Matrice u SAGE-u

verzija: SageMath 9.4

Važna napomena

SAGE kao i python počinje numeraciju s brojem nula u listama i matricama. Stoga, prema našoj definiciji element na poziciji $(1,3)$ je u SAGE-u zapravo na poziciji $(0,2)$. Općenito, element na poziciji $(i,j)$ je u SAGE-u na poziciji $(i-1,j-1)$. To treba imati na umu kada radimo s matricama u SAGE-u, što već možete vidjeti u sljedećem zadatku.

1. zadatak

Napišite matricu $A=\big[a_{ij}\big]$ tipa $(3,4)$ ako je $a_{ij}=\begin{cases}\cos{\frac{i\pi}{2}},&\text{ za }i>j\\ \log_2{(i+j)},&\text{ za }i\leq j\end{cases}.$

Rješenje

1. način. Možemo najprije definirati python listu, a zatim iz nje dobiti matricu pomoću naredbe matrix. Python lista nije matrica i ne podržava metode na matricama (poput računanja determinanti, binarne operacije s matricama i slično). Stoga je uvijek važno upotrijebiti ključnu riječ matrix kako bismo zaista u SAGE-u kreirali matricu koja će podržavati sve standardne operacije s matricama.

a je python lista

In [1]:
a=[[cos((i*pi)/2) if i>j else log(i+j,2) for j in range(1,5)] for i in range(1,4)]
In [2]:
a
Out[2]:
[[1, log(3)/log(2), 2, log(5)/log(2)],
 [-1, 2, log(5)/log(2), log(6)/log(2)],
 [0, 0, log(6)/log(2), log(7)/log(2)]]

A je matrica

In [3]:
A=matrix(a)
A
Out[3]:
[            1 log(3)/log(2)             2 log(5)/log(2)]
[           -1             2 log(5)/log(2) log(6)/log(2)]
[            0             0 log(6)/log(2) log(7)/log(2)]
In [4]:
show(A)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1 & \frac{\log\left(3\right)}{\log\left(2\right)} & 2 & \frac{\log\left(5\right)}{\log\left(2\right)} \\ -1 & 2 & \frac{\log\left(5\right)}{\log\left(2\right)} & \frac{\log\left(6\right)}{\log\left(2\right)} \\ 0 & 0 & \frac{\log\left(6\right)}{\log\left(2\right)} & \frac{\log\left(7\right)}{\log\left(2\right)} \end{array}\right)\]

A je matrica nad simboličkim prstenom (zbog logaritamske funkcije)

In [5]:
A.parent()
Out[5]:
Full MatrixSpace of 3 by 4 dense matrices over Symbolic Ring
In [6]:
A.base_ring()
Out[6]:
Symbolic Ring

element na poziciji $(1,1)$

In [7]:
a[0][0]
Out[7]:
1
In [8]:
A[0,0]
Out[8]:
1

element na poziciji $(2,3)$

In [9]:
a[1][2]
Out[9]:
log(5)/log(2)
In [10]:
A[1,2]
Out[10]:
log(5)/log(2)

dimenzije matrice $A$

In [11]:
A.nrows(),A.ncols()
Out[11]:
(3, 4)

2. način. Možemo odmah kreirati matricu bez da prethodno definiramo python listu. Imajte ovdje na umu spomenutu važnu napomenu. Na prvi pogled malo drukčije izgleda definicija matrice A1, ali to je zbog toga što naša pozicija $(i,j)$ odgovara u SAGE-u poziciji $(i-1,j-1)$. Kada smo gore matricu definirali preko python liste, tu činjenicu smo imali na umu u naredbi range.

In [12]:
A1=matrix(3,4,lambda i,j: cos((i+1)*pi/2) if i>j else log(i+j+2,2))
show(A1)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1 & \frac{\log\left(3\right)}{\log\left(2\right)} & 2 & \frac{\log\left(5\right)}{\log\left(2\right)} \\ -1 & 2 & \frac{\log\left(5\right)}{\log\left(2\right)} & \frac{\log\left(6\right)}{\log\left(2\right)} \\ 0 & 0 & \frac{\log\left(6\right)}{\log\left(2\right)} & \frac{\log\left(7\right)}{\log\left(2\right)} \end{array}\right)\]

želite li saznati više detalja o naredbi matrix

2. zadatak

Zadane su matrice

$$A=\begin{bmatrix}1&2\\ 0&-3\\ 5&4 \end{bmatrix} \quad\text{i}\quad  B=\begin{bmatrix}1&0&-2&5\\ 8&4&-1&3\end{bmatrix}.$$

Odredite $A^T$, $AB$ i $BA$.

Rješenje

In [13]:
A=matrix([[1,2],[0,-3],[5,4]])
B=matrix([[1,0,-2,5],[8,4,-1,3]])

$A^T$

In [14]:
A.transpose()
Out[14]:
[ 1  0  5]
[ 2 -3  4]

$AB$

In [15]:
A*B
Out[15]:
[ 17   8  -4  11]
[-24 -12   3  -9]
[ 37  16 -14  37]

SAGE nas upozorava da $BA$ nije definirano

In [16]:
B*A

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
/tmp/ipykernel_15441/3451148411.py in <module>
----> 1 B*A

/usr/lib/python3.9/site-packages/sage/structure/element.pyx in sage.structure.element.Matrix.__mul__ (build/cythonized/sage/structure/element.c:23966)()
   3813 
   3814         if BOTH_ARE_ELEMENT(cl):
-> 3815             return coercion_model.bin_op(left, right, mul)
   3816 
   3817         cdef long value

/usr/lib/python3.9/site-packages/sage/structure/coerce.pyx in sage.structure.coerce.CoercionModel.bin_op (build/cythonized/sage/structure/coerce.c:11723)()
   1246         # We should really include the underlying error.
   1247         # This causes so much headache.
-> 1248         raise bin_op_exception(op, x, y)
   1249 
   1250     cpdef canonical_coercion(self, x, y):

TypeError: unsupported operand parent(s) for *: 'Full MatrixSpace of 2 by 4 dense matrices over Integer Ring' and 'Full MatrixSpace of 3 by 2 dense matrices over Integer Ring'

3. zadatak

Odredite matricu $3AB-7BA$ ako je

$$A=\begin{bmatrix}3&1&-4\\ -4&6&-2\\ 5&8&5\end{bmatrix} \quad\text{i}\quad B=\begin{bmatrix}3&7&-4\\ 2&1&0\\ -5&3&2\end{bmatrix}.$$

Rješenje

In [17]:
A=matrix([[3,1,-4],[-4,6,-2],[5,8,5]])
B=matrix([[3,7,-4],[2,1,0],[-5,3,2]])

$AB$

In [18]:
A*B
Out[18]:
[ 31  10 -20]
[ 10 -28  12]
[  6  58 -10]

$3AB$

In [19]:
3*A*B
Out[19]:
[ 93  30 -60]
[ 30 -84  36]
[ 18 174 -30]

$BA$

In [20]:
B*A
Out[20]:
[-39  13 -46]
[  2   8 -10]
[-17  29  24]

$7BA$

In [21]:
7*B*A
Out[21]:
[-273   91 -322]
[  14   56  -70]
[-119  203  168]

$3AB-7BA$

In [22]:
3*A*B-7*B*A
Out[22]:
[ 366  -61  262]
[  16 -140  106]
[ 137  -29 -198]

4. zadatak

Zadana je matrica $A=\begin{bmatrix}3&-2\\ -1&5\end{bmatrix}$ i polinom $f(x)=x^3+2x^2+3$. Odredite $f(A)$.

Rješenje

In [23]:
A=matrix([[3,-2],[-1,5]])

$A$ je matrica nad prstenom $\mathbb{Z}$

In [24]:
A.base_ring()
Out[24]:
Integer Ring

jedinična matrica reda 2

In [25]:
identity_matrix(2)
Out[25]:
[1 0]
[0 1]

jedinična matrica reda 5

In [26]:
identity_matrix(5)
Out[26]:
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]

$A^2$

In [27]:
A*A
Out[27]:
[ 11 -16]
[ -8  27]
In [28]:
A^2
Out[28]:
[ 11 -16]
[ -8  27]

$A^3$

In [29]:
A^3
Out[29]:
[  49 -102]
[ -51  151]

$f(A)=A^3+2A^2+3I$

In [30]:
A^3+2*A^2+3*identity_matrix(2)
Out[30]:
[  74 -134]
[ -67  208]

5. zadatak

Izračunajte $\begin{bmatrix}2&-2&0\\ 1&-2&1\end{bmatrix}\begin{bmatrix}2\\ 1\\ 3\end{bmatrix}$.

Rješenje

In [31]:
matrix([[2,-2,0],[1,-2,1]])*matrix([[2],[1],[3]])
Out[31]:
[2]
[3]

6. zadatak

Izračunajte $AB$ ako je $A=\begin{bmatrix}2&3\\ 1&5\end{bmatrix}$,  $B=\dfrac{9}{5}\begin{bmatrix}1&0&4\\ 3&6&8\end{bmatrix}$.

Rješenje

In [32]:
A=matrix([[2,3],[1,5]])
B=9/5*matrix([[1,0,4],[3,6,8]])
In [33]:
A*B
Out[33]:
[ 99/5 162/5 288/5]
[144/5    54 396/5]
In [34]:
show(A*B)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} \frac{99}{5} & \frac{162}{5} & \frac{288}{5} \\ \frac{144}{5} & 54 & \frac{396}{5} \end{array}\right)\]

7. zadatak

Odredite $A^n$ za proizvoljni $n\in\mathbb{N}$ ako je $A=\begin{bmatrix}1&3\\ 0&1\end{bmatrix}$.

Rješenje

In [35]:
A=matrix([[1,3],[0,1]])
In [36]:
for n in range(1,21):
    print("A^%d="%n)
    print(A^n)
    print("\n")

A^1=
[1 3]
[0 1]


A^2=
[1 6]
[0 1]


A^3=
[1 9]
[0 1]


A^4=
[ 1 12]
[ 0  1]


A^5=
[ 1 15]
[ 0  1]


A^6=
[ 1 18]
[ 0  1]


A^7=
[ 1 21]
[ 0  1]


A^8=
[ 1 24]
[ 0  1]


A^9=
[ 1 27]
[ 0  1]


A^10=
[ 1 30]
[ 0  1]


A^11=
[ 1 33]
[ 0  1]


A^12=
[ 1 36]
[ 0  1]


A^13=
[ 1 39]
[ 0  1]


A^14=
[ 1 42]
[ 0  1]


A^15=
[ 1 45]
[ 0  1]


A^16=
[ 1 48]
[ 0  1]


A^17=
[ 1 51]
[ 0  1]


A^18=
[ 1 54]
[ 0  1]


A^19=
[ 1 57]
[ 0  1]


A^20=
[ 1 60]
[ 0  1]

Gledajući prvih 20 potencija matrice $A$ razumno je da donesemo hipotezu $A^n=\left[\begin{smallmatrix}1&3n\\ 0&1\end{smallmatrix}\right]$. Naravno, mi tu hipotezu moramo dokazati matematičkom indukcijom. Ovdje smo zapravo proveli nepotpunu indukciju koju matematika ne priznaje kao dokaz.

8. zadatak

Odredite sve matrice koje komutiraju s matricom $A$ s obzirom na množenje ako je $A=\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$.

Rješenje

In [37]:
var('a b c d')
A=matrix([[1,1],[0,1]])
X=matrix([[a,b],[c,d]])
In [38]:
A*X-X*A
Out[38]:
[     c -a + d]
[     0     -c]
In [39]:
solve([c==0,-a+d==0,-c==0],a,b,c,d)
Out[39]:
[[a == r1, b == r2, c == 0, d == r1]]

Možemo definirati svoju funkciju koja će direktno raditi s matricama

In [40]:
def solve_mat(lijevo,desno,varijable):
    jednadzbe=[]
    for i in range(lijevo.nrows()):
        for j in range(lijevo.ncols()):
            jednadzbe.append(lijevo[i,j]==desno[i,j])
    rj=solve(jednadzbe,*varijable)
    return rj
In [41]:
solve_mat(A*X,X*A,(a,b,c,d))
Out[41]:
[[a == r3, b == r4, c == 0, d == r3]]

9. zadatak

Izračunajte determinante:

$$\begin{vmatrix}2&5\\ 1&-3\end{vmatrix},\quad \begin{vmatrix}x-a&-a\\ a&x+a\end{vmatrix},\quad \begin{vmatrix}\sin{\alpha}&\cos{\alpha}\\ \sin{\beta}&\cos{\beta}\end{vmatrix}.$$

Rješenje

In [42]:
det(matrix([[2,5],[1,-3]]))
Out[42]:
-11
In [43]:
matrix([[2,5],[1,-3]]).det()
Out[43]:
-11
In [44]:
expand(det(matrix([[x-a,-a],[a,x+a]])))
Out[44]:
x^2
In [45]:
matrix([[x-a,-a],[a,x+a]]).det().expand()
Out[45]:
x^2
In [46]:
det(matrix([[sin(a),cos(a)],[sin(b),cos(b)]]))
Out[46]:
cos(b)*sin(a) - cos(a)*sin(b)

10. zadatak

Izračunajte determinantu matrice $A=\begin{bmatrix}9&4&-5\\ 8&7&-2\\ 2&-1&8\end{bmatrix}$.

Rješenje

In [47]:
A=matrix(3,3,[9,4,-5,8,7,-2,2,-1,8])
In [48]:
A
Out[48]:
[ 9  4 -5]
[ 8  7 -2]
[ 2 -1  8]
In [49]:
det(A)
Out[49]:
324
In [50]:
A.det()
Out[50]:
324

11. zadatak

Izračunajte determinantu matrice $A=\begin{bmatrix}2&-5&1&2\\ -3&7&-1&4\\ 5&-9&2&7\\ 4&-6&1&2\end{bmatrix}$.

Rješenje

In [51]:
A=matrix(4,4,[2,-5,1,2,-3,7,-1,4,5,-9,2,7,4,-6,1,2])
In [52]:
A
Out[52]:
[ 2 -5  1  2]
[-3  7 -1  4]
[ 5 -9  2  7]
[ 4 -6  1  2]
In [53]:
det(A)
Out[53]:
-9
In [54]:
A.det()
Out[54]:
-9

12. zadatak

Zadana je matrica $A=\begin{bmatrix}4+x&2&2\\ 7&x-1&2\\ x+1&5&5\end{bmatrix}$.

  1. Odredite sve $x\in\mathbb{N}$ za koje je $\det{A}=0$.
  2. Za $x=-1$ izračunajte $\det{\big(A^T\big)}+5\det{\big(A^3\big)}-2\det{\big(\frac{1}{2}A\big)}$.

Rješenje

In [55]:
A=matrix(3,3,[4+x,2,2,7,x-1,2,x+1,5,5])

a) dio

In [56]:
expand(det(A))
Out[56]:
3*x^2 + 9*x - 54
In [57]:
solve(det(A)==0,x)
Out[57]:
[x == -6, x == 3]

b) dio

In [58]:
det(A.transpose())+5*det(A^3)-2*det(1/2*A)
Out[58]:
5*((2*(x + 4)*(x + 1) + 34*x + 309)*(((x - 1)^2 + 24)*(x - 1) + 42*x + 156) - 2*(2*(x + 4)*(x + 1) + (7*x + 22)*(x - 1) + 20*x + 265)*((x - 1)^2 + 21*x + 102))*(((x + 4)^2 + 2*x + 16)*(x + 4) + 2*(x + 11)*(x + 1) + 28*x + 112) - 10*(((x + 4)^2 + 11*x + 87)*(((x - 1)^2 + 24)*(x - 1) + 42*x + 156) - 2*((x + 4)^2 + 2*(x + 4)*(x - 1) + 7*x + 71)*((x - 1)^2 + 21*x + 102))*(((x + 4)*(x + 1) + 5*x + 40)*(x + 4) + (2*x + 37)*(x + 1) + 49*x + 154) + 10*(((x + 4)^2 + 11*x + 87)*(2*(x + 4)*(x + 1) + (7*x + 22)*(x - 1) + 20*x + 265) - ((x + 4)^2 + 2*(x + 4)*(x - 1) + 7*x + 71)*(2*(x + 4)*(x + 1) + 34*x + 309))*((16*x + 23)*(x + 4) + 2*(x + 11)*(x + 1) + 7*(x - 1)^2 + 168) - 2*(x + 1)*(x - 1) + 15/4*(x + 4)*(x - 3) + 1/2*(x + 1)*(x - 3) + 4*x + 4
In [59]:
expand(det(A.transpose())+5*det(A^3)-2*det(1/2*A)).show()
\[\newcommand{\Bold}[1]{\mathbf{#1}}135 \, x^{6} + 1215 \, x^{5} - 3645 \, x^{4} - 40095 \, x^{3} + \frac{262449}{4} \, x^{2} + \frac{1574667}{4} \, x - \frac{1574721}{2}\]
In [60]:
expand(det(A.transpose())+5*det(A^3)-2*det(1/2*A)).subs(x==-1)
Out[60]:
-1080045

ili da najprije $x=-1$ uvrstimo u matricu $A$

In [61]:
A.subs(x==-1)
Out[61]:
[ 3  2  2]
[ 7 -2  2]
[ 0  5  5]
In [62]:
det(A.subs(x==-1))
Out[62]:
-60
In [63]:
expand(det(A.subs(x==-1).transpose())+5*det(A.subs(x==-1)^3)-2*det(1/2*A.subs(x==-1)))
Out[63]:
-1080045

13. zadatak

Izračunajte determinantu $\begin{vmatrix}3&a&3&a\\ 6&3&6&3\\ 7&7&6&6\\ a&5&a&5\end{vmatrix}$.

Rješenje

In [64]:
det(matrix(4,4,[3,a,3,a,6,3,6,3,7,7,6,6,a,5,a,5]))
Out[64]:
0
In [65]:
matrix(4,4,[3,a,3,a,6,3,6,3,7,7,6,6,a,5,a,5]).det()
Out[65]:
0

14. zadatak

Izračunajte determinantu $\begin{vmatrix}a&a&a&a&a\\ -2&a&a&a&a\\ 2&0&a&a&a\\ -4&0&0&a&a\\ 4&0&0&0&a\end{vmatrix}$.

Rješenje

In [66]:
det(matrix(5,5,[a,a,a,a,a,-2,a,a,a,a,2,0,a,a,a,-4,0,0,a,a,4,0,0,0,a]))
Out[66]:
a^5 + 2*a^4
In [67]:
det(matrix(5,5,[a,a,a,a,a,-2,a,a,a,a,2,0,a,a,a,-4,0,0,a,a,4,0,0,0,a])).show()
\[\newcommand{\Bold}[1]{\mathbf{#1}}a^{5} + 2 \, a^{4}\]
In [68]:
matrix(5,5,[a,a,a,a,a,-2,a,a,a,a,2,0,a,a,a,-4,0,0,a,a,4,0,0,0,a]).det().show()
\[\newcommand{\Bold}[1]{\mathbf{#1}}a^{5} + 2 \, a^{4}\]

15. zadatak

Riješite jednadžbu $\begin{vmatrix}x-3&3&3&3\\ 3&2x+3&3&3\\ 3&3&x-3&3\\ 3&3&3&2x+3\end{vmatrix}=0.$

Rješenje

In [69]:
det(matrix(4,4,[x-3,3,3,3,3,2*x+3,3,3,3,3,x-3,3,3,3,3,2*x+3])).show()
\[\newcommand{\Bold}[1]{\mathbf{#1}}4 \, x^{4} - 12 \, x^{3} - 144 \, x^{2} + 432 \, x\]
In [70]:
det(matrix(4,4,[x-3,3,3,3,3,2*x+3,3,3,3,3,x-3,3,3,3,3,2*x+3])).factor().show()
\[\newcommand{\Bold}[1]{\mathbf{#1}}4 \, {\left(x + 6\right)} {\left(x - 3\right)} {\left(x - 6\right)} x\]
In [71]:
solve(det(matrix(4,4,[x-3,3,3,3,3,2*x+3,3,3,3,3,x-3,3,3,3,3,2*x+3]))==0,x)
Out[71]:
[x == 3, x == -6, x == 6, x == 0]
In [72]:
show(solve(det(matrix(4,4,[x-3,3,3,3,3,2*x+3,3,3,3,3,x-3,3,3,3,3,2*x+3]))==0,x))
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left[x = 3, x = \left(-6\right), x = 6, x = 0\right]\]

16. zadatak

Odredite inverzne matrice matrica $A=\begin{bmatrix}2&1\\ -5&4\end{bmatrix}$  i  $B=\dfrac{2}{3}\begin{bmatrix}3&5\\ 1&-3\end{bmatrix}$.

Rješenje

In [73]:
A=matrix([[2,1],[-5,4]])
B=2/3*matrix([[3,5],[1,-3]])
In [74]:
A^-1
Out[74]:
[ 4/13 -1/13]
[ 5/13  2/13]
In [75]:
show(A^-1)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} \frac{4}{13} & -\frac{1}{13} \\ \frac{5}{13} & \frac{2}{13} \end{array}\right)\]
In [76]:
A.inverse()
Out[76]:
[ 4/13 -1/13]
[ 5/13  2/13]
In [77]:
~A
Out[77]:
[ 4/13 -1/13]
[ 5/13  2/13]
In [78]:
B^-1
Out[78]:
[ 9/28 15/28]
[ 3/28 -9/28]
In [79]:
show(B^-1)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} \frac{9}{28} & \frac{15}{28} \\ \frac{3}{28} & -\frac{9}{28} \end{array}\right)\]
In [80]:
B.inverse()
Out[80]:
[ 9/28 15/28]
[ 3/28 -9/28]
In [81]:
~B
Out[81]:
[ 9/28 15/28]
[ 3/28 -9/28]

17. zadatak

Odredite inverznu matricu matrice $E=\begin{bmatrix}-3&4&-5\\ 4&-3&2\\ 1&-3&4\end{bmatrix}$.

Rješenje

In [82]:
E=matrix(3,3,[-3,4,-5,4,-3,2,1,-3,4])
In [83]:
E^-1
Out[83]:
[-6/7 -1/7   -1]
[  -2   -1   -2]
[-9/7 -5/7   -1]
In [84]:
show(E^-1)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} -\frac{6}{7} & -\frac{1}{7} & -1 \\ -2 & -1 & -2 \\ -\frac{9}{7} & -\frac{5}{7} & -1 \end{array}\right)\]
In [85]:
E.inverse()
Out[85]:
[-6/7 -1/7   -1]
[  -2   -1   -2]
[-9/7 -5/7   -1]
In [86]:
~E
Out[86]:
[-6/7 -1/7   -1]
[  -2   -1   -2]
[-9/7 -5/7   -1]

adjunkta matrice $E$

In [87]:
E.adjugate()
Out[87]:
[ -6  -1  -7]
[-14  -7 -14]
[ -9  -5  -7]

18. zadatak

Za koje vrijednosti parametara $a,b\in\mathbb{R}$ je matrica $AB$ regularna ako je

$$A=\begin{bmatrix}a&1&-1\\ -1&1&1\\ -2&0&0\end{bmatrix},\quad B=\begin{bmatrix}1&0&0\\ 0&a^2-1&1\\ b&0&1\end{bmatrix}.$$

Rješenje

In [88]:
A=matrix(3,3,[a,1,-1,-1,1,1,-2,0,0])
B=matrix(3,3,[1,0,0,0,a^2-1,1,b,0,1])
In [89]:
det(A*B)
Out[89]:
-4*a^2 + 4
In [90]:
solve(det(A*B)==0,a)
Out[90]:
[a == -1, a == 1]

Dakle, $\det{(AB)}=0$ jedino za $a=1$ i $a=-1$. Stoga je $AB$ regularna matrica za svaki $b\in\mathbb{R}$ i za sve $a\in\mathbb{R}\setminus\{-1,1\}$.

19. zadatak

Odredite matricu $X$ tako da vrijedi jednakost $AX=B$ ako je $A=\begin{bmatrix}1&2\\ -2&0\end{bmatrix}$,  $B=\begin{bmatrix}2&0&-1\\ 0&-2&1\end{bmatrix}$.

Rješenje

In [91]:
A=matrix(2,2,[1,2,-2,0])
B=matrix(2,3,[2,0,-1,0,-2,1])

1. način

In [92]:
A^-1*B
Out[92]:
[   0    1 -1/2]
[   1 -1/2 -1/4]
In [93]:
show(A^-1*B)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & 1 & -\frac{1}{2} \\ 1 & -\frac{1}{2} & -\frac{1}{4} \end{array}\right)\]

2. način

In [94]:
A\B
Out[94]:
[   0    1 -1/2]
[   1 -1/2 -1/4]
In [95]:
show(A\B)
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & 1 & -\frac{1}{2} \\ 1 & -\frac{1}{2} & -\frac{1}{4} \end{array}\right)\]

3. način

In [96]:
A.solve_right(B)
Out[96]:
[   0    1 -1/2]
[   1 -1/2 -1/4]
In [97]:
show(A.solve_right(B))
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & 1 & -\frac{1}{2} \\ 1 & -\frac{1}{2} & -\frac{1}{4} \end{array}\right)\]

4. način

In [98]:
var('x1,x2,x3,x4,x5,x6')
X=matrix(2,3,[x1,x2,x3,x4,x5,x6])
In [99]:
solve_mat(A*X,B,(x1,x2,x3,x4,x5,x6))
Out[99]:
[[x1 == 0, x2 == 1, x3 == (-1/2), x4 == 1, x5 == (-1/2), x6 == (-1/4)]]

20. zadatak

Riješite matričnu jednadžbu $XA=B$ ako je $A=\begin{bmatrix}1&2\\ -2&0\end{bmatrix}$,  $B=\begin{bmatrix}2&0\\ -1&0\\ -2&1\end{bmatrix}$.

Rješenje

In [100]:
A=matrix(2,2,[1,2,-2,0])
B=matrix(3,2,[2,0,-1,0,-2,1])

1. način

In [101]:
B*A^-1
Out[101]:
[  0  -1]
[  0 1/2]
[1/2 5/4]

2. način

In [102]:
A.solve_left(B)
Out[102]:
[  0  -1]
[  0 1/2]
[1/2 5/4]

3. način

In [103]:
X=matrix(3,2,[x1,x2,x3,x4,x5,x6])
In [104]:
solve_mat(X*A,B,(x1,x2,x3,x4,x5,x6))
Out[104]:
[[x1 == 0, x2 == -1, x3 == 0, x4 == (1/2), x5 == (1/2), x6 == (5/4)]]

21. zadatak

Riješite matričnu jednadžbu  $AX=B$ ako je  $A=\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$,  $B=\begin{bmatrix}1&2\\ 3&0\end{bmatrix}$.

Rješenje

In [105]:
A=matrix(2,2,[1,1,1,1])
B=matrix(2,2,[1,2,3,0])
In [106]:
det(A)
Out[106]:
0
In [107]:
A\B

---------------------------------------------------------------------------
NotFullRankError                          Traceback (most recent call last)
/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix.solve_right (build/cythonized/sage/matrix/matrix2.c:8537)()
    853             try:
--> 854                 X = self._solve_right_nonsingular_square(C, check_rank=True)
    855             except NotFullRankError:

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix._solve_right_nonsingular_square (build/cythonized/sage/matrix/matrix2.c:8906)()
    899         if check_rank and self.rank() < self.nrows():
--> 900             raise NotFullRankError
    901         D = self.augment(B)

NotFullRankError: 

During handling of the above exception, another exception occurred:

ValueError                                Traceback (most recent call last)
/tmp/ipykernel_15441/1346290215.py in <module>
----> 1 A * BackslashOperator() * B

/usr/lib/python3.9/site-packages/sage/misc/misc.py in __mul__(self, right)
    831             (0.0, 0.5, 1.0, 1.5, 2.0)
    832         """
--> 833         return self.left._backslash_(right)
    834 
    835 

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix._backslash_ (build/cythonized/sage/matrix/matrix2.c:5667)()
    151             True
    152         """
--> 153         return self.solve_right(B)
    154 
    155     def subs(self, *args, **kwds):

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix.solve_right (build/cythonized/sage/matrix/matrix2.c:8604)()
    854                 X = self._solve_right_nonsingular_square(C, check_rank=True)
    855             except NotFullRankError:
--> 856                 X = self._solve_right_general(C, check=check)
    857 
    858         if b_is_vec:

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix._solve_right_general (build/cythonized/sage/matrix/matrix2.c:9825)()
    972             # Have to check that we actually solved the equation.
    973             if self*X != B:
--> 974                 raise ValueError("matrix equation has no solutions")
    975         return X
    976 

ValueError: matrix equation has no solutions
In [108]:
A.solve_right(B)

---------------------------------------------------------------------------
NotFullRankError                          Traceback (most recent call last)
/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix.solve_right (build/cythonized/sage/matrix/matrix2.c:8537)()
    853             try:
--> 854                 X = self._solve_right_nonsingular_square(C, check_rank=True)
    855             except NotFullRankError:

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix._solve_right_nonsingular_square (build/cythonized/sage/matrix/matrix2.c:8906)()
    899         if check_rank and self.rank() < self.nrows():
--> 900             raise NotFullRankError
    901         D = self.augment(B)

NotFullRankError: 

During handling of the above exception, another exception occurred:

ValueError                                Traceback (most recent call last)
/tmp/ipykernel_15441/790423984.py in <module>
----> 1 A.solve_right(B)

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix.solve_right (build/cythonized/sage/matrix/matrix2.c:8604)()
    854                 X = self._solve_right_nonsingular_square(C, check_rank=True)
    855             except NotFullRankError:
--> 856                 X = self._solve_right_general(C, check=check)
    857 
    858         if b_is_vec:

/usr/lib/python3.9/site-packages/sage/matrix/matrix2.pyx in sage.matrix.matrix2.Matrix._solve_right_general (build/cythonized/sage/matrix/matrix2.c:9825)()
    972             # Have to check that we actually solved the equation.
    973             if self*X != B:
--> 974                 raise ValueError("matrix equation has no solutions")
    975         return X
    976 

ValueError: matrix equation has no solutions
In [109]:
X=matrix(2,2,[x1,x2,x3,x4])
In [110]:
solve_mat(A*X,B,(x1,x2,x3,x4))
Out[110]:
[]

Zaključujemo da zadana matrična jednadžba nema rješenja.

22. zadatak

Riješite matričnu jednadžbu  $AX=B$ ako je  $A=\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$,  $B=\begin{bmatrix}2&5\\ 2&5\end{bmatrix}$.

Rješenje

In [111]:
A=matrix(2,2,[1,1,1,1])
B=matrix(2,2,[2,5,2,5])
In [112]:
det(A)
Out[112]:
0
In [113]:
A\B
Out[113]:
[2 5]
[0 0]
In [114]:
A.solve_right(B)
Out[114]:
[2 5]
[0 0]

želimo li dobiti sva rješenja

In [115]:
X=matrix(2,2,[x1,x2,x3,x4])
solve_mat(A*X,B,(x1,x2,x3,x4))
Out[115]:
[[x1 == -r6 + 2, x2 == -r5 + 5, x3 == r6, x4 == r5]]

ako želimo da varijable x1 i x2 budu parametri

In [116]:
solve_mat(A*X,B,(x3,x4,x1,x2))
Out[116]:
[[x3 == -r8 + 2, x4 == -r7 + 5, x1 == r8, x2 == r7]]

23. zadatak

Odredite matricu $X$ tako da vrijedi  $XB+A=AXB$ ako je  $A=\dfrac{1}{3}\begin{bmatrix}4&-5\\ 2&1\end{bmatrix}$  i  $A+B=I$.

Rješenje

In [117]:
A=1/3*matrix(2,2,[4,-5,2,1])
B=identity_matrix(2)-A
X=matrix(2,2,[a,b,c,d])
In [118]:
solve_mat(X*B+A,A*X*B,(a,b,c,d))
Out[118]:
[[a == (51/32), b == (-75/64), c == (15/32), d == (57/64)]]
In [ ]: