Tutorial 3 (AY24/25 Sem 1)¶

In [1]:

import sympy as sym
from ma1522 import Matrix

import sympy as sym from ma1522 import Matrix

Question 1¶

Let $\mathbf{A}$ be the $4 \times 4$ matrix obtained from $\mathbf{I}$ by the following sequence of elementary row operations:

$$\mathbf{I} \xrightarrow{\frac{1}{2}R_2} \xrightarrow{R_1 - R_2} \xrightarrow{R_2 \leftrightarrow R_4} \xrightarrow{R_3 + 3R_1} \mathbf{A}$$

Write $\mathbf{A}^{-1}$ as a product of four elementary matrices.

In [2]:

E_1 = Matrix.eye(4).scale_row(1, sym.Rational(1, 2))
E_2 = Matrix.eye(4).reduce_row(0, 1, 1)
E_3 = Matrix.eye(4).swap_row(1, 3)
E_4 = Matrix.eye(4).reduce_row(2, -3, 0)

E_1 = Matrix.eye(4).scale_row(1, sym.Rational(1, 2)) E_2 = Matrix.eye(4).reduce_row(0, 1, 1) E_3 = Matrix.eye(4).swap_row(1, 3) E_4 = Matrix.eye(4).reduce_row(2, -3, 0)

$\displaystyle \left(\frac{1}{2}\right) R_2 \rightarrow R_2$

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 0 & 0\\0 & \frac{1}{2} & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]$

$\displaystyle R_1 - \left(1\right)R_2 \rightarrow R_1$

$\displaystyle \left[\begin{array}{cccc}1 & -1 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]$

$\displaystyle R_2 \leftrightarrow R_4$

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 0 & 0\\0 & 0 & 0 & 1\\0 & 0 & 1 & 0\\0 & 1 & 0 & 0\end{array}\right]$

$\displaystyle R_3 - \left(-3\right)R_1 \rightarrow R_3$

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\3 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]$

In [3]:

# The matrix A^-1 is the product of these elementary matrices
E_1**-1, E_2**-1, E_3**-1, E_4**-1

# The matrix A^-1 is the product of these elementary matrices E_1**-1, E_2**-1, E_3**-1, E_4**-1

Out[3]:

$\displaystyle \left( \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 2 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right], \ \left[\begin{matrix}1 & 1 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right], \ \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 0 & 0 & 1\\0 & 0 & 1 & 0\\0 & 1 & 0 & 0\end{matrix}\right], \ \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\-3 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]\right)$

Question 2¶

Find an LU factorization for the matrices $\mathbf{A}$, and solve the equation $\mathbf{A}x = b$.

(a)¶

$\mathbf{A} = \begin{pmatrix} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ 0 \\ 4 \end{pmatrix}$.

In [4]:

A = Matrix.from_str("2 -1 2; -6 0 -2; 8 -1 5")
b = Matrix.from_str("1; 0; 4")

aug_mat = A.row_join(b)
aug_mat

A = Matrix.from_str("2 -1 2; -6 0 -2; 8 -1 5") b = Matrix.from_str("1; 0; 4") aug_mat = A.row_join(b) aug_mat

Out[4]:

$\displaystyle \left[\begin{array}{ccc|c}2 & -1 & 2 & 1\\-6 & 0 & -2 & 0\\8 & -1 & 5 & 4\end{array}\right]$

In [5]:

plu = A.ref(verbosity=2)
plu # The LU factorization of A is given by the matrices L and U
# You may ignore P as it is the identity matrix, which suggest that it is LU factorisable.

plu = A.ref(verbosity=2) plu # The LU factorization of A is given by the matrices L and U # You may ignore P as it is the identity matrix, which suggest that it is LU factorisable.

$\displaystyle R_2 - \left(-3\right)R_1 \rightarrow R_2$

$\displaystyle \left[\begin{array}{ccc}2 & -1 & 2\\0 & -3 & 4\\8 & -1 & 5\end{array}\right]$

$\displaystyle R_3 - \left(4\right)R_1 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc}2 & -1 & 2\\0 & -3 & 4\\0 & 3 & -3\end{array}\right]$

$\displaystyle R_3 - \left(-1\right)R_2 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc}2 & -1 & 2\\0 & -3 & 4\\0 & 0 & 1\end{array}\right]$

Out[5]:

$\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0\\-3 & 1 & 0\\4 & -1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & -1 & 2\\0 & -3 & 4\\0 & 0 & 1\end{array}\right]$

In [6]:

y = plu.L.solve(b)
y

y = plu.L.solve(b) y

Out[6]:

$\displaystyle \left[ \left[\begin{array}{c}1\\3\\3\end{array}\right]\right]$

In [7]:

x = plu.U.solve(y)
x

x = plu.U.solve(y) x

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
Cell In[7], line 1
----> 1 x = plu.U.solve(y)
      2 x

File ~/work/linear-algebra/linear-algebra/src/ma1522/symbolic.py:1852, in Matrix.solve(self, rhs)
   1850 # Use sympy's solve function directly
   1851 x = Matrix.create_unk_matrix(r=self.cols, c=1)
-> 1852 solution = sym.solve(self @ x - rhs, x.free_symbols, dict=True)
   1854 if len(solution) == 0:
   1855     # If no solution is found (e.g., inconsistent system or empty list), raise an error
   1856     display(self.row_join(rhs, aug_line=True).rref())

File ~/work/linear-algebra/linear-algebra/.venv/lib/python3.10/site-packages/sympy/core/decorators.py:106, in call_highest_priority.<locals>.priority_decorator.<locals>.binary_op_wrapper(self, other)
    104         if f is not None:
    105             return f(self)
--> 106 return func(self, other)

File ~/work/linear-algebra/linear-algebra/.venv/lib/python3.10/site-packages/sympy/matrices/matrixbase.py:3047, in MatrixBase.__sub__(self, a)
   3045 @call_highest_priority('__rsub__')
   3046 def __sub__(self, a):
-> 3047     return self + (-a)

TypeError: bad operand type for unary -: 'list'

(b)¶

$\mathbf{A} = \begin{pmatrix} 2 & -4 & 4 & -2 \\ 6 & -9 & 7 & -3 \\ -1 & -4 & 8 & 0 \end{pmatrix}$ and $b = \begin{pmatrix} 0 \\ 0 \\ 5 \end{pmatrix}$.

In [8]:

A = Matrix.from_str("2 -4 4 -2; 6 -9 7 -3; -1 -4 8 0")
b = Matrix.from_str("0; 0; 5")

aug_mat = A.row_join(b)
aug_mat

A = Matrix.from_str("2 -4 4 -2; 6 -9 7 -3; -1 -4 8 0") b = Matrix.from_str("0; 0; 5") aug_mat = A.row_join(b) aug_mat

Out[8]:

$\displaystyle \left[\begin{array}{cccc|c}2 & -4 & 4 & -2 & 0\\6 & -9 & 7 & -3 & 0\\-1 & -4 & 8 & 0 & 5\end{array}\right]$

In [9]:

plu = A.ref(verbosity=2)
plu

plu = A.ref(verbosity=2) plu

$\displaystyle R_2 - \left(3\right)R_1 \rightarrow R_2$

$\displaystyle \left[\begin{array}{cccc}2 & -4 & 4 & -2\\0 & 3 & -5 & 3\\-1 & -4 & 8 & 0\end{array}\right]$

$\displaystyle R_3 - \left(- \frac{1}{2}\right)R_1 \rightarrow R_3$

$\displaystyle \left[\begin{array}{cccc}2 & -4 & 4 & -2\\0 & 3 & -5 & 3\\0 & -6 & 10 & -1\end{array}\right]$

$\displaystyle R_3 - \left(-2\right)R_2 \rightarrow R_3$

$\displaystyle \left[\begin{array}{cccc}2 & -4 & 4 & -2\\0 & 3 & -5 & 3\\0 & 0 & 0 & 5\end{array}\right]$

Out[9]:

$\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0\\3 & 1 & 0\\- \frac{1}{2} & -2 & 1\end{array}\right]\left[\begin{array}{cccc}2 & -4 & 4 & -2\\0 & 3 & -5 & 3\\0 & 0 & 0 & 5\end{array}\right]$

In [10]:

y = plu.L.solve(b)
y

y = plu.L.solve(b) y

Out[10]:

$\displaystyle \left[ \left[\begin{array}{c}0\\0\\5\end{array}\right]\right]$

In [11]:

x = plu.U.solve(y)
x

x = plu.U.solve(y) x

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
Cell In[11], line 1
----> 1 x = plu.U.solve(y)
      2 x

File ~/work/linear-algebra/linear-algebra/src/ma1522/symbolic.py:1852, in Matrix.solve(self, rhs)
   1850 # Use sympy's solve function directly
   1851 x = Matrix.create_unk_matrix(r=self.cols, c=1)
-> 1852 solution = sym.solve(self @ x - rhs, x.free_symbols, dict=True)
   1854 if len(solution) == 0:
   1855     # If no solution is found (e.g., inconsistent system or empty list), raise an error
   1856     display(self.row_join(rhs, aug_line=True).rref())

File ~/work/linear-algebra/linear-algebra/.venv/lib/python3.10/site-packages/sympy/core/decorators.py:106, in call_highest_priority.<locals>.priority_decorator.<locals>.binary_op_wrapper(self, other)
    104         if f is not None:
    105             return f(self)
--> 106 return func(self, other)

File ~/work/linear-algebra/linear-algebra/.venv/lib/python3.10/site-packages/sympy/matrices/matrixbase.py:3047, in MatrixBase.__sub__(self, a)
   3045 @call_highest_priority('__rsub__')
   3046 def __sub__(self, a):
-> 3047     return self + (-a)

TypeError: bad operand type for unary -: 'list'

Question 3¶

Let $\mathbf{A} = \begin{pmatrix} 2 & -6 & 6 \\ -4 & 5 & -7 \\ 3 & 5 & -1 \\ -6 & 4 & -8 \\ 8 & -3 & 9 \end{pmatrix}$.

(a)¶

Find an LU factorization of $\mathbf{A}$.

In [12]:

A = Matrix.from_str("2 -6 6; -4 5 -7; 3 5 -1; -6 4 -8; 8 -3 9")
A

A = Matrix.from_str("2 -6 6; -4 5 -7; 3 5 -1; -6 4 -8; 8 -3 9") A

Out[12]:

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\-4 & 5 & -7\\3 & 5 & -1\\-6 & 4 & -8\\8 & -3 & 9\end{array}\right]$

In [13]:

plu = A.ref(verbosity=2)
plu

plu = A.ref(verbosity=2) plu

$\displaystyle R_2 - \left(-2\right)R_1 \rightarrow R_2$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\3 & 5 & -1\\-6 & 4 & -8\\8 & -3 & 9\end{array}\right]$

$\displaystyle R_3 - \left(\frac{3}{2}\right)R_1 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 14 & -10\\-6 & 4 & -8\\8 & -3 & 9\end{array}\right]$

$\displaystyle R_4 - \left(-3\right)R_1 \rightarrow R_4$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 14 & -10\\0 & -14 & 10\\8 & -3 & 9\end{array}\right]$

$\displaystyle R_5 - \left(4\right)R_1 \rightarrow R_5$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 14 & -10\\0 & -14 & 10\\0 & 21 & -15\end{array}\right]$

$\displaystyle R_3 - \left(-2\right)R_2 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 0 & 0\\0 & -14 & 10\\0 & 21 & -15\end{array}\right]$

$\displaystyle R_4 - \left(2\right)R_2 \rightarrow R_4$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 0 & 0\\0 & 0 & 0\\0 & 21 & -15\end{array}\right]$

$\displaystyle R_5 - \left(-3\right)R_2 \rightarrow R_5$

$\displaystyle \left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]$

Out[13]:

$\left[\begin{array}{ccccc}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccccc}1 & 0 & 0 & 0 & 0\\-2 & 1 & 0 & 0 & 0\\\frac{3}{2} & -2 & 1 & 0 & 0\\-3 & 2 & 0 & 1 & 0\\4 & -3 & 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}2 & -6 & 6\\0 & -7 & 5\\0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]$

Question 4¶

Let $\mathbf{A} = \begin{pmatrix} -x & 1 & 0 \\ 0 & -x & 1 \\ 2 & -5 & 4-x \end{pmatrix}$. Compute the determinant of $\mathbf{A}$ and find all the values of $x$ such that $\mathbf{A}$ is singular.

In [14]:

A = Matrix.from_str("-x 1 0; 0 -x 1; 2 -5 4-x")
A

A = Matrix.from_str("-x 1 0; 0 -x 1; 2 -5 4-x") A

Out[14]:

$\displaystyle \left[\begin{array}{ccc}- x & 1 & 0\\0 & - x & 1\\2 & -5 & 4 - x\end{array}\right]$

In [15]:

A.det().factor()

A.det().factor()

Out[15]:

$\displaystyle - \left(x - 2\right) \left(x - 1\right)^{2}$

Question 5¶

Show that $$\begin{vmatrix} a + px & b + qx & c + rx \\ p + ux & q + vx & r + wx \\ u + ax & v + bx & w + cx \end{vmatrix} = (1 + x^3) \begin{vmatrix} a & b & c \\ p & q & r \\ u & v & w \end{vmatrix}$$

In [16]:

a, b, c, p, q, r, u, v, w, x = sym.symbols("a b c p q r u v w x")

lhs = Matrix([[a + p*x, b + q*x, c + r*x],
              [p + u*x, q + v*x, r + w*x],
              [u + a*x, v + b*x, w + c*x]])

lhs

a, b, c, p, q, r, u, v, w, x = sym.symbols("a b c p q r u v w x") lhs = Matrix([[a + p*x, b + q*x, c + r*x], [p + u*x, q + v*x, r + w*x], [u + a*x, v + b*x, w + c*x]]) lhs

Out[16]:

$\displaystyle \left[\begin{array}{ccc}a + p x & b + q x & c + r x\\p + u x & q + v x & r + w x\\a x + u & b x + v & c x + w\end{array}\right]$

In [17]:

lhs.det()

lhs.det()

Out[17]:

$\displaystyle a q w x^{3} + a q w - a r v x^{3} - a r v - b p w x^{3} - b p w + b r u x^{3} + b r u + c p v x^{3} + c p v - c q u x^{3} - c q u$

In [18]:

rhs = Matrix.from_str("a b c; p q r; u v w")
rhs

rhs = Matrix.from_str("a b c; p q r; u v w") rhs

Out[18]:

$\displaystyle \left[\begin{array}{ccc}a & b & c\\p & q & r\\u & v & w\end{array}\right]$

In [19]:

rhs.det()

rhs.det()

Out[19]:

$\displaystyle a q w - a r v - b p w + b r u + c p v - c q u$

In [20]:

sym.simplify(lhs.det() / rhs.det())

sym.simplify(lhs.det() / rhs.det())

Out[20]:

$\displaystyle \frac{a q w x^{3} + a q w - a r v x^{3} - a r v - b p w x^{3} - b p w + b r u x^{3} + b r u + c p v x^{3} + c p v - c q u x^{3} - c q u}{a q w - a r v - b p w + b r u + c p v - c q u}$

In [21]:

# Alternative way: less brute force

(lhs.copy()
 .reduce_row(1, x, 2)
 .reduce_row(0, x, 1)
 .scale_row(0, 1/(1 + x**3))
 .reduce_row(1, -x**2, 0)
 .reduce_row(2, x, 0)
 )

# Alternative way: less brute force (lhs.copy() .reduce_row(1, x, 2) .reduce_row(0, x, 1) .scale_row(0, 1/(1 + x**3)) .reduce_row(1, -x**2, 0) .reduce_row(2, x, 0) )

$\displaystyle R_2 - \left(x\right)R_3 \rightarrow R_2$

$\displaystyle \left[\begin{array}{ccc}a + p x & b + q x & c + r x\\- a x^{2} + p & - b x^{2} + q & - c x^{2} + r\\a x + u & b x + v & c x + w\end{array}\right]$

$\displaystyle R_1 - \left(x\right)R_2 \rightarrow R_1$

$\displaystyle \left[\begin{array}{ccc}a x^{3} + a & b x^{3} + b & c x^{3} + c\\- a x^{2} + p & - b x^{2} + q & - c x^{2} + r\\a x + u & b x + v & c x + w\end{array}\right]$

$\displaystyle \left(\frac{1}{x^{3} + 1}\right) R_1 \rightarrow R_1$

$\displaystyle \left[\begin{array}{ccc}a & b & c\\- a x^{2} + p & - b x^{2} + q & - c x^{2} + r\\a x + u & b x + v & c x + w\end{array}\right]$

$\displaystyle R_2 - \left(- x^{2}\right)R_1 \rightarrow R_2$

$\displaystyle \left[\begin{array}{ccc}a & b & c\\p & q & r\\a x + u & b x + v & c x + w\end{array}\right]$

$\displaystyle R_3 - \left(x\right)R_1 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc}a & b & c\\p & q & r\\u & v & w\end{array}\right]$

Out[21]:

$\displaystyle \left[\begin{array}{ccc}a & b & c\\p & q & r\\u & v & w\end{array}\right]$

In [22]:

# Therefore, det has only been scaled by a factor of 1 + x**3

# Therefore, det has only been scaled by a factor of 1 + x**3

In [23]:

Matrix.from_str("1 5 3; 0 2 -2; 0 1 3").cramer_solve(Matrix.from_str("1; 2; 0"))

Matrix.from_str("1 5 3; 0 2 -2; 0 1 3").cramer_solve(Matrix.from_str("1; 2; 0"))

Modified matrix for column 1:

$\displaystyle \left[\begin{array}{ccc}1 & 5 & 3\\2 & 2 & -2\\0 & 1 & 3\end{array}\right]$

$\displaystyle \text{Determinant for column }1: -2$

Modified matrix for column 2:

$\displaystyle \left[\begin{array}{ccc}1 & 1 & 3\\0 & 2 & -2\\0 & 0 & 3\end{array}\right]$

$\displaystyle \text{Determinant for column }2: \frac{3}{4}$

Modified matrix for column 3:

$\displaystyle \left[\begin{array}{ccc}1 & 5 & 1\\0 & 2 & 2\\0 & 1 & 0\end{array}\right]$

$\displaystyle \text{Determinant for column }3: - \frac{1}{4}$

Out[23]:

$\displaystyle \left[\begin{array}{c}-2\\\frac{3}{4}\\- \frac{1}{4}\end{array}\right]$

Question 7¶

Use Cramer's rule to solve $$\begin{cases} x &+& 5y &+& 3z &=& 1 \\ && 2y &-& 2z &=& 2 \\ && y &+& 3z &=& 0 \end{cases}$$

In [24]:

mat = Matrix.from_str("1 5 3; 0 2 -2; 0 1 3")
rhs = Matrix.from_str("1; 2; 0")

mat.cramer_solve(rhs, verbosity=2)

mat = Matrix.from_str("1 5 3; 0 2 -2; 0 1 3") rhs = Matrix.from_str("1; 2; 0") mat.cramer_solve(rhs, verbosity=2)

Modified matrix for column 1:

$\displaystyle \left[\begin{array}{ccc}1 & 5 & 3\\2 & 2 & -2\\0 & 1 & 3\end{array}\right]$

$\displaystyle \text{Determinant for column }1: -2$

Modified matrix for column 2:

$\displaystyle \left[\begin{array}{ccc}1 & 1 & 3\\0 & 2 & -2\\0 & 0 & 3\end{array}\right]$

$\displaystyle \text{Determinant for column }2: \frac{3}{4}$

Modified matrix for column 3:

$\displaystyle \left[\begin{array}{ccc}1 & 5 & 1\\0 & 2 & 2\\0 & 1 & 0\end{array}\right]$

$\displaystyle \text{Determinant for column }3: - \frac{1}{4}$

Out[24]:

$\displaystyle \left[\begin{array}{c}-2\\\frac{3}{4}\\- \frac{1}{4}\end{array}\right]$

Question 8¶

Compute the adjoint of $\mathbf{A} = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 2 & 1 \\ 3 & 0 & 6 \end{pmatrix}$, and use it to compute $\mathbf{A}^{-1}$.

In [25]:

A = Matrix.from_str("1 -1 2; 0 2 1; 3 0 6")

A

A = Matrix.from_str("1 -1 2; 0 2 1; 3 0 6") A

Out[25]:

$\displaystyle \left[\begin{array}{ccc}1 & -1 & 2\\0 & 2 & 1\\3 & 0 & 6\end{array}\right]$

In [26]:

# Note: Please use `adj` method instead of `adjoint` or `adjugate` 
# even if they are aliased correctly based on MA1522 definitions as
# they are not the same as the SymPy's `adjoint` or `adjugate` methods.

A.adj()

# Note: Please use `adj` method instead of `adjoint` or `adjugate` # even if they are aliased correctly based on MA1522 definitions as # they are not the same as the SymPy's `adjoint` or `adjugate` methods. A.adj()

Out[26]:

$\displaystyle \left[\begin{matrix}12 & 6 & -5\\3 & 0 & -1\\-6 & -3 & 2\end{matrix}\right]$

In [27]:

A.adjoint()

A.adjoint()

/tmp/ipykernel_2323/2803142297.py:1: DeprecationWarning: The classical adjoint of the matrix is computed rather than the conjugate transpose.
            Please use self.adj() instead to remove ambiguity.
  A.adjoint()

Out[27]:

$\displaystyle \left[\begin{matrix}12 & 6 & -5\\3 & 0 & -1\\-6 & -3 & 2\end{matrix}\right]$

In [28]:

A_inv = A.adj() / A.det()
A_inv

A_inv = A.adj() / A.det() A_inv

Out[28]:

$\displaystyle \left[\begin{array}{ccc}-4 & -2 & \frac{5}{3}\\-1 & 0 & \frac{1}{3}\\2 & 1 & - \frac{2}{3}\end{array}\right]$