System of Linear Equations

A system of linear equations is a set of 2 or more linear equations involving the same set of variables. 

• 3x – y = 2 and x + 4y = 1 are linear equations in 2 variables

Together, they are called a system of linear equations.

Similarly,

• 2x + 5y – z = 7 and x – y + 2z = 3 are linear equations in 3 variables

These equations represent straight lines in a coordinate system, and their solutions correspond to points where these lines intersect. 

Formula

The general form of a linear equation in n variables is expressed as:

a₁₁x₁ + a₁₂x₂ + … + a_1nx_n = b₁

a₂₁x₁ + a₂₂x₂ + … + a_2nx_n = b₂

… … … … … … … 

a_n1x₁ + a_n2x₂ + … + a_nnx_n = b_n

Here,

• x₁, x₂, …, x_n are the variables
• a₁, a₂, …, a_n are real numbers (called the coefficients of the variables)
• b is the constant term

The solutions to a system are the values of x₁, x₂, … that satisfy all equations simultaneously.

Types of Solutions

A system of linear equations normally has a single solution, but sometimes, it can have no solution or infinite solutions depending on the relationship between the equations. We can verify these solutions either algebraically or graphically:

Unique Solution (Consistent and Independent System)

A system is said to have a unique solution if it has exactly one solution. This occurs when the equations represent two lines that intersect at a single point.

For example, the system 2x + y = 5, x – y = 1 has the solution (2, 1) 

No Solution (Inconsistent System)

A system has no solution if the equations represent parallel lines and thus can never intersect.

For example, the system 2x + y = 3, 4x + 2y = 8 has no solution 

Infinite Solutions (Consistent and Dependent System)

A system has infinitely many solutions when a system of equations is represented by the same line, which means every point on the line is a solution.

For example, 2x + 4y = 8, x + 2y = 4 are identical, and thus the system has infinitely many solutions.

Solving

We use different methods to solve a system of linear equations. It depends on the complexity of the system and the way we want to solve it.

By Graphing

This method provides a visual way to find the solution. Each equation is plotted as a straight line on a coordinate plane. The point where the lines intersect represents the solution to the system. 

Let us consider the system:

y = 2x + 1 …..(i)

y = -x + 4 …..(ii)

Step 1: Graphing the First Equation

From equation (i),

y = 2x + 1

Here, the slope m is 2, and the y-intercept c is 1

Thus, the coordinates of the y-intercept are (0, 1)

Now, using the slope and y-intercept, we get 

x	2x + 1	y
-1	2(-1) + 1	-1
0	2(0) + 1	1
1	2(1) + 1	3

After plotting these points, we get

Step 2: Graphing the Second Equation

From equation (ii),

y = -x + 4

Here, the slope m is -1, and the y-intercept c is 4

Thus, the coordinates of the y-intercept are (0, 4)

Now, using the slope and y-intercept, we get 

x	-x + 4	y
-1	-(-1) + 4	5
0	-(0) + 4	4
1	-(1) + 4	3

After plotting these points, we get

Step 3: Identifying the Intersection Point

Now, observing the graphs of these two lines, we conclude

The two lines intersect at (1, 3).

This means the solution to the system is x = 1 and y = 3

Here, since two lines intersect at one point (1, 3), the given system has one unique solution.

Note: Graphing may not always give accurate solutions, especially when working with decimals or fractions.

By Substitution 

In this method, we solve one equation for a variable in terms of the other and substitute it into the second equation. This transforms the system into a single equation with one variable which makes the system easier to solve. Once we find the value of the first variable, we substitute it back into the original equation to determine the second variable.

Let us consider the system:

x + y = 6 …..(i)

2x – y = 3 …..(ii)

Step 1: Solving for one variable in Terms of the Other

First, we solve for x in terms of y.

From equation (i), 

x = 6 – y …..(iii)

Step 2: Substituting 

Now, substituting (iii) into the equation (ii),

2(6 – y) – y = 3

⇒ 12 – 2y – y = 3

⇒ 12 – 3y = 3

⇒ 3y = 12 – 3

⇒ 3y = 9

⇒ y = 3

Step 3: Solving for x

Substituting y = 3 back into (iii), we get

x = 6 – 3

⇒ x = 3

Thus, the solution is x = 3 and y = 3

By Elimination (Adding or Subtracting)

This method involves adding or subtracting the given equations to eliminate one variable, simplifying the system into a single equation with one variable. Once we solve for that variable, we substitute its value back into one of the original equations to find the remaining variable(s).

Let us consider the system:

3x + 2y = 12 …..(i)

5x – 2y = 8 …..(ii)

Step 1: Adding the Equations

(3x + 2y) + (5x – 2y) = 12 + 8

⇒ 3x + 2y + 5x – 2y = 20

Step 2: Simplifying

⇒ 3x + 5x = 20

⇒ 8x = 20

⇒ x = ${\dfrac{20}{8}}$

⇒ x = ${\dfrac{5}{2}}$

Step 3: Substituting

Substituting x = ${\dfrac{5}{2}}$ back into equation (i), we get

⇒ ${3\left( \dfrac{5}{2}\right) +2y=12}$

⇒ ${\dfrac{15}{2}+2y=12}$

⇒ ${2y=12-\dfrac{15}{2}}$

⇒ ${2y=\dfrac{24-15}{2}}$

⇒ ${2y=\dfrac{9}{2}}$

⇒ ${y=\dfrac{9}{4}}$

Thus, the solution is x = ${\dfrac{5}{2}}$ and ${y=\dfrac{9}{4}}$

By Inverse Matrix Method 

The inverse matrix method is particularly useful for solving complex or larger systems of linear equations. In this approach, the system is rewritten in matrix form as:

AX = B

To solve the system, we find the matrix X, which represents the values of the unknowns.

Here,

• A is the coefficient matrix containing the coefficients of the variables
• X is the variable matrix representing the unknown variables
• B is the constant matrix containing the constant terms from the right-hand side of the equations

The solution can be found using the formula:

• X = A^-1B

Note: Since A^-1 exists, A is an invertible matrix.

Now, let us solve the system of linear equations:

2x + 3y = 8 …..(i)

4x + y = 5 …..(ii)

Step 1: Expressing the System in Matrix Form

To organize the equations in a standard way, we rewrite the equations (i) and (ii) in the matrix form as:

AX = B

⇒ ${\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 8 \\ 5 \end{bmatrix}}$

Step 2: Finding the Inverse of Matrix A

To compute the inverse matrix of A, we first find the determinant A. If A is invertible (i.e., whether an inverse matrix exists):

• If det(A) = 0, the system has no unique solution, which means the system may be inconsistent or have infinitely many solutions.
• If det(A) ≠ 0, we proceed to the next step.

Now, for the given matrix A = ${\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}}$, 

det(A) = (2)(1) – (3)(4) = 2 – 12 = -10 ≠ 0, the inverse exists.

Since the inverse of a 2 × 2 matrix A = ${\begin{bmatrix} a & b \\ c & d \end{bmatrix}}$

A^-1 = ${\dfrac{1}{\det \left( A\right) }\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}}$, where det(A) = (ad – bc)

⇒ A^-1 = ${\dfrac{1}{-10}\begin{bmatrix} 1 & -3 \\ -4 & 2 \end{bmatrix}}$ = ${\begin{bmatrix} -0.1 & 0.3 \\ 0.4 & -0.2 \end{bmatrix}}$

Step 3: Multiplying A⁻¹ by B to Find X

Thus, by calculating the variable matrix, we get

X = A^-1B

⇒ ${\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} -0.1 & 0.3 \\ 0.4 & -0.2 \end{bmatrix}\begin{bmatrix} 8 \\ 5 \end{bmatrix}}$

Multiplying the matrices, we get

x = (-0.1 × 8) + (0.3 × 5) = -0.8 + 1.5 = 0.7

y = (0.4 × 8) + (-0.2 × 5) = 3.2 – 1 = 2.2

Thus, the solution to the system is x = 0.7 and y = 2.2

By Gaussian Elimination Method

The Gaussian elimination method is often used for larger systems with three or more equations. It involves converting the system into row echelon form and then using back-substitution to find the variable values.

Let us solve the system of equations:

2x + y – z ​= 8 …..(i)

-3x – y + 2z = -11 …..(ii)

-2x + y + 2z = -3 …..(iii)

Step 1: Converting the System of Equations into an Augmented Matrix

To organize the equations systematically, we express the system as an augmented matrix. It is a matrix where the column of constants is added to the right side of the coefficient matrix. 

${\left[\begin{array}{ccc|c}2 & 1 & -1 & 8 \\-3 & -1 & 2 & -11 \\-2 & 1 & 2 & -3\end{array}\right]}$

Step 2: Forwarding Elimination to Convert to Row Echelon Form

To find a solution for one variable, we eliminate the values of the other variables by replacing the rows of the augmented matrix. 

Here, we create zeros below the main diagonal, converting the matrix into an upper triangular form.

By replacing rows into the forms:

${R_{2}\leftarrow R_{2}+\dfrac{3}{2}R_{1}}$

${R_{3}\leftarrow R_{3}+R_{1}}$

After performing these operations, we get

${\left[\begin{array}{ccc|c}2 & 1 & -1 & 8 \\0 & 0.5 & 0.5 & -7 \\0 & 2 & 1 & 5\end{array}\right]}$

Eliminating values until we reach row echelon form,

${\left[\begin{array}{ccc|c}2 & 1 & -1 & 8 \\0 & 2 & 1 & 5 \\0 & 0 & 1 & -1\end{array}\right]}$

Step 3: Back-Substituting to Solve for Variables

If we have the system in upper triangular form, we solve for variables starting from the last row,

From the last row,

z = -1

Substituting z = -1 into the second row,

2y + (-1) = 5

⇒ 2y = 6

⇒ y = 3

Substituting y = 3 and z = -1 into the first row,

2x + 3 – (-1) = 8

⇒ 2x + 4 = 8

⇒ x = 2

Thus, the solution is (x, y, z) = (2, 3, -1)

By Cramer’s Rule

This method is used when the system of equations has the same number of equations as unknowns, and the determinant of the coefficient matrix is nonzero.

To solve a system of n linear equations with n variables, we use the formula:

${x_{i}=\dfrac{\det \left( A_{i}\right) }{\det A}}$

Here, 

• A is the coefficient matrix
• A_i is the matrix formed by replacing the i^th column of A with the constant matrix B
• det(A) is the determinant of the coefficient matrix

Let us solve the equations:

2x + 3y = 5 …..(i)

4x – y = 1 …..(ii)

Step 1: Expressing the System in Matrix Form

In this method, we first express the system in organized matrix form AX = B

Here, 

A = ${\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}}$, X = ${\begin{bmatrix} x \\ y \end{bmatrix}}$, and B = ${\begin{bmatrix} 5 \\ 1 \end{bmatrix}}$

Step 2: Calculating the Determinant of A 

• If det(A) = 0, Cramer’s rule does not work
• If det (A) ≠ 0, Cramer’s rule applies, and we will get a unique solution

Here, 

det(A) = (2)(-1) – (3)(4) = -2 – 12 = -14

Step 3: Calculating the Determinants of Modified Matrices A_x and A_y 

Here, we replace columns to isolate the effect of each variable on the final solution.

To find x, we replace the first column of A with B:

A_x = ${\begin{bmatrix} 5 & 3 \\ 1 & -1 \end{bmatrix}}$

Thus, det(A_x) = (5)(-1) – (3)(1) = -5 – 3 = -8

To find y, replace the second column of A with B:

A_y = ${\begin{bmatrix} 2 & 5 \\ 4 & 1 \end{bmatrix}}$

Thus, det(A_y) = (2)(1) – (5)(4) = 2 – 20 = -18

Step 4: Calculating the Values of x and y

By using Cramer’s Rule, we get

x = ${\dfrac{\det \left( A_{x}\right) }{\det A}}$ = ${\dfrac{-8}{-14}}$ = ${\dfrac{4}{7}}$

y = ${\dfrac{\det \left( A_{y}\right) }{\det A}}$ = ${\dfrac{-18}{-14}}$ = ${\dfrac{9}{7}}$

Thus, the solutions are: x = ${\dfrac{4}{7}}$ and y = ${\dfrac{9}{7}}$