 \\ 1.50000000000000 & 1.50000000000000 & 3.00000000000000 & 2.00000000000000 \end{array}\right)$

ili:

show (B.T)

MPZI_vj05_s_izradom

2444 days ago by fresl