Matrix creation and operations in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a 3x3 identity matrix using NumPy.

SciPy

from numpy import eye
matrix = eye([1])

Drag options to blanks, or click blank then click option'

A5

B2

C4

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using a size other than 3 will create a matrix of wrong dimensions.

Confusing eye with ones or zeros.

The eye function creates an identity matrix of the given size. Here, a 3x3 identity matrix is created by passing 3.

2fill in blank

medium

Complete the code to create a 2x4 matrix of zeros using NumPy.

SciPy

from numpy import zeros
matrix = zeros(([1]))

Drag options to blanks, or click blank then click option'

A2, 2

B4, 2

C2, 4

D3, 4

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping rows and columns in the tuple.

Passing separate arguments instead of a tuple.

The zeros function creates a matrix filled with zeros. Passing the shape as a tuple (2, 4) creates a 2 rows by 4 columns matrix.

3fill in blank

hard

Fix the error in the code to multiply two matrices using NumPy.

SciPy

from numpy import array
A = array([[1, 2], [3, 4]])
B = array([[5, 6], [7, 8]])
result = A [1] B

Drag options to blanks, or click blank then click option'

A@

B*

Cdot

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using * instead of @ for matrix multiplication.

Trying to use dot as an operator instead of a method.

Matrix multiplication in Python with arrays uses the @ operator. Using * does element-wise multiplication, which is not matrix multiplication.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

SciPy

words = ['data', 'science', 'is', 'fun']
lengths = {word: [1] for word in words if [2]

Drag options to blanks, or click blank then click option'

Alen(word)

Blen(word) > 3

Cword.startswith('s')

Dword.upper()

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length.

Using a wrong condition that filters incorrectly.

The dictionary comprehension maps each word to its length using len(word). The condition len(word) > 3 filters words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their lengths only if the length is greater than 2.

SciPy

words = ['cat', 'dog', 'elephant']
result = { [1]: [2] for word in words if [3] }

Drag options to blanks, or click blank then click option'

Aword.upper()

Blen(word)

Clen(word) > 2

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase instead of uppercase for keys.

Incorrect condition that filters out all words.

The dictionary comprehension uses word.upper() as the key, len(word) as the value, and filters words with length greater than 2 using len(word) > 2.

Practice

1. What does the scipy.linalg.inv() function do when applied to a square matrix?

easy

A. Transposes the matrix

B. Calculates the determinant of the matrix

C. Calculates the inverse of the matrix

D. Creates a zero matrix of the same size

Solution

1. Step 1: Understand the function purpose

   The scipy.linalg.inv() function is designed to find the inverse of a square matrix if it exists.

2. Step 2: Differentiate from other matrix operations

   Determinant calculation uses scipy.linalg.det(), transpose uses .T, and zero matrix creation uses numpy.zeros().

3. Final Answer:

   Calculates the inverse of the matrix -> Option C

4. Quick Check:

   Inverse = inv() [OK]

Hint: Inverse means flipping matrix; use inv() for that [OK]

Common Mistakes:

• Confusing inverse with determinant
• Thinking inv() transposes matrix
• Assuming inv() creates zero matrix

2. Which of the following is the correct way to create a 3x3 identity matrix using NumPy?

easy

A. scipy.identity(3)

B. scipy.eye(3)

C. scipy.linalg.eye(3)

D. np.identity(3)

Solution

1. Step 1: Identify the correct module and function

   The identity matrix function is in numpy (imported as np) as identity().

2. Step 2: Check other options for correctness

   scipy.eye() and scipy.linalg.eye() do not exist; scipy.identity() is not valid.

3. Final Answer:

   np.identity(3) -> Option D

4. Quick Check:

   Identity matrix = np.identity() [OK]

Hint: Use np.identity() for identity matrix [OK]

Common Mistakes:

• Using scipy.eye() which does not exist
• Calling identity() from wrong module
• Confusing identity with eye from numpy

3. What is the output of the following code?

import numpy as np
from scipy.linalg import det
A = np.array([[1, 2], [3, 4]])
print(round(det(A), 2))

medium

A. -2.00

B. 2.00

C. 10.00

D. -10.00

Solution

1. Step 1: Calculate determinant of matrix A

   Determinant formula for 2x2 matrix [[a,b],[c,d]] is ad - bc = 1*4 - 2*3 = 4 - 6 = -2.

2. Step 2: Round the result to 2 decimals

   Rounding -2 to 2 decimals remains -2.00.

3. Final Answer:

   -2.00 -> Option A

4. Quick Check:

   Determinant = -2.00 [OK]

Hint: Determinant 2x2 = ad - bc [OK]

Common Mistakes:

• Calculating ad + bc instead of ad - bc
• Forgetting negative sign
• Not rounding output

4. Identify the error in this code snippet:

import numpy as np
from scipy.linalg import inv
B = np.array([[1, 2], [2, 4]])
B_inv = inv(B)
print(B_inv)

medium

A. Matrix B is singular and cannot be inverted

B. Syntax error in array creation

C. inv() function is not imported correctly

D. Matrix B is not square

Solution

1. Step 1: Check matrix properties

   Matrix B is square (2x2), but rows are linearly dependent (second row is twice the first).

2. Step 2: Understand invertibility

   Such a matrix is singular (determinant zero), so inverse does not exist, causing an error when calling inv().

3. Final Answer:

   Matrix B is singular and cannot be inverted -> Option A

4. Quick Check:

   Singular matrix = no inverse [OK]

Hint: Check determinant; zero means no inverse [OK]

Common Mistakes:

• Thinking non-square matrix causes error here
• Assuming inv() import is wrong
• Missing linear dependence in matrix

5. You have two matrices:

M = np.array([[1, 2], [3, 4]])
N = np.array([[2, 0], [1, 2]])

Which code correctly computes the product P = M x N using scipy?

hard

A. from scipy.linalg import dot
P = dot(M, N)

B. import numpy as np
P = np.dot(M, N)

C. from scipy.linalg import matmul
P = matmul(M, N)

D. from scipy.linalg import inv
P = inv(M) * N

Solution

1. Step 1: Identify correct matrix multiplication function

   Scipy.linalg does not have dot or matmul functions; numpy provides np.dot() for matrix multiplication.

2. Step 2: Check options for correctness

   import numpy as np P = np.dot(M, N) uses np.dot(M, N) which correctly multiplies matrices M and N. from scipy.linalg import inv P = inv(M) * N incorrectly uses inverse and element-wise multiplication.

3. Final Answer:

   import numpy as np P = np.dot(M, N) -> Option B

4. Quick Check:

   Matrix product = np.dot() [OK]

Hint: Use np.dot() for matrix multiplication [OK]

Common Mistakes:

• Using inv() instead of multiplication
• Assuming scipy.linalg has dot or matmul
• Using * for matrix multiplication