Notes of Linear Algebra for Preparing Mathematical Modeling

Chapter 1: Linear Equation

Author: Kenneth, S.K. Cheng

Table of Contents

• Notes of Linear Algebra for Preparing Mathematical Modeling
  • Chapter 1: Linear Equation
    • Author: Kenneth, S.K. Cheng
  • Table of Contents
    • 1.1 Introduction
    • 1.2 Gaussian Elimination
      • Using Python to Solve Linear Equation
      • Exercises
    • 1.3 Gauss-Jordan Elimination
    • 1.4 Ill-Conditioned System
    • 1.5 Matrix Operations
      • 1.5.1 Matrix Addition
      • 1.5.2 Matrix Multiplication
    • 1.6 Matrix Inversion
      • Using Python to Find the Inverse of a Matrix
      • Exercises
    • 1.7 Rank and Nullity of a Matrix
      • Using Python to Find the Rank and Nullity of a Matrix
      • Exercises
    • 1.8 Determinant of a Matrix
      • Properties of Determinants
        • Using Python to Find the Determinant of a Matrix
        • Exercises
    • References

1.1 Introduction

A fundamental problem that surfaces in all mathematical sciences is that of analyzing and solving $m$ algebraic equations in $n$ unknowns. The study of a system of simultaneous linear equations is in a natural and indivisible alliance with the study of the rectangular array of numbers defined by the coefficients of the equations.

Today, this problem would be formulated as three equations in three unknowns by writing

$$\begin{align*} a_{11}x_1 + \dots + a_{1n}x_n &= b_1, \\ a_{21}x_1 + \dots + a_{2n}x_n &= b_2, \\ &\vdots \\ a_{m1}x_1 + \dots + a_{mn}x_n &= b_m. \end{align*}$$

The coefficients $a_{ij}$ and the constants $b_i$ are real or complex numbers. The unknowns $x_1, \dots, x_n$ are the variables of the system. The system is said to be linear if the functions $f_i(x_1, \dots, x_n) = a_{i1}x_1 + \dots + a_{in}x_n$ are linear functions of the variables $x_1, \dots, x_n$. The system is said to be homogeneous if the constants $b_i$ are all zero.

The system of equations can be written in matrix form as $A\mathbf{x} = \mathbf{b}$, where

$$A = \begin{bmatrix} a_{11} & \dots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \dots & a_{mn} \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} b_1 \\ \vdots \\ b_m \end{bmatrix}.$$

In the above notation, the matrix $A$ is called the coefficient matrix, the vector $\mathbf{x}$ is called the unknown vector, and the vector $\mathbf{b}$ is called the constant vector. The system of equations is said to be consistent if there is at least one solution. The system of equations is said to be inconsistent if there is no solution. The system of equations is said to be dependent if there are infinitely many solutions.

As secondary students, we are familiar with the method of substitution and the method of elimination. The method of substitution is to solve one of the equations for one of the variables and substitute the result into the other equations. The method of elimination is to add or subtract multiples of the equations to eliminate one of the variables.

But in here, we will introduce the method of Gaussian elimination and the method of matrix inversion. The method of Gaussian elimination is similar to what we have learned in secondary school, but it is more systematic and efficient.

1.2 Gaussian Elimination

Before we introduce the method of Gaussian elimination, we need to first observe the linear system.

$$\begin{align*} a_{11}x_1 + \dots + a_{1n}x_n &= b_1, \\ a_{21}x_1 + \dots + a_{2n}x_n &= b_2, \\ &\vdots \\ a_{m1}x_1 + \dots + a_{mn}x_n &= b_m. \end{align*}$$

For such system, there are three possibilities:

• The system has no solution: There is no such value of $x_1, \dots, x_n$ that satisfies all the equations.
• The system has a unique solution: There is only one set value of $x_1, \dots, x_n$ that satisfies all the equations.
• The system has infinitely many solutions: There are infinitely many sets of values of $x_1, \dots, x_n$ that satisfy all the equations.

Then we may introduce the idea of augmented matrix. The augmented matrix of the system is the matrix obtained by adjoining the column of constants to the coefficient matrix. For the above system, we may take the augmented matrix as $[A|\mathbf{b}]$.

$$[A|\mathbf{b}] = \begin{bmatrix} a_{11} & \dots & a_{1n} & & b_1 \\ a_{21} & \dots & a_{2n} & & b_2 \\ \vdots & \ddots & \vdots & & \vdots \\ a_{m1} & \dots & a_{mn} & & b_m \end{bmatrix}.$$

If this augmented matrix has the form

$$\begin{bmatrix} a_{11} & \dots & a_{1n} & & b_1 \\ 0 & \dots & a_{2n} & & b_2' \\ \vdots & \ddots & \vdots & & \vdots \\ 0 & \dots & a_{mn} & & b_m' \end{bmatrix},$$

then the system has a unique solution. We called this form as the row echelon form. The system has no solution if the row echelon form has a row of the form $[0 \dots 0 | b]$, where $b \neq 0$. The system has infinitely many solutions if the row echelon form has a row of the form $[0 \dots 0 | 0]$.

Gaussian elimination is a methodical process of systematically transforming one system into another simpler, but equivalent, system (two systems are called equivalent if they possess equal solution sets) by successively eliminating unknowns and eventually arriving at a system that is easily solvable. The elimination process relies on three simple operations by which to transform one system to another equivalent system.

To describe these operations, let $E_k$ denote the $k$th equation:

$$E_k: a_{k1}x_1 + \dots + a_{kn}x_n = b_k.$$

and write the system of equations as

$$S = \begin{cases} E_1, \\ E_2, \\ \vdots \\ E_m. \end{cases}$$

For a linear system $S$, each of the following operations is called an elementary row operation:

1. Interchange two equations: Interchanging two equations $E_i$ and $E_j$ in $S$ results in a new system $S'$.

$$S = \begin{cases} E_1, \\ \vdots \\ E_i, \\ \vdots \\ E_j, \\ \vdots \\ E_m. \end{cases} \quad \text{then,} \quad S' = \begin{cases} E_1, \\ \vdots \\ E_j, \\ \vdots \\ E_i, \\ \vdots \\ E_m. \end{cases}$$

2. Multiply an equation by a nonzero constant: Multiplying an equation $E_i$ by a nonzero constant $c$ results in a new system $S'$.

$$S = \begin{cases} E_1, \\ \vdots \\ E_i, \\ \vdots \\ E_m. \end{cases} \quad \text{then,} \quad S' = \begin{cases} E_1, \\ \vdots \\ cE_i, \\ \vdots \\ E_m. \end{cases}$$

3. Add a multiple of one equation to another: Adding a multiple of equation $E_i$ to equation $E_j$ results in a new system $S'$.

$$S = \begin{cases} E_1, \\ \vdots \\ E_i, \\ \vdots \\ E_j, \\ \vdots \\ E_m. \end{cases} \quad \text{then,} \quad S' = \begin{cases} E_1, \\ \vdots \\ E_i, \\ \vdots \\ E_j + cE_i, \\ \vdots \\ E_m. \end{cases}$$

The reason for applying these operations cannot change the solution set of the system is that each operation is reversible. For example, if we interchange two equations, we can interchange them back. If we multiply an equation by a nonzero constant, we can multiply it by the reciprocal of the constant. If we add a multiple of one equation to another, we can subtract the multiple of one equation from another.

The most common problem encounted in practice is the one which there are exactly $n$ equations and $n$ unknowns. In this case, we called the system as a square system. The method of Gaussian elimination is particularly efficient for solving square systems.

Take the following system as an example:

$$\begin{align*} 2x_1 + x_2 + x_3 &= 1, \\ 6x_1 + 2x_2 + x_3 &= -1, \\ -2x_1 + 2x_2 + x_3 &= 7. \end{align*}$$

We may write the augmented matrix as

$$\begin{bmatrix} 2 & 1 & 1 & | & 1 \\ 6 & 2 & 1 & | & -1 \\ -2 & 2 & 1 & | & 7 \end{bmatrix}.$$

The first step is to eliminate the $x_1$ in the second and third equations. We may multiply the first equation by $3$ and subtract it from the second equation, and multiply the first equation by $-1$ and add it to the third equation. The augmented matrix becomes

$$\begin{bmatrix} 2 & 1 & 1 & | & 1 \\ 0 & -1 & -2 & | & -4 \\ 0 & 3 & 2 & | & 8 \end{bmatrix}.$$

The second step is to eliminate the $x_2$ in the third equation. We may multiply the second equation by $3$ and add it to the third equation. The augmented matrix becomes

$$\begin{bmatrix} 2 & 1 & 1 & | & 1 \\ 0 & -1 & -2 & | & -4 \\ 0 & 0 & -4 & | & -4 \end{bmatrix}.$$

The third step is to eliminate the $x_3$ in the second equation. We may multiply the third equation by $-2$ and add it to the second equation. The augmented matrix becomes

$$\begin{bmatrix} 2 & 1 & 1 & | & 1 \\ 0 & -1 & 0 & | & -2 \\ 0 & 0 & -4 & | & -4 \end{bmatrix}.$$

The fourth step is to time the second equation by $-1$ and divide the third equation by $-2$, then we get the row echelon form

$$\begin{bmatrix} 2 & 1 & 1 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 1 \end{bmatrix}.$$

For simplicity, we may also eliminate the $x_2$ and $x_3$ in the first equation. The augmented matrix becomes

$$\begin{bmatrix} 2 & 0 & 0 & | & -2 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 1 \end{bmatrix}.$$

Then we may solve the system by back substitution. We have $x_3 = 1$, $x_2 = 2$, and $x_1 = -2$. The solution is unique.

Using Python to Solve Linear Equation

We may use the numpy library to solve the linear equation. The numpy.linalg.solve function can be used to solve the linear equation.

import numpy as np

A = np.array([[2, 1, 1], [6, 2, 1], [-2, 2, 1]]) # Coefficient matrix
b = np.array([1, -1, 7]) # Constants
x = np.linalg.solve(A, b) # Solve the linear equation
print(x) # Print the solution

Exercises

1. Solve the following system of linear equations.

$$\begin{align*} x_1 + 2x_2 + 3x_3 &= 6, \\ 2x_1 + 5x_2 + 2x_3 &= 4, \\ 6x_1 - 3x_2 + x_3 &= 2. \end{align*}$$

You first need to write the augmented matrix and then use the method of Gaussian elimination to solve the system. Finally, use Python to verify your answer.

2. Solve the following system of linear equations.

$$\begin{align*} 4x_1 - 2x_2 + 3x_3 &= 1, \\ 2x_1 + 1x_2 + 3x_3 &= 5, \\ 3x_1 + 2x_2 + 4x_3 &= 7. \end{align*}$$

You first need to write the augmented matrix and then use the method of Gaussian elimination to solve the system. Finally, use Python to verify your answer.

1.3 Gauss-Jordan Elimination

The method of Gaussian elimination is to transform the augmented matrix into row echelon form. The method of Gauss-Jordan elimination is to transform the augmented matrix into reduced row echelon form. The reduced row echelon form is a form that is even simpler than the row echelon form.

The reduced row echelon form is a form that has the following properties:

1. The leading entry (some may called it as the pivot) in each row is $1$.
2. The leading $1$ in each row is the only nonzero entry in its column.

The method of Gauss-Jordan elimination is similar to the method of Gaussian elimination. The only difference is that we need to make the leading entry in each row to be $1$ and make the leading $1$ in each row to be the only nonzero entry in its column.

1.4 Ill-Conditioned System

Take the following system as an example:

$$\begin{align*} 0.835x_1 + 0.667x_2 &= 0.168, \\ 0.333x_1 + 0.266x_2 &= 0.067. \end{align*}$$

For they are floating-point numbers, we better to use Python to solve the system. We may use the numpy.linalg.solve function to solve the system.

import numpy as np

A = np.array([[0.835, 0.667], [0.333, 0.266]]) # Coefficient matrix
b = np.array([0.168, 0.067]) # Constants
x = np.linalg.solve(A, b) # Solve the linear equation
print(x) # Print the solution

The solution is $x_1 = 0.1$ and $x_2 = 0.1$. But if we perturb the constants a little bit, the solution may change. For example, if we change $0.067$ to $0.066$, the solution may then change to $x_1 = -666$ and $x_2 = 834$. This is called the ill-conditioned system.

This is an example of an ill-conditioned system. An ill-conditioned system is a system that is very sensitive to small changes in the constants. The condition number of a matrix is a measure of how sensitive the system is to small changes in the constants.

1.5 Matrix Operations

1.5.1 Matrix Addition

If $A$ and $B$ are two matrices of the same size, then the sum $A + B$ is the matrix obtained by adding the corresponding entries of $A$ and $B$.

For example, if

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix},$$

then

$$A + B = \begin{bmatrix} 1 + 5 & 2 + 6 \\ 3 + 7 & 4 + 8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}.$$

1.5.2 Matrix Multiplication

If $A$ is a matrix of size $m \times n$ and $B$ is a matrix of size $n \times p$, then the product $AB$ is a matrix of size $m \times p$.

For example, if

$$A = \begin{bmatrix} 3 & 2\\ 1 & 8\\ 4 & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix},$$

then

$$AB = \begin{bmatrix} 3 \times 1 + 2 \times 4 & 3 \times 2 + 2 \times 5 & 3 \times 3 + 2 \times 6\\ 1 \times 1 + 8 \times 4 & 1 \times 2 + 8 \times 5 & 1 \times 3 + 8 \times 6\\ 4 \times 1 + 6 \times 4 & 4 \times 2 + 6 \times 5 & 4 \times 3 + 6 \times 6 \end{bmatrix} = \begin{bmatrix} 11 & 16 & 21\\ 33 & 42 & 51\\ 22 & 32 & 42 \end{bmatrix}.$$

However, one may note that

$$BA = \begin{bmatrix} 1 \times 3 + 2 \times 1 + 3 \times 4 & 1 \times 2 + 2 \times 8 + 3 \times 6\\ 4 \times 3 + 5 \times 1 + 6 \times 4 & 4 \times 2 + 5 \times 8 + 6 \times 6 \end{bmatrix} = \begin{bmatrix} 11 & 28\\ 31 & 64 \end{bmatrix}.$$

1.6 Matrix Inversion

If $A$ is a square matrix, then we can form the product $AA$, which is a square matrix of the same size as $A$. It is denoted as $A^2$. Similarly, for any general matrix $A$, we can form the product $A^n$ for any positive integer $n$.

We define the unit $n \times n$ matrix $I_n$ to be the matrix with $1$'s on the diagonal and $0$'s elsewhere. For example, the $2 \times 2$ unit matrix is

$$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

In general, the $n \times n$ unit matrix is

$$I_n = \begin{bmatrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix}.$$

One may note that if $A = I$, then $Ax = b$ is equivalent to $x = b$. This is because $I_nx = x$ for any vector $x$. The matrix $I_n$ is called the identity matrix.

We can define $A^0 = I_n$ for any square matrix $A$. We can also define $A^{-1}$ to be the matrix such that $AA^{-1} = A^{-1}A = I_n$. The matrix $A^{-1}$ is called the inverse of $A$. If $A$ has an inverse, then $A$ is said to be invertible.

To find the inverse of a matrix, we may use the method of matrix inversion. The method of matrix inversion is to transform the augmented matrix $[A|I_n]$ into $[I_n|A^{-1}]$. The method of matrix inversion is similar to the method of Gaussian elimination.

For example, if

$$A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix},$$

then we may write the augmented matrix as

$$\begin{bmatrix} 4 & 7 & | & 1 & 0 \\ 2 & 6 & | & 0 & 1 \end{bmatrix}.$$

By using the method of matrix inversion, we may transform the augmented matrix into

$$\begin{bmatrix} 1 & 0 & | & \frac{3}{5} & -\frac{7}{10} \\ 0 & 1 & | & -\frac{1}{5} & \frac{2}{5} \end{bmatrix}.$$

Therefore, the inverse of the matrix $A$ is

$$A^{-1} = \begin{bmatrix} \frac{3}{5} & -\frac{7}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}.$$

Using Python to Find the Inverse of a Matrix

We may use the numpy library to find the inverse of a matrix. The numpy.linalg.inv function can be used to find the inverse of a matrix.

import numpy as np

A = np.array([[4, 7], [2, 6]]) # Coefficient matrix
A_inv = np.linalg.inv(A) # Find the inverse of the matrix
print(A_inv) # Print the inverse of the matrix

Exercises

1. Find the inverse of the following matrix.

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}.$$

You first need to write the augmented matrix and then use the method of matrix inversion to find the inverse of the matrix. Finally, use Python to verify your answer.

2. Find the inverse of the following matrix.

$$A = \begin{bmatrix} 14 & 12 \\ 21 & 18 \end{bmatrix}.$$

You first need to write the augmented matrix and then use the method of matrix inversion to find the inverse of the matrix. Finally, use Python to verify your answer.

1.7 Rank and Nullity of a Matrix

Recall what we have learned in the method of Gaussian elimination. The method of Gaussian elimination is to transform the augmented matrix into row echelon form. The row echelon form is the following:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} & | & b_1 \\ 0 & a_{22} & a_{23} & | & b_2 \\ 0 & 0 & a_{33} & | & b_3 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}.$$

for a $3 \times 3$ matrix. Remember that when we talk about the possibility of the system, we have mentioned that the system has no solution if the row echelon form has a row of the form $[0 \dots 0 | b]$, where $b \neq 0$. To determine whether the system has solution, one may think of observing the leading $1$ in each row. If the leading $1$ in each row is the only nonzero entry in its column, then the system has a unique solution. If the leading $1$ in each row is not the only nonzero entry in its column, then the system has infinitely many solutions.

The rank of a matrix is the number of leading $1$ in the row echelon form. For example, the rank of the following matrix is $3$.

$$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}.$$

However, the following $3 \times 3$ matrix has rank $2$.

$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 0 & 0 & 1 \end{bmatrix}.$$

One may wonder the reason why the rank of the matrix is $2$. The reason is that the second row is a multiple of the first row. Therefore, the second row can be eliminated. The rank of the matrix is then $2$.

Therefore, we may popose the following definition.

Definition(Linear Independence): A set of vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ is said to be linearly independent if the only solution to the equation $c_1\mathbf{v}_1 + \dots + c_n\mathbf{v}_n = \mathbf{0}$ is $c_1 = \dots = c_n = 0$. Otherwise, the set of vectors is said to be linearly dependent.

To illustrate the concept of linear independence, we may consider the following example. The set of vectors $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is linearly independent. This is because the only solution to the equation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$ is $c_1 = c_2 = 0$. However, the set of vectors $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is linearly dependent. This is because the equation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$ has infinitely many solutions.

The rank of a matrix is the number of linearly independent rows in the matrix. The rank of a matrix is also the number of linearly independent columns in the matrix. The rank of a matrix is also the number of linearly independent vectors in the matrix.

while on the other hand, the nullity of a matrix is the number of free variables in the system. The nullity of a matrix is also the number of dependent rows in the matrix. The nullity of a matrix is also the number of dependent columns in the matrix. The nullity of a matrix is also the number of dependent vectors in the matrix.

For example, the following matrix has rank $2$ and nullity $1$.

$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 0 & 0 & 1 \end{bmatrix}.$$

The following matrix has rank $3$ and nullity $0$.

$$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}.$$

Using Python to Find the Rank and Nullity of a Matrix

We may use the numpy library to find the rank and nullity of a matrix. The numpy.linalg.matrix_rank function can be used to find the rank of a matrix.

import numpy as np

A = np.array([[1, 2, 3], [2, 4, 6], [0, 0, 1]]) # Coefficient matrix
rank = np.linalg.matrix_rank(A) # Find the rank of the matrix
print(rank) # Print the rank of the matrix

The rank of the matrix is $2$. The nullity of the matrix is then $1$.

Exercises

1. Find the rank and nullity of the following matrix.

$$A = \begin{bmatrix} 3 & 4 & 8 \\ 1 & 2 & 3 \\ 5 & 9 & 7 \end{bmatrix}.$$

You first need to transform the matrix into row echelon form and then find the rank and nullity of the matrix. Finally, use Python to verify your answer.

1.8 Determinant of a Matrix

The determinant of a square matrix is a scalar value that is a function of the entries of the matrix. The determinant of a matrix is denoted as $\det(A)$ or $|A|$. The determinant of a $2 \times 2$ matrix is

$$\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc.$$

The determinant of a $3 \times 3$ matrix is

$$\det\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = aei + bfg + cdh - ceg - bdi - afh.$$

For a more general $n \times n$ matrix, the determinant can be found by expanding along any row or column. It is the the Laplace expansion, which is given by

$$\det(A) = \sum_{j=1}^n (-1)^{i+j}a_{ij}\det(A_{ij}),$$

where $A_{ij}$ is the matrix obtained by deleting the $i$th row and $j$th column of $A$.

Example: Find the determinant of the following matrix.

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}.$$

By applying the Laplace expansion along the first row, we have

$$\begin{align*} \det(A) &= 1 \times \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \times \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \times \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \\ &= 1 \times (5 \times 9 - 6 \times 8) - 2 \times (4 \times 9 - 6 \times 7) + 3 \times (4 \times 8 - 5 \times 7) \\ &= 1 \times (-3) - 2 \times (-6) + 3 \times (-3) \\ &= -3 + 12 - 9 \\ &= 0. \end{align*}$$

Properties of Determinants

1. The determinant of the identity matrix is $1$.
2. The determinant of a matrix is zero if the matrix has a row of zeros.
3. The determinant of a matrix is zero if the matrix has two identical rows.
4. The determinant of a matrix is zero if the matrix is singular.

Using Python to Find the Determinant of a Matrix

We may use the numpy library to find the determinant of a matrix. The numpy.linalg.det function can be used to find the determinant of a matrix.

import numpy as np

A = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) # Coefficient matrix
det = np.linalg.det(A) # Find the determinant of the matrix
print(det) # Print the determinant of the matrix

The determinant of the matrix is $0$.

Exercises

1. Find the determinant of the following matrix.

$$A = \begin{bmatrix} -1 & -2 & 2\\ 2 & 1 & 1\\ 3 & 4 & 5 \end{bmatrix}.$$

You first need to apply the Laplace expansion along the first row and then find the determinant of the matrix. Finally, use Python to verify your answer.

References

• Meyer, C. D. (2000). Matrix Analysis and Applied Linear Algebra. SIAM.
• Horn, R. A., & Johnson, C. R. (2013). Matrix Analysis. Cambridge University Press.
• Lang, S. (2002). Linear Algebra. Springer.