Matrix multiplication with @ operator in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this matrix multiplication?

Given two numpy arrays A and B, what is the result of A @ B?

NumPy

import numpy as np
A = np.array([[1, 2], [3, 4]])
B = np.array([[2, 0], [1, 2]])
result = A @ B
print(result)

A

[[4 2]
 [7 8]]

B

[[4 4]
 [10 8]]

C

[[2 4]
 [3 8]]

D

[[2 0]
 [1 2]]

💡 Hint

Remember matrix multiplication sums the products of rows of A with columns of B.

Matrix multiplication with @ multiplies rows of A by columns of B and sums the products. For example, element (0,0) is 1*2 + 2*1 = 4.

❓ data_output

intermediate

What is the shape of the result after matrix multiplication?

If A has shape (3, 4) and B has shape (4, 2), what is the shape of A @ B?

NumPy

import numpy as np
A = np.zeros((3,4))
B = np.zeros((4,2))
result = A @ B
print(result.shape)

A (3, 4)

B (4, 4)

C (2, 3)

D (3, 2)

💡 Hint

The resulting matrix shape is rows of A by columns of B.

Matrix multiplication results in a matrix with the number of rows of the first matrix and the number of columns of the second matrix.

🔧 Debug

advanced

Why does this matrix multiplication raise an error?

What error will this code raise and why?

NumPy

import numpy as np
A = np.array([[1, 2, 3], [4, 5, 6]])
B = np.array([[7, 8], [9, 10]])
result = A @ B

A IndexError: index out of bounds

B TypeError: unsupported operand type(s) for @: 'list' and 'ndarray'

C ValueError: shapes (2,3) and (2,2) not aligned: 3 (dim 1) != 2 (dim 0)

D No error, outputs a (2,2) matrix

💡 Hint

Check if the inner dimensions of the matrices match for multiplication.

Matrix multiplication requires the number of columns in A to equal the number of rows in B. Here, A has 3 columns but B has 2 rows, so it raises a ValueError.

🚀 Application

advanced

Calculate the product of a matrix and its transpose

Given matrix M, what is the result of M @ M.T?

NumPy

import numpy as np
M = np.array([[1, 2], [3, 4], [5, 6]])
result = M @ M.T
print(result)

A

[[ 1 3 5]
 [ 2 4 6]]

B

[[ 5 11 17]
 [11 25 39]
 [17 39 61]]

C

[[ 1 2 3]
 [3 4 5]
 [5 6 7]]

D

[[ 5 11 17]
 [11 25 41]
 [17 39 61]]

💡 Hint

Multiplying a matrix by its transpose results in a symmetric matrix with dot products of rows.

M @ M.T computes dot products of rows of M with each other, resulting in a symmetric matrix with shape (3,3).

🧠 Conceptual

expert

Which statement about the @ operator in numpy is TRUE?

Select the correct statement about the @ operator for numpy arrays.

A The @ operator performs matrix multiplication and requires the inner dimensions to match.

B The @ operator performs element-wise multiplication of two numpy arrays.

C The @ operator can multiply any two numpy arrays regardless of their shapes.

D The @ operator is only used for multiplying numpy arrays with scalars.

💡 Hint

Think about how matrix multiplication works and what the @ operator does.

The @ operator performs matrix multiplication, which requires the number of columns in the first array to match the number of rows in the second array.