Tutorial 1 (AY24/25 Sem 1)¶

In [1]:

# Required imports
import sympy as sym
from ma1522 import Matrix

# Required imports import sympy as sym from ma1522 import Matrix

Question 3¶

Solve the following linear systems.

(a)¶

$$\begin{cases} 3x_1 & + & 2x_2 & - & 4x_3 & = & 3 \\ 2x_1 & + & 3x_2 & + & 3x_3 & = & 15 \\ 5x_1 & - & 3x_2 & + & x_3 & = & 14 \end{cases}$$

In [2]:

mat = Matrix([[3, 2, -4],
              [2, 3, 3],
              [5, -3, 1]])

aug = Matrix([3, 15, 14])

aug_mat = mat.row_join(aug)
aug_mat

mat = Matrix([[3, 2, -4], [2, 3, 3], [5, -3, 1]]) aug = Matrix([3, 15, 14]) aug_mat = mat.row_join(aug) aug_mat

Out[2]:

$\displaystyle \left[\begin{array}{ccc|c}3 & 2 & -4 & 3\\2 & 3 & 3 & 15\\5 & -3 & 1 & 14\end{array}\right]$

In [3]:

plu = aug_mat.ref() # Show steps to get to Row Echelon Form

plu = aug_mat.ref() # Show steps to get to Row Echelon Form

$\displaystyle R_2 - \left(\frac{2}{3}\right)R_1 \rightarrow R_2$

$\displaystyle \left[\begin{array}{ccc|c}3 & 2 & -4 & 3\\0 & \frac{5}{3} & \frac{17}{3} & 13\\5 & -3 & 1 & 14\end{array}\right]$

$\displaystyle R_3 - \left(\frac{5}{3}\right)R_1 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc|c}3 & 2 & -4 & 3\\0 & \frac{5}{3} & \frac{17}{3} & 13\\0 & - \frac{19}{3} & \frac{23}{3} & 9\end{array}\right]$

$\displaystyle R_3 - \left(- \frac{19}{5}\right)R_2 \rightarrow R_3$

$\displaystyle \left[\begin{array}{ccc|c}3 & 2 & -4 & 3\\0 & \frac{5}{3} & \frac{17}{3} & 13\\0 & 0 & \frac{146}{5} & \frac{292}{5}\end{array}\right]$

In [4]:

plu.U.rref()  # Reduced Row Echelon Form from Upper Triangular Matrix

plu.U.rref() # Reduced Row Echelon Form from Upper Triangular Matrix

Out[4]:

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

(b)¶

$$\begin{cases} a &+& b &-& c &-& 2d &=& 0 \\ 2a &+& b &-& c &+& d &=& -2 \\ -a &+& b &-& 3c &+& d &=& 4 \end{cases}$$

In [5]:

mat = Matrix([[1, 1, -1, -2],
              [2, 1, -1, 1],
              [-1, 1, -3, 1]])

aug = Matrix([0, -2, 4])

aug_mat = mat.row_join(aug)
aug_mat

mat = Matrix([[1, 1, -1, -2], [2, 1, -1, 1], [-1, 1, -3, 1]]) aug = Matrix([0, -2, 4]) aug_mat = mat.row_join(aug) aug_mat

Out[5]:

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

In [6]:

aug_mat.rref() # Find Reduced Row Echelon Form

aug_mat.rref() # Find Reduced Row Echelon Form

Out[6]:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc|c}1 & 0 & 0 & 3 & -2\\0 & 1 & 0 & - \frac{19}{2} & 2\\0 & 0 & 1 & - \frac{9}{2} & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

In [7]:

mat.solve(aug)[0]  # Solve the system of equations represented by the augmented matrix

mat.solve(aug)[0] # Solve the system of equations represented by the augmented matrix

Out[7]:

$\displaystyle \left[\begin{array}{c}- 3 z - 2\\\frac{19 z}{2} + 2\\\frac{9 z}{2}\\z\end{array}\right]$

(c)¶

$$\begin{cases} x &-& 4y &+& 2z &=& -2 \\ x &+& 2y &-& 2z &=& -3 \\ x &-& y && &=& 4 \end{cases}$$

In [8]:

# add aug_pos to specify the position of the augmented column
aug_mat = Matrix([[1, -4, 2, -2],
                  [1, 2, -2, -3],
                  [1, -1, 0, 4]], aug_pos = 2)

aug_mat

# add aug_pos to specify the position of the augmented column aug_mat = Matrix([[1, -4, 2, -2], [1, 2, -2, -3], [1, -1, 0, 4]], aug_pos = 2) aug_mat

Out[8]:

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

In [9]:

aug_mat.rref()

aug_mat.rref()

Out[9]:

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

In [10]:

mat = aug_mat.select_cols(0, 1, 2)  # Remove the augmented column
aug = aug_mat.select_cols(3)  # Select the augmented column
# Solve the system of equations represented by the augmented matrix
try:
    mat.solve(aug)
except ValueError as e:
    print("Error raised due to inconsistent system:", e)

mat = aug_mat.select_cols(0, 1, 2) # Remove the augmented column aug = aug_mat.select_cols(3) # Select the augmented column # Solve the system of equations represented by the augmented matrix try: mat.solve(aug) except ValueError as e: print("Error raised due to inconsistent system:", e)

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

Error raised due to inconsistent system: No solution found for the linear system. The system may be inconsistent.

Question 4¶

Determine the values of $a$ and $b$ so that the linear system

$$\begin{cases} ax && &+& bz &=& 2 \\ ax &+& ay &+& 4z &=& 4 \\ && ay &+& 2z &=& b \end{cases}$$

In [11]:

a, b = sym.symbols('a b') # Define symbolic variables

mat = Matrix([[a, 0, b, 2],
              [a, a, 4, 4],
              [0, a, 2, b]], aug_pos=2)

mat

a, b = sym.symbols('a b') # Define symbolic variables mat = Matrix([[a, 0, b, 2], [a, a, 4, 4], [0, a, 2, b]], aug_pos=2) mat

Out[11]:

$\displaystyle \left[\begin{array}{ccc|c}a & 0 & b & 2\\a & a & 4 & 4\\0 & a & 2 & b\end{array}\right]$

In [12]:

plu = mat.ref()  # Reduced Row Echelon Form using PLU decomposition

plu = mat.ref() # Reduced Row Echelon Form using PLU decomposition

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

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

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

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

(a)¶

has no solution

In [13]:

# Case 1: b != 2, a = 0

plu.U.subs({a: 0}).rref()  # substitute a = 0

# Case 1: b != 2, a = 0 plu.U.subs({a: 0}).rref() # substitute a = 0

Out[13]:

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

(b)¶

has only one solution

In [14]:

# Case 2: b != 2, a != 0

plu.U.rref()

# Case 2: b != 2, a != 0 plu.U.rref()

Out[14]:

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

(c)¶

has infinitely many solutions and a general solution has one arbitrary parameter

In [15]:

# Case 3: b = 2, a != 0
plu.U.subs({b: 2}).rref()  # substitute b = 2

# Case 3: b = 2, a != 0 plu.U.subs({b: 2}).rref() # substitute b = 2

Out[15]:

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

(d)¶

has infinitely many solutions and a general solution has two arbitrary parameters.

In [16]:

# Case 4: b = 2, a = 0
plu.U.subs({a: 0, b: 2}).rref()  # substitute a = 0 and b = 2

# Case 4: b = 2, a = 0 plu.U.subs({a: 0, b: 2}).rref() # substitute a = 0 and b = 2

Out[16]:

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

Question 6¶

Solve the following system of non-linear equations:

$$\begin{cases} x^2 &-& y^2 &+& 2z^2 &=& 6 \\ 2x^2 &+& 2y^2 &-& 5z^2 &=& 3 \\ 2x^2 &+& 5y^2 &+& z^2 &=& 9 \end{cases}$$

In [17]:

mat = Matrix([[1, -1, 2, 6],
              [2, 2, -5, 3],
              [2, 5, 1, 9]], aug_pos=2)

mat

mat = Matrix([[1, -1, 2, 6], [2, 2, -5, 3], [2, 5, 1, 9]], aug_pos=2) mat

Out[17]:

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

In [18]:

mat.rref()

mat.rref()

Out[18]:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc|c}1 & 0 & 0 & 4\\0 & 1 & 0 & 0\\0 & 0 & 1 & 1\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

In [19]:

# Alternative Method: Using sym.solve

x, y, z = sym.symbols('x y z')  # Define symbolic variables for the solution

vec = Matrix([x**2, y**2, z**2])  # Define a vector of symbolic variables

mat = Matrix([[1, -1, 2],
              [2, 2, -5],
              [2, 5, 1]])

aug = Matrix([6, 3, 9])

mat, vec, aug

# Alternative Method: Using sym.solve x, y, z = sym.symbols('x y z') # Define symbolic variables for the solution vec = Matrix([x**2, y**2, z**2]) # Define a vector of symbolic variables mat = Matrix([[1, -1, 2], [2, 2, -5], [2, 5, 1]]) aug = Matrix([6, 3, 9]) mat, vec, aug

Out[19]:

$\displaystyle \left( \left[\begin{array}{ccc}1 & -1 & 2\\2 & 2 & -5\\2 & 5 & 1\end{array}\right], \ \left[\begin{array}{c}x^{2}\\y^{2}\\z^{2}\end{array}\right], \ \left[\begin{array}{c}6\\3\\9\end{array}\right]\right)$

In [20]:

sols = sym.solve(mat @ vec - aug)  # Solve the system of equations
for sol in sols:
    display(sol)  # Print the solutions for each variable

sols = sym.solve(mat @ vec - aug) # Solve the system of equations for sol in sols: display(sol) # Print the solutions for each variable

$\displaystyle \left\{ x : -2, \ y : 0, \ z : -1\right\}$

$\displaystyle \left\{ x : -2, \ y : 0, \ z : 1\right\}$

$\displaystyle \left\{ x : 2, \ y : 0, \ z : -1\right\}$

$\displaystyle \left\{ x : 2, \ y : 0, \ z : 1\right\}$

Question 7¶

A network of one-way streets of a downtown section can be represented by the diagram below, with traffic flowing in the direction indicated. The average hourly volume of traffic entering and leaving this section during rush hour is given in the diagram.

(a)¶

Do we have enough information to find the traffic volumes $x_1$, $x_2$, $x_3$, $x_4$, $x_5$, $x_6$, and $x_7$?

In [21]:

aug_mat = Matrix([[1, 0, 1, 0, 0, 0, 0, 800],
                  [1, -1, 0, 1, 0, 0, 0, 200],
                  [0, 1, 0, 0, -1, 0, 0, 500],
                  [0, 0, 1, 0, 0, 1, 0, 750],
                  [0, 0, 0, -1, 0, -1, 1, -600],
                  [0, 0, 0, 0, 1, 0, -1, -50]], aug_pos=6)

aug_mat

aug_mat = Matrix([[1, 0, 1, 0, 0, 0, 0, 800], [1, -1, 0, 1, 0, 0, 0, 200], [0, 1, 0, 0, -1, 0, 0, 500], [0, 0, 1, 0, 0, 1, 0, 750], [0, 0, 0, -1, 0, -1, 1, -600], [0, 0, 0, 0, 1, 0, -1, -50]], aug_pos=6) aug_mat

Out[21]:

$\displaystyle \left[\begin{array}{ccccccc|c}1 & 0 & 1 & 0 & 0 & 0 & 0 & 800\\1 & -1 & 0 & 1 & 0 & 0 & 0 & 200\\0 & 1 & 0 & 0 & -1 & 0 & 0 & 500\\0 & 0 & 1 & 0 & 0 & 1 & 0 & 750\\0 & 0 & 0 & -1 & 0 & -1 & 1 & -600\\0 & 0 & 0 & 0 & 1 & 0 & -1 & -50\end{array}\right]$

In [22]:

aug_mat.rref()

aug_mat.rref()

Out[22]:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccccccc|c}1 & 0 & 0 & 0 & 0 & -1 & 0 & 50\\0 & 1 & 0 & 0 & 0 & 0 & -1 & 450\\0 & 0 & 1 & 0 & 0 & 1 & 0 & 750\\0 & 0 & 0 & 1 & 0 & 1 & -1 & 600\\0 & 0 & 0 & 0 & 1 & 0 & -1 & -50\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2, 3, 4)\right\}$

In [23]:

mat = aug_mat.select_cols(*range(7))
aug = aug_mat.select_cols(7)

sol = mat.solve(aug)[0]
sol

mat = aug_mat.select_cols(*range(7)) aug = aug_mat.select_cols(7) sol = mat.solve(aug)[0] sol

Out[23]:

$\displaystyle \left[\begin{array}{c}y + 50\\z + 450\\750 - y\\- y + z + 600\\z - 50\\y\\z\end{array}\right]$

(b)¶

Suppose $x_6 = 50$ and $x_7 = 100$. What is $x_1$, $x_2$, $x_3$, $x_4$, and $x_5$?

In [24]:

y = sol[5]
z = sol[6]
sol.subs({y: 50, z: 100})  # Substitute y and z with specified values

y = sol[5] z = sol[6] sol.subs({y: 50, z: 100}) # Substitute y and z with specified values

Out[24]:

$\displaystyle \left[\begin{array}{c}100\\550\\700\\650\\50\\50\\100\end{array}\right]$

(c)¶

Can the road between junction A and B be closed for construction while still keeping the traffic flowing in the same directions on the other streets? Explain.

In [25]:

sol.subs({y: -50}) # Set sol[0] = 0

sol.subs({y: -50}) # Set sol[0] = 0

Out[25]:

$\displaystyle \left[\begin{array}{c}0\\z + 450\\800\\z + 650\\z - 50\\-50\\z\end{array}\right]$