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Basic Operations: Linear Algebra

Linear Algebra Operations

TensorFlow offers a suite of functions dedicated to linear algebra operations, making matrix operations straightforward.

Matrix Multiplication

Here's a quick reminder of how matrix multiplication works.

There are two equivalent approaches for matrix multiplication:

The tf.matmul() function;
Using the @ operator.


              1234567891011121314
            
import tensorflow as tf

# Create two matrices
matrix1 = tf.constant([[1, 2], [3, 4], [2, 1]])
matrix2 = tf.constant([[2, 0, 2, 5], [2, 2, 1, 3]])

# Multiply the matrices
product1 = tf.matmul(matrix1, matrix2)
product2 = matrix1 @ matrix2

# Display tensors
print(product1)
print('-' * 50)
print(product2)

Note

Multiplying matrices of size 3x2 and 2x4 will give a matrix of 3x4.

Matrix Inversion

You can obtain the inverse of a matrix using the tf.linalg.inv() function. Additionally, let's verify a fundamental property of the inverse matrix.


              123456789101112131415
            
import tensorflow as tf

# Create 2x2 matrix
matrix = tf.constant([[1., 2.], [3., 4.]])

# Compute the inverse of a matrix
inverse_mat = tf.linalg.inv(matrix)

# Check the result
identity = matrix @ inverse_mat

# Display tensors
print(inverse_mat)
print('-' * 50)
print(identity)

Note

Multiplying a matrix with its inverse should yield an identity matrix, which has ones on its main diagonal and zeros everywhere else. Additionally, the tf.linalg module offers a wide range of linear algebra functions. For further details or more advanced operations, you might want to refer to its official documentation.

Transpose

You can obtain a transposed matrix using the tf.transpose() function.


              123456789101112
            
import tensorflow as tf

# Create a matrix 3x2
matrix = tf.constant([[1, 2], [3, 4], [2, 1]])

# Get the transpose of a matrix
transposed = tf.transpose(matrix)

# Display tensors
print(matrix)
print('-' * 40)
print(transposed)

Dot Product

You can obtain a dot product using the tf.tensordot() function. By setting up an axes argument you can choose along which axes to calculate a dot product. E.g. for two vectors by setting up axes=1 you will get the classic dot product between vectors. But when setting axes=0 you will get broadcasted matrix along 0 axes:


              1234567891011121314
            
import tensorflow as tf

# Create two vectors
matrix1 = tf.constant([1, 2, 3, 4])
matrix2 = tf.constant([2, 0, 2, 5])

# Compute the dot product of two tensors
dot_product_axes1 = tf.tensordot(matrix1, matrix2, axes=1)
dot_product_axes0 = tf.tensordot(matrix1, matrix2, axes=0)

# Display tensors
print(dot_product_axes1)
print('-' * 40)
print(dot_product_axes0)

Note

If you take two matrices with appropriate dimensions (NxM @ MxK, where NxM represents the dimensions of the first matrix and MxK the second), and compute the dot product along axes=1, it essentially performs matrix multiplication.

Task

Swipe to start coding

Background

A system of linear equations can be represented in matrix form using the equation:

AX = B

Where:

A is a matrix of coefficients.
X is a column matrix of variables.
B is a column matrix representing the values on the right side of the equations.

The solution to this system can be found using the formula:

X = A^-1 B

Where A^-1 is the inverse of matrix A.

Objective

Given a system of linear equations, use TensorFlow to solve it. You are given the following system of linear equations:

2x + 3y - z = 1.
4x + y + 2z = 2.
-x + 2y + 3z = 3.

Represent the system of equations in matrix form (separate it into matrices A and B).
Using TensorFlow, find the inverse of matrix A.
Multiply the inverse of matrix A by matrix B to find the solution matrix X, which contains the values of x, y, and z.

Note

Slicing in TensorFlow operates similarly to NumPy. Therefore, X[:, 0] will retrieve all elements from the column at index 0. We will get to the slicing later in the course.

Solution

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Section 1. Chapter 9

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Basic Operations: Linear Algebra

Linear Algebra Operations

TensorFlow offers a suite of functions dedicated to linear algebra operations, making matrix operations straightforward.

Matrix Multiplication

Here's a quick reminder of how matrix multiplication works.

There are two equivalent approaches for matrix multiplication:

The tf.matmul() function;
Using the @ operator.


              1234567891011121314
            
import tensorflow as tf

# Create two matrices
matrix1 = tf.constant([[1, 2], [3, 4], [2, 1]])
matrix2 = tf.constant([[2, 0, 2, 5], [2, 2, 1, 3]])

# Multiply the matrices
product1 = tf.matmul(matrix1, matrix2)
product2 = matrix1 @ matrix2

# Display tensors
print(product1)
print('-' * 50)
print(product2)

Note

Multiplying matrices of size 3x2 and 2x4 will give a matrix of 3x4.

Matrix Inversion

You can obtain the inverse of a matrix using the tf.linalg.inv() function. Additionally, let's verify a fundamental property of the inverse matrix.


              123456789101112131415
            
import tensorflow as tf

# Create 2x2 matrix
matrix = tf.constant([[1., 2.], [3., 4.]])

# Compute the inverse of a matrix
inverse_mat = tf.linalg.inv(matrix)

# Check the result
identity = matrix @ inverse_mat

# Display tensors
print(inverse_mat)
print('-' * 50)
print(identity)

Note

Multiplying a matrix with its inverse should yield an identity matrix, which has ones on its main diagonal and zeros everywhere else. Additionally, the tf.linalg module offers a wide range of linear algebra functions. For further details or more advanced operations, you might want to refer to its official documentation.

Transpose

You can obtain a transposed matrix using the tf.transpose() function.


              123456789101112
            
import tensorflow as tf

# Create a matrix 3x2
matrix = tf.constant([[1, 2], [3, 4], [2, 1]])

# Get the transpose of a matrix
transposed = tf.transpose(matrix)

# Display tensors
print(matrix)
print('-' * 40)
print(transposed)

Dot Product


              1234567891011121314
            
import tensorflow as tf

# Create two vectors
matrix1 = tf.constant([1, 2, 3, 4])
matrix2 = tf.constant([2, 0, 2, 5])

# Compute the dot product of two tensors
dot_product_axes1 = tf.tensordot(matrix1, matrix2, axes=1)
dot_product_axes0 = tf.tensordot(matrix1, matrix2, axes=0)

# Display tensors
print(dot_product_axes1)
print('-' * 40)
print(dot_product_axes0)

Note

If you take two matrices with appropriate dimensions (NxM @ MxK, where NxM represents the dimensions of the first matrix and MxK the second), and compute the dot product along axes=1, it essentially performs matrix multiplication.

Task

Swipe to start coding

Background

A system of linear equations can be represented in matrix form using the equation:

AX = B

Where:

A is a matrix of coefficients.
X is a column matrix of variables.
B is a column matrix representing the values on the right side of the equations.

The solution to this system can be found using the formula:

X = A^-1 B

Where A^-1 is the inverse of matrix A.

Objective

Given a system of linear equations, use TensorFlow to solve it. You are given the following system of linear equations:

2x + 3y - z = 1.
4x + y + 2z = 2.
-x + 2y + 3z = 3.

Represent the system of equations in matrix form (separate it into matrices A and B).
Using TensorFlow, find the inverse of matrix A.
Multiply the inverse of matrix A by matrix B to find the solution matrix X, which contains the values of x, y, and z.

Note

Slicing in TensorFlow operates similarly to NumPy. Therefore, X[:, 0] will retrieve all elements from the column at index 0. We will get to the slicing later in the course.

Solution

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Thanks for your feedback!

Swipe to show menu

Basic Operations: Linear Algebra

Linear Algebra Operations

Matrix Multiplication

Matrix Inversion

Transpose

Dot Product

Background

Objective

Solution

Awesome!

Basic Operations: Linear Algebra

Linear Algebra Operations

Matrix Multiplication

Matrix Inversion

Transpose

Dot Product

Background

Objective

Solution

Awesome!