System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. In general, a system with \(m\) equations and \(n\) variables can be represented as:

\[ \begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m \end{cases} \]

This is called a rectangular system if the number of equations (\(m\)) is not equal to the number of variables (\(n\)). However, we are particularly interested in square systems, where the number of equations equals the number of variables (\(m = n\)).

In a square system, the equations can be represented as:

\[ \begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\ \vdots \\ a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n \end{cases} \]

For example, you may recall from class 10 how to solve a system of two linear equations with two variables, such as:

\[ \begin{cases} a_1x + b_1y = d_1 \\ a_2x + b_2y = d_2 \end{cases} \]

where \(x\) and \(y\) are the variables, and \(a_1\), \(b_1\), \(d_1\), etc., are constants.

In this section, we will primarily focus on solving square systems with three variables, which involves three linear equations of the form:

\[ \begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases} \]

where \(x\), \(y\), and \(z\) are the variables we aim to solve for, and the constants \(a_i\), \(b_i\), \(c_i\), and \(d_i\) are known values.

Solving a system of linear equations in three variables can be done using methods such as substitution, elimination, or matrix methods (like Cramer's Rule or Gaussian elimination (that we already discussed in the beginning)), which provide systematic approaches to find the values of \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.'

Consistency

A consistent system of linear equations is a system that has at least one solution. This means that there exists at least one set of values for the unknowns that satisfies all the equations simultaneously. Consistent systems can be further classified into:

Uniquely consistent: A system that has exactly one solution.
Infinitely consistent: A system that has infinitely many solutions.

For example:

\[ x + y = 2 \]

\[ 2x + 2y = 4 \]

This system is consistent with infinitely many solutions because the second equation is a multiple of the first, meaning the two equations represent the same line.

An inconsistent system of linear equations is a system that has no solution. In such a system, there are no values for the unknowns that can simultaneously satisfy all the equations. This typically occurs when the equations represent parallel lines or planes that do not intersect.

For example:

\[ x + y = 2 \]

\[ x + y = 3 \]

This system is inconsistent because the two equations represent parallel lines that never intersect. Therefore, there are no values of \(x\) and \(y\) that satisfy both equations simultaneously.

Cramer's Rule

Let's write down the steps more mathematically. We start with the system:

\[ \begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases} \]

Let the determinant \(\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}\).

To proceed, we multiply each equation by the cofactor of the respective \(a_i\) in \(\Delta\):

Multiply the first equation by the cofactor \(C_{11}\) of \(a_1\):

\[ C_{11} (a_1 x + b_1 y + c_1 z) = C_{11} d_1 \]
Multiply the second equation by the cofactor \(C_{21}\) of \(a_2\):

\[ C_{21} (a_2 x + b_2 y + c_2 z) = C_{21} d_2 \]
Multiply the third equation by the cofactor \(C_{31}\) of \(a_3\):

\[ C_{31} (a_3 x + b_3 y + c_3 z) = C_{31} d_3 \]

Now, add these three equations:

\[ C_{11} (a_1 x + b_1 y + c_1 z) + C_{21} (a_2 x + b_2 y + c_2 z) + C_{31} (a_3 x + b_3 y + c_3 z) = C_{11} d_1 + C_{21} d_2 + C_{31} d_3 \]

Expanding the left-hand side:

\[ (C_{11} a_1 + C_{21} a_2 + C_{31} a_3) x + (C_{11} b_1 + C_{21} b_2 + C_{31} b_3) y + (C_{11} c_1 + C_{21} c_2 + C_{31} c_3) z \]

By properties of determinants:

\((C_{11} a_1 + C_{21} a_2 + C_{31} a_3) = \Delta\),
\((C_{11} b_1 + C_{21} b_2 + C_{31} b_3) = 0\),
\((C_{11} c_1 + C_{21} c_2 + C_{31} c_3) = 0\).

Thus, the equation reduces to:

\[ \Delta x = C_{11} d_1 + C_{21} d_2 + C_{31} d_3 \]

The right-hand side, \(C_{11} d_1 + C_{21} d_2 + C_{31} d_3\), can be recognized as the determinant \(\Delta_1\), obtained by replacing the first column of \(\Delta\) with \(d_1\), \(d_2\), and \(d_3\):

\[ \Delta x = \Delta_1 \]

Similarly, by replacing the second column with \(d_1\), \(d_2\), and \(d_3\), we get \(\Delta y = \Delta_2\), and by replacing the third column, we obtain \(\Delta z = \Delta_3\).

Case 1: If \(\Delta \neq 0\), then we can solve for \(x\), \(y\), and \(z\) as follows:

\[ x = \frac{\Delta_1}{\Delta} = \frac{\begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}} \]

Similarly, we have:

\[ y = \frac{\Delta_2}{\Delta} = \frac{\begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}} \]

\[ z = \frac{\Delta_3}{\Delta} = \frac{\begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}} \]

Thus, when \(\Delta \neq 0\), the system has a unique solution given by:

\[ x = \frac{\Delta_1}{\Delta}, \quad y = \frac{\Delta_2}{\Delta}, \quad z = \frac{\Delta_3}{\Delta} \]

Case 2: If \(\Delta = 0\) but at least one of \(\Delta_1\), \(\Delta_2\), or \(\Delta_3\) is not zero, the system has no solution. Here’s why:

Suppose \(\Delta_1 \neq 0\). Then, according to Cramer’s Rule:

\[ \Delta \cdot x = \Delta_1 \]

Since \(\Delta = 0\), this implies:

\[ 0 \cdot x = \Delta_1 \]

which simplifies to:

\[ 0 = \Delta_1 \]

However, this leads to a contradiction because we assumed \(\Delta_1 \neq 0\). This contradiction indicates that there are no values of \(x\), \(y\), and \(z\) that satisfy all three equations. Thus, the system is inconsistent, and no solution exists.

The same reasoning applies if \(\Delta_2 \neq 0\) or \(\Delta_3 \neq 0\). In each case, we get a similar contradiction, confirming that the system cannot have a solution when \(\Delta = 0\) but any of \(\Delta_1\), \(\Delta_2\), or \(\Delta_3\) is non-zero.

Case 3: If \(\Delta = 0\), \(\Delta_1 = 0\), \(\Delta_2 = 0\), and \(\Delta_3 = 0\), then the system may have infinitely many solutions or no solution, depending on whether the equations are consistent or inconsistent.

To see this, let's start by considering that Cramer's Rule would give us:

\[ \Delta \cdot x = \Delta_1, \quad \Delta \cdot y = \Delta_2, \quad \Delta \cdot z = \Delta_3 \]

With \(\Delta = 0\), \(\Delta_1 = 0\), \(\Delta_2 = 0\), and \(\Delta_3 = 0\), we get:

\[ 0 \cdot x = 0, \quad 0 \cdot y = 0, \quad 0 \cdot z = 0 \]

These equations hold true for any values of \(x\), \(y\), and \(z\), meaning there could be infinitely many solutions. This situation suggests that the equations are dependent, allowing multiple values of \(x\), \(y\), and \(z\) to satisfy the system.

However, this is not always the case. In some instances, even with \(\Delta = 0\), \(\Delta_1 = 0\), \(\Delta_2 = 0\), and \(\Delta_3 = 0\), the system may still be inconsistent. For example, consider the system:

\[ x + y + z = 1 \]

\[ x + y + z = 2 \]

\[ x + y + z = 3 \]

For this system, we calculate:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0, \quad \Delta_1 = 0, \quad \Delta_2 = 0, \quad \Delta_3 = 0 \]

At first glance, this might suggest infinite solutions. However, let’s examine the equations more closely. If \((x_0, y_0, z_0)\) is a solution to this system, then:

\[ x_0 + y_0 + z_0 = 1 \]

\[ x_0 + y_0 + z_0 = 2 \]

\[ x_0 + y_0 + z_0 = 3 \]

These conditions imply that \(1 = 2 = 3\), which is a clear contradiction. This means that there are no values of \(x\), \(y\), and \(z\) that can satisfy all three equations simultaneously. Therefore, in this example, the system has no solution.

Why does a system has infinite solutions?

A system of linear equations has infinitely many solutions when the equations are dependent on each other, meaning that at least one equation can be derived from the others and does not provide new information. This situation typically arises when the number of equations is less than the number of unknowns.

For example, consider the following two equations with three unknowns \(x\), \(y\), and \(z\):

\[ x + y + z = 1 \]

\[ x + 2y + 3z = 3 \]

Let's analyze this system by expressing one variable in terms of a parameter. Suppose we let \(z = \lambda\), where \(\lambda \in \mathbb{R}\). Substituting \(z = \lambda\) into the equations, we get:

From the first equation:

\[ x + y = 1 - \lambda \]
From the second equation:

\[ x + 2y = 3 - 3\lambda \]

Now, subtract the first equation from the second to solve for \(y\):

\[ (x + 2y) - (x + y) = (3 - 3\lambda) - (1 - \lambda) \]

\[ y = 2 - 2\lambda \]

Substitute \(y = 2 - 2\lambda\) back into \(x + y = 1 - \lambda\) to find \(x\):

\[ x + (2 - 2\lambda) = 1 - \lambda \]

\[ x = -1 + \lambda \]

Thus, the solutions are:

\[ x = -1 + \lambda, \quad y = 2 - 2\lambda, \quad z = \lambda \]

where \(\lambda\) can take any real value, giving infinitely many solutions for \(x\), \(y\), and \(z\).

This occurs because the system is underdetermined (fewer independent equations than unknowns). If we try to create a third equation by adding the two given equations, we obtain:

\[ (x + y + z) + (x + 2y + 3z) = 1 + 3 \]

\[ 2x + 3y + 4z = 4 \]

This new equation, \(2x + 3y + 4z = 4\), contains no additional information about \(x\), \(y\), and \(z\), as it is simply a linear combination of the first two equations. We call this equation dependent on the first two.

Thus, our system:

\[ x + y + z = 1, \quad x + 2y + 3z = 3, \quad 2x + 3y + 4z = 4 \]

has three equations but infinitely many solutions, because the equations are not independent. We refer to this as a dependent system of linear equations.

How to remove the ambiguity in case III

To distinguish whether a system has infinitely many solutions or no solution, we examine the cofactors of \(\Delta\).

If at least one of the cofactors of \(\Delta\) is non-zero, then the system has infinitely many solutions.
If all the cofactors of \(\Delta\) are zero, then we proceed by examining the cofactors of \(\Delta_1\), \(\Delta_2\), and \(\Delta_3\):
- If all the cofactors of \(\Delta_1\), \(\Delta_2\), and \(\Delta_3\) are zero, then the system has infinitely many solutions.
- If at least one of the cofactors of \(\Delta_1\), \(\Delta_2\), or \(\Delta_3\) is non-zero, then the system has no solution.

Summary

Case 1: If \(\Delta \neq 0\), the system has a unique solution given by \( x = \frac{\Delta_1}{\Delta}\), \( y = \frac{\Delta_2}{\Delta}\), and \( z = \frac{\Delta_3}{\Delta}\).
Case 2: If \(\Delta = 0\) but at least one of \(\Delta_1\), \(\Delta_2\), or \(\Delta_3\) is non-zero, the system has no solution (inconsistent).
Case 3: If \(\Delta = 0\) and \(\Delta_1 = \Delta_2 = \Delta_3 = 0\), the system may have infinitely many solutions (consistent and dependent) or no solution. To distinguish, examine the cofactors:
- If any cofactor of \(\Delta\) is non-zero, there are infinitely many solutions.
- If all cofactors of \(\Delta\) are zero, check the cofactors of \(\Delta_1\), \(\Delta_2\), and \(\Delta_3\). If they’re all zero, there are infinitely many solutions; if not, the system has no solution.

Cramer's Rule for a Homogeneous System:

For a homogeneous system of linear equations, such as:

\[ a_1 x + b_1 y + c_1 z = 0 \]

\[ a_2 x + b_2 y + c_2 z = 0 \]

\[ a_3 x + b_3 y + c_3 z = 0 \]

the solution \(x = y = z = 0\) is always a solution. This is known as the trivial solution, and it can be easily verified. Thus, a homogeneous system is always consistent.

The behavior of the system depends on the value of \(\Delta\), the determinant of the coefficient matrix:

Case 1: If \(\Delta \neq 0\), then the only solution is the trivial solution (\(x = y = z = 0\)), making it the unique solution.
Case 2: If \(\Delta = 0\), the trivial solution still exists, but the system has infinitely many non-zero solutions (non-trivial solutions). This means there are infinitely many values of \(x\), \(y\), and \(z\) that satisfy the system, in addition to the trivial solution.

Solving System of Linear Equations using Matrices

Consider the system of linear equations:

\[ a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n = b_1 \]

\[ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n = b_2 \]

\[ \vdots \]

\[ a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nn} x_n = b_n \]

This system can be written in matrix form as \( \mathbf{A} \mathbf{x} = \mathbf{b} \), where:

\[ \mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} \]

Since this is a system of \(n\) equations in \(n\) unknowns, \(\mathbf{A}\) is a square matrix.

If \(\mathbf{A}\) is non-singular (\(|\mathbf{A}| \neq 0\)):
- The inverse of \(\mathbf{A}\) exists.
- Multiply both sides of the equation \(\mathbf{A} \mathbf{x} = \mathbf{b}\) by \(\mathbf{A}^{-1}\) to get:
  
  \[ \mathbf{x} = \mathbf{A}^{-1} \mathbf{b} \]
- Thus, the solution is given by \(\mathbf{x} = \mathbf{A}^{-1} \mathbf{b}\).
If \(\mathbf{A}\) is singular (\(|\mathbf{A}| = 0\)):
- Calculate \(\operatorname{adj}(\mathbf{A})\), the adjugate of \(\mathbf{A}\).
- Check the result of \(\operatorname{adj}(\mathbf{A}) \mathbf{b}\):
  - If \(\operatorname{adj}(\mathbf{A}) \mathbf{b} \neq \mathbf{0}\): No solution exists, and the system is inconsistent.
  - If \(\operatorname{adj}(\mathbf{A}) \mathbf{b} = \mathbf{0}\): The system may have infinitely many solutions or no solution.