Tutorial 2 (AY24/25 Sem 1)¶

In [1]:

# Required imports
import sympy as sym
from ma1522 import Matrix, PartGen

# Required imports import sympy as sym from ma1522 import Matrix, PartGen

Question 2¶

(a)¶

Let $A = \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}$. Find a $4 \times 3$ matrix $X$ such that $AX = I_3$.

In [2]:

A = Matrix.from_str("1 1 0 1; 0 1 1 0; 0 0 1 1", col_sep=" ", row_sep=";")
A

A = Matrix.from_str("1 1 0 1; 0 1 1 0; 0 0 1 1", col_sep=" ", row_sep=";") A

Out[2]:

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

In [3]:

X = A.inverse("right", matrices=2, verbosity=1) 
X
# matrices=2 separates X into two matrices, the particular and general solution to see the results clearly

X = A.inverse("right", matrices=2, verbosity=1) X # matrices=2 separates X into two matrices, the particular and general solution to see the results clearly

Before RREF: [self | eye]

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

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc|c|c|c}1 & 0 & 0 & 2 & 1 & -1 & 1\\0 & 1 & 0 & -1 & 0 & 1 & -1\\0 & 0 & 1 & 1 & 0 & 0 & 1\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Out[3]:

$\left(\left[\begin{array}{ccc}1 & -1 & 1\\0 & 1 & -1\\0 & 0 & 1\\0 & 0 & 0\end{array}\right] + \left[\begin{array}{ccc}- 2 x_{4,1} & - 2 x_{4,2} & - 2 x_{4,3}\\x_{4,1} & x_{4,2} & x_{4,3}\\- x_{4,1} & - x_{4,2} & - x_{4,3}\\x_{4,1} & x_{4,2} & x_{4,3}\end{array}\right]\right)$

In [4]:

# Further factorisation
scalar_factor = X.gen_sol.scalar_factor(column=True)
scalar_factor

# Further factorisation scalar_factor = X.gen_sol.scalar_factor(column=True) scalar_factor

Out[4]:

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

In [5]:

# Therefore, X can be expressed as:
PartGen(X.part_sol, scalar_factor)

# Therefore, X can be expressed as: PartGen(X.part_sol, scalar_factor)

Out[5]:

$\left(\left[\begin{array}{ccc}1 & -1 & 1\\0 & 1 & -1\\0 & 0 & 1\\0 & 0 & 0\end{array}\right] + \left[\begin{array}{ccc}-2 & -2 & -2\\1 & 1 & 1\\-1 & -1 & -1\\1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}x_{4,1} & 0 & 0\\0 & x_{4,2} & 0\\0 & 0 & x_{4,3}\end{array}\right]\right)$

(b)¶

Let $B = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$. Find a $3 \times 4$ matrix $Y$ such that $YB = I_3$.

In [6]:

B = Matrix.from_str("1 0 1; 1 1 0; 0 1 1; 0 0 1", col_sep=" ", row_sep=";")
B

B = Matrix.from_str("1 0 1; 1 1 0; 0 1 1; 0 0 1", col_sep=" ", row_sep=";") B

Out[6]:

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

In [7]:

Y = B.inverse("left", matrices=2, verbosity=1)  # Find the right inverse of B
Y

Y = B.inverse("left", matrices=2, verbosity=1) # Find the right inverse of B Y

Before RREF: [self^T | eye]

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

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc|c|c|c}1 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & \frac{1}{2}\\0 & 1 & 0 & - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & - \frac{1}{2}\\0 & 0 & 1 & \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Out[7]:

$\left(\left[\begin{array}{cccc}\frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & 0\\- \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0\\\frac{1}{2} & - \frac{1}{2} & \frac{1}{2} & 0\end{array}\right] + \left[\begin{array}{cccc}- \frac{x_{1,4}}{2} & \frac{x_{1,4}}{2} & - \frac{x_{1,4}}{2} & x_{1,4}\\- \frac{x_{2,4}}{2} & \frac{x_{2,4}}{2} & - \frac{x_{2,4}}{2} & x_{2,4}\\- \frac{x_{3,4}}{2} & \frac{x_{3,4}}{2} & - \frac{x_{3,4}}{2} & x_{3,4}\end{array}\right]\right)$

In [8]:

PartGen(Y.part_sol, Y.gen_sol.scalar_factor(column=False))

PartGen(Y.part_sol, Y.gen_sol.scalar_factor(column=False))

Out[8]:

$\left(\left[\begin{array}{cccc}\frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & 0\\- \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0\\\frac{1}{2} & - \frac{1}{2} & \frac{1}{2} & 0\end{array}\right] + \left[\begin{array}{ccc}x_{1,4} & 0 & 0\\0 & x_{2,4} & 0\\0 & 0 & x_{3,4}\end{array}\right]\left[\begin{array}{cccc}- \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & 1\\- \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & 1\\- \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & 1\end{array}\right]\right)$

Question 3¶

(i) Reduce the following matrices $A$ to its reduced row-echelon form $R$.

(ii) For each of the elementary row operation, write the corresponding elementary matrix.

(iii) Write the matrices $A$ in the form $E_1 E_2 \ldots E_n R$ where $E_1, E_2, \ldots, E_n$ are elementary matrices and $R$ is the reduced row-echelon form of $A$.

(a)¶

$A = \begin{pmatrix} 5 & -2 & 6 & 0 \\ -2 & 1 & 3 & 1 \end{pmatrix}$.

In [9]:

A = Matrix.from_str("5 -2 6 0; -2 1 3 1")
A

A = Matrix.from_str("5 -2 6 0; -2 1 3 1") A

Out[9]:

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

In [10]:

# Tip: Use `copy()` to avoid modifying the original matrix so that 
# the same cell can be ran multiple times without side effects.

(A.copy()
 .reduce_row(1, -2/5, 0)
 .scale_row(0, 1/5)
 .scale_row(1, 5)
 .reduce_row(0, -2/5, 1)
)

# Tip: Use `copy()` to avoid modifying the original matrix so that # the same cell can be ran multiple times without side effects. (A.copy() .reduce_row(1, -2/5, 0) .scale_row(0, 1/5) .scale_row(1, 5) .reduce_row(0, -2/5, 1) )

$\displaystyle R_2 - \left(-0.4\right)R_1 \rightarrow R_2$

$\displaystyle \left[\begin{array}{cccc}5 & -2 & 6 & 0\\0 & \frac{1}{5} & \frac{27}{5} & 1\end{array}\right]$

$\displaystyle \left(0.2\right) R_1 \rightarrow R_1$

$\displaystyle \left[\begin{array}{cccc}1 & - \frac{2}{5} & \frac{6}{5} & 0\\0 & \frac{1}{5} & \frac{27}{5} & 1\end{array}\right]$

$\displaystyle \left(5\right) R_2 \rightarrow R_2$

$\displaystyle \left[\begin{array}{cccc}1 & - \frac{2}{5} & \frac{6}{5} & 0\\0 & 1 & 27 & 5\end{array}\right]$

$\displaystyle R_1 - \left(-0.4\right)R_2 \rightarrow R_1$

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 12 & 2\\0 & 1 & 27 & 5\end{array}\right]$

Out[10]:

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 12 & 2\\0 & 1 & 27 & 5\end{array}\right]$

(b)¶

$A = \begin{pmatrix} -1 & 3 & -4 \\ 2 & 4 & 1 \\ -4 & 2 & -9 \end{pmatrix}$.

In [11]:

A = Matrix.from_str("-1 3 -4; 2 4 1; -4 2 -9")
A

A = Matrix.from_str("-1 3 -4; 2 4 1; -4 2 -9") A

Out[11]:

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

In [12]:

(A.copy()
 .reduce_row(1, -2, 0)
 .reduce_row(2, 4, 0)
 .reduce_row(2, -1, 1)
 .scale_row(0, -1)
 .scale_row(1, 1/10)
 .reduce_row(0, -3, 1)
)

(A.copy() .reduce_row(1, -2, 0) .reduce_row(2, 4, 0) .reduce_row(2, -1, 1) .scale_row(0, -1) .scale_row(1, 1/10) .reduce_row(0, -3, 1) )

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

$\displaystyle \left[\begin{array}{ccc}-1 & 3 & -4\\0 & 10 & -7\\-4 & 2 & -9\end{array}\right]$

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

$\displaystyle \left[\begin{array}{ccc}-1 & 3 & -4\\0 & 10 & -7\\0 & -10 & 7\end{array}\right]$

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

$\displaystyle \left[\begin{array}{ccc}-1 & 3 & -4\\0 & 10 & -7\\0 & 0 & 0\end{array}\right]$

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

$\displaystyle \left[\begin{array}{ccc}1 & -3 & 4\\0 & 10 & -7\\0 & 0 & 0\end{array}\right]$

$\displaystyle \left(0.1\right) R_2 \rightarrow R_2$

$\displaystyle \left[\begin{array}{ccc}1 & -3 & 4\\0 & 1 & - \frac{7}{10}\\0 & 0 & 0\end{array}\right]$

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

$\displaystyle \left[\begin{array}{ccc}1 & 0 & \frac{19}{10}\\0 & 1 & - \frac{7}{10}\\0 & 0 & 0\end{array}\right]$

Out[12]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & \frac{19}{10}\\0 & 1 & - \frac{7}{10}\\0 & 0 & 0\end{array}\right]$

(c)¶

$A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & -2 & 1 \\ 1 & 2 & 3 \end{pmatrix}$.

In [13]:

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

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

Out[13]:

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

In [14]:

(A.copy()
 .reduce_row(1, 2, 0)
 .reduce_row(2, 1, 0)
 .swap_row(1, 2)
 .scale_row(1, sym.Rational(1, 3)) # To avoid floating point errors
 .reduce_row(1, 1, 2)
 .reduce_row(0, -1, 1)
)

(A.copy() .reduce_row(1, 2, 0) .reduce_row(2, 1, 0) .swap_row(1, 2) .scale_row(1, sym.Rational(1, 3)) # To avoid floating point errors .reduce_row(1, 1, 2) .reduce_row(0, -1, 1) )

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

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

$\displaystyle R_3 - \left(1\right)R_1 \rightarrow R_3$

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

$\displaystyle R_2 \leftrightarrow R_3$

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

$\displaystyle \left(\frac{1}{3}\right) R_2 \rightarrow R_2$

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

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

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

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

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

Out[14]:

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

Question 4¶

Determine if the following matrices are invertible. If the matrix is invertible, find its inverse.

(a)¶

$\begin{pmatrix} -1 & 3 \\ 3 & -2 \end{pmatrix}$.

In [15]:

mat = Matrix.from_str("-1 3; 3 -2")
mat.inverse(verbosity=2)

mat = Matrix.from_str("-1 3; 3 -2") mat.inverse(verbosity=2)

Left inverse found!
Right inverse found!
Before RREF: [self | eye]

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

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc|c|c}1 & 0 & \frac{2}{7} & \frac{3}{7}\\0 & 1 & \frac{3}{7} & \frac{1}{7}\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Out[15]:

$\displaystyle \left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7}\\\frac{3}{7} & \frac{1}{7}\end{array}\right]$

(b)¶

$\begin{pmatrix} -1 & 3 & -4 \\ 2 & 4 & 1 \\ -4 & 2 & -9 \end{pmatrix}$.

In [16]:

mat = Matrix.from_str("-1 3 -4; 2 4 1; -4 2 -9")

try:
    mat.inverse(verbosity=2) # without specifying the option, no rref will be shown
except ValueError as e:
    print(f"Error: {e}")

mat = Matrix.from_str("-1 3 -4; 2 4 1; -4 2 -9") try: mat.inverse(verbosity=2) # without specifying the option, no rref will be shown except ValueError as e: print(f"Error: {e}")

Error: No inverse found! Rank: 2, Rows: 3, Columns: 3. Try pseudo-inverse: .pinv()

In [17]:

try:
    mat.inverse("both", verbosity=2)
except ValueError as e:
    print(f"Error: {e}")

try: mat.inverse("both", verbosity=2) except ValueError as e: print(f"Error: {e}")

Before RREF: [self | eye]

$\displaystyle \left[\begin{array}{ccc|c|c|c}-1 & 3 & -4 & 1 & 0 & 0\\2 & 4 & 1 & 0 & 1 & 0\\-4 & 2 & -9 & 0 & 0 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc|c|c|c}1 & 0 & \frac{19}{10} & 0 & \frac{1}{10} & - \frac{1}{5}\\0 & 1 & - \frac{7}{10} & 0 & \frac{1}{5} & \frac{1}{10}\\0 & 0 & 0 & 1 & - \frac{1}{2} & - \frac{1}{2}\end{array}\right], \quad\text{pivots} = (0, 1, 3)\right\}$

Error: No both inverse found! Try pseudo-inverse: .pinv()

Question 5¶

Write down the conditions so that the matrix $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix}$ is invertible.

In [18]:

a, b, c = sym.symbols("a b c")
mat = Matrix.from_str("1 1 1; a b c; a**2 b**2 c**2")
mat

a, b, c = sym.symbols("a b c") mat = Matrix.from_str("1 1 1; a b c; a**2 b**2 c**2") mat

Out[18]:

$\displaystyle \left[\begin{array}{ccc}1 & 1 & 1\\a & b & c\\a^{2} & b^{2} & c^{2}\end{array}\right]$

In [19]:

inv = mat.inverse(verbosity=2)
inv

inv = mat.inverse(verbosity=2) inv

Left inverse found!
Right inverse found!
Before RREF: [self | eye]

$\displaystyle \left[\begin{array}{ccc|c|c|c}1 & 1 & 1 & 1 & 0 & 0\\a & b & c & 0 & 1 & 0\\a^{2} & b^{2} & c^{2} & 0 & 0 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc|c|c|c}1 & 0 & 0 & \frac{b c}{a^{2} - a b - a c + b c} & \frac{- b - c}{a^{2} - a b - a c + b c} & \frac{1}{a^{2} - a b - a c + b c}\\0 & 1 & 0 & - \frac{a c}{a b - a c - b^{2} + b c} & \frac{a + c}{a b - a c - b^{2} + b c} & - \frac{1}{a b - a c - b^{2} + b c}\\0 & 0 & 1 & \frac{a b}{a b - a c - b c + c^{2}} & \frac{- a - b}{a b - a c - b c + c^{2}} & \frac{1}{a b - a c - b c + c^{2}}\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Out[19]:

$\displaystyle \left[\begin{array}{ccc}\frac{b c}{a^{2} - a b - a c + b c} & \frac{- b - c}{a^{2} - a b - a c + b c} & \frac{1}{a^{2} - a b - a c + b c}\\- \frac{a c}{a b - a c - b^{2} + b c} & \frac{a + c}{a b - a c - b^{2} + b c} & - \frac{1}{a b - a c - b^{2} + b c}\\\frac{a b}{a b - a c - b c + c^{2}} & \frac{- a - b}{a b - a c - b c + c^{2}} & \frac{1}{a b - a c - b c + c^{2}}\end{array}\right]$

In [20]:

# For inverse to exist, all the divisors must be non-zero

divisors = 1
for entry in inv.select_cols(0): # same divisor along columns
    _, divisor = sym.fraction(entry)
    divisors *= divisor

sym.solve(divisors)
# This means that a != b and a != c and b != c for the inverse to exist

# For inverse to exist, all the divisors must be non-zero divisors = 1 for entry in inv.select_cols(0): # same divisor along columns _, divisor = sym.fraction(entry) divisors *= divisor sym.solve(divisors) # This means that a != b and a != c and b != c for the inverse to exist

Out[20]:

$\displaystyle \left[ \left\{ a : b\right\}, \ \left\{ a : c\right\}, \ \left\{ b : c\right\}\right]$

In [21]:

# Alternative Method via determinant

det = mat.det()
det.factor()

# Alternative Method via determinant det = mat.det() det.factor()

Out[21]:

$\displaystyle - \left(a - b\right) \left(a - c\right) \left(b - c\right)$