System of Linear Equations

Overview

A system of linear equations represents one of the most fundamental and widely applied mathematical constructs in engineering and the computational sciences. Its principles form the bedrock of numerous advanced topics, from the analysis of electrical circuits and data networks to the core algorithms of machine learning, computer graphics, and optimization. For the Computer Science engineer, a mastery of linear systems is not merely an academic exercise; it is an indispensable tool for modeling and solving a vast array of computational problems. The ability to systematically analyze and solve these systems is a prerequisite for understanding the theoretical underpinnings of many modern technologies.

In the context of the GATE examination, questions derived from this chapter are a consistent feature, testing both conceptual understanding and procedural skill. This chapter is designed to provide a rigorous and structured approach to the topic, focusing squarely on the competencies required for the examination. We shall first investigate the conditions under which a system of equations possesses a solution—a concept known as consistency. We will establish the criteria for determining whether a system has a unique solution, infinitely many solutions, or no solution at all, primarily through the powerful concept of matrix rank. Subsequently, we will turn our attention to the algorithmic methods for finding these solutions, equipping you with the practical techniques to solve problems efficiently and accurately.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Consistency and Solutions | Analyzing the existence and nature of solutions. |
| 2 | Solving Linear Systems | Applying methods to find explicit system solutions. |

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Learning Objectives

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We now turn our attention to Consistency and Solutions...

Part 1: Consistency and Solutions

Introduction

A system of linear equations represents a collection of linear relationships between multiple variables. In engineering and computer science, such systems arise in a vast array of applications, from circuit analysis and network flow problems to machine learning models and computer graphics. The central task is not merely to solve these systems, but to first understand the nature of their solution sets. A system may possess exactly one solution, an infinite number of solutions, or no solution at all.

This chapter provides a rigorous framework for analyzing any system of linear equations. We will move beyond ad-hoc methods of substitution and elimination to employ the powerful concepts of matrix rank. By understanding the relationship between the rank of the coefficient matrix and the augmented matrix, we can determine the existence and uniqueness of solutions without explicitly computing them. We will also explore the special properties of homogeneous systems and introduce the fundamental Rank-Nullity Theorem, a cornerstone of linear algebra with significant implications for understanding solution spaces.

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Key Concepts

1. The Augmented Matrix and Rank

To analyze a system, we combine the coefficient matrix $A$ and the constant vector $\mathbf{b}$ into a single matrix.

The concept of rank is central to determining the nature of the solution. The rank of a matrix is the maximum number of linearly independent rows (or columns) in the matrix. It can be found by reducing the matrix to its row echelon form and counting the number of non-zero rows. We denote the rank of a matrix $A$ as $\operatorname{rank}(A)$ or $\rho(A)$.

2. The Rouché–Capelli Theorem: A Unified Framework

The relationship between the rank of the coefficient matrix $A$ and the augmented matrix $[A|\mathbf{b}]$ completely characterizes the solution set of the system $A\mathbf{x} = \mathbf{b}$. This fundamental result is summarized by the Rouché–Capelli theorem.

Let us consider a system with $m$ variables. The conditions are as follows:

No Solution (Inconsistent System): If the rank of the coefficient matrix is less than the rank of the augmented matrix, the system is inconsistent.

This situation arises when the row reduction process leads to a row of the form $[0 \ 0 \ \cdots \ 0 \ | \ c]$ where $c \neq 0$, which corresponds to the absurd equation $0 = c$.

Consistent System: If the ranks are equal, the system is consistent and has at least one solution.

Within this case, there are two possibilities:
* Unique Solution: If the common rank $r$ is equal to the number of variables $m$, the system has exactly one solution.

* Infinite Solutions: If the common rank $r$ is less than the number of variables $m$, the system has infinitely many solutions.

In this case, there will be $m-r$ free variables (or parameters), which can be chosen arbitrarily, and the remaining $r$ variables will be expressed in terms of them.

The following diagram illustrates this decision process:

Start with system $A\mathbf{x}=\mathbf{b}$


Is $\operatorname{rank}(A) = \operatorname{rank}([A|\mathbf{b}])$?
Let common rank be $r$.

Yes
No


No Solution



Is $r = m$?
($m$ = number of variables)

Yes
No


Unique Solution



Infinite Solutions

Worked Example:

Problem: For what value of $\lambda$ will the following system of equations have no solution?

Solution:

Step 1: Write the augmented matrix $[A|\mathbf{b}]$.

Step 2: Reduce the matrix to row echelon form using elementary row operations.

Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$.

Step 3: Continue the row reduction.

Apply $R_3 \rightarrow R_3 - R_2$.

Step 4: Analyze the ranks based on the value of $\lambda$.

For the system to have no solution, we require $\operatorname{rank}(A) < \operatorname{rank}([A|\mathbf{b}])$.
From the echelon form, we observe that if the term $\lambda-3$ becomes zero, the last row of the coefficient matrix part will be all zeros.

Let $\lambda - 3 = 0$, which means $\lambda = 3$.

If $\lambda = 3$, the matrix becomes:

In this case, $\operatorname{rank}(A) = 2$ (two non-zero rows in the $A$ part), but $\operatorname{rank}([A|\mathbf{b}]) = 3$ (three non-zero rows in the full augmented matrix).
Since $\operatorname{rank}(A) < \operatorname{rank}([A|\mathbf{b}])$, the system is inconsistent.

Answer: The system has no solution when $\lambda = 3$.

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## 3. Homogeneous Systems ($A\mathbf{x} = \mathbf{0}$)

Homogeneous systems are a special case of profound importance. Since $\mathbf{b} = \mathbf{0}$, the augmented matrix is $\left[ A \middle| \mathbf{0} \right]$. Adding a column of zeros never changes the rank of a matrix, so we always have $\operatorname{rank}(A) = \operatorname{rank}\left[ A \middle| \mathbf{0} \right]$.

The critical question for a homogeneous system is whether a non-trivial solution (a solution where at least one variable is non-zero) exists. Applying the Rouché-Capelli conditions:

• Trivial Solution Only: A unique solution exists if $\operatorname{rank}(A) = m$ (number of variables). Since the trivial solution is always one possibility, this must be the only one.


• Infinite (Non-trivial) Solutions: Infinite solutions exist if $\operatorname{rank}(A) < m$. This is the condition for the existence of non-trivial solutions.

The set of all solutions to a homogeneous system $A\mathbf{x} = \mathbf{0}$ forms a vector space called the null space or kernel of matrix $A$.

#
## 4. The Rank-Nullity Theorem

This theorem provides a direct relationship between the rank of a matrix and the dimension of its null space.

A key consequence of this theorem arises when a system has more variables than equations ($m > n$).
For an $n \times m$ matrix $A$, we know that $\operatorname{rank}(A) \le \min(n, m)$. If $m > n$, then $\operatorname{rank}(A) \le n$.
Using the Rank-Nullity Theorem:

Since $\operatorname{rank}(A) \le n$, it follows that:

Because $m > n$, we have $m-n \ge 1$. Therefore, the $\operatorname{nullity}$ must be at least 1. This proves that any homogeneous system with more variables than equations must have a non-trivial solution.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Consider the system of linear equations $2x - y + 3z = a$, $3x + 2y + 5z = b$, and $x - 4y + z = c$. This system is consistent if:" options=[" $c = 2a - b$ "," $a = 2b - c$ "," $b = 2a - c$ "," $c = a - 2b$ "] answer="$c = 2a - b$" hint="A system is consistent if the same row operations that make a row of $A$ zero also make the corresponding element in $b$ zero. Check for linear dependence in the rows of the coefficient matrix." solution="
Step 1: Write the augmented matrix for the system.

Step 2: Observe the relationship between the rows of the coefficient matrix $A$. Let the rows be $R_1, R_2, R_3$. We test for linear dependence. Let's see if $R_3$ can be written as a combination of $R_1$ and $R_2$.
Suppose $k_1 R_1 + k_2 R_2 = R_3$.
For the first column: $2k_1 + 3k_2 = 1$.
For the second column: $-k_1 + 2k_2 = -4$.
Solving these two equations: Multiply the second by 2: $-2k_1 + 4k_2 = -8$. Add this to the first equation: $7k_2 = -7 \implies k_2 = -1$.
Substitute $k_2 = -1$ into the first equation: $2k_1 - 3 = 1 \implies 2k_1 = 4 \implies k_1 = 2$.
So, we have found that $2R_1 - R_2 = R_3$. Let's check this for the third column: $2(3) - 5 = 6 - 5 = 1$. This matches the third element of $R_3$.
Thus, the third row of $A$ is linearly dependent on the first two: $2R_1 - R_2 = R_3$.

Step 3: Apply the consistency condition.
For the system to be consistent, the same linear dependency must hold for the vector $\mathbf{b}$. The $\operatorname{rank}$ of $A$ will be equal to the $\operatorname{rank}$ of $[A|\mathbf{b}]$ only if the relationship holds for the last column as well.

Result: The condition for consistency is $c = 2a - b$.
Answer: $\boxed{c = 2a - b}$
"
:::

:::question type="NAT" question="Let $M$ be a $5 \times 7$ real matrix. If the dimension of the solution space of $M\mathbf{x} = \mathbf{0}$ is $3$, then the rank of $M$ is _________." answer="4" hint="Use the Rank-Nullity Theorem. Identify the number of columns and the given $\operatorname{nullity}$." solution="
Step 1: Identify the given parameters from the problem statement.
The matrix $M$ is of size $n \times m$, where $n=5$ and $m=7$.
The number of columns is $m = 7$.
The dimension of the solution space of $M\mathbf{x} = \mathbf{0}$ is the $\operatorname{nullity}$ of $M$.
So, $\operatorname{nullity}(M) = 3$.

Step 2: Apply the Rank-Nullity Theorem.
The theorem states: $\operatorname{rank}(M) + \operatorname{nullity}(M) = m$.

Step 3: Substitute the known values and solve for the rank.

Result: The $\operatorname{rank}$ of the matrix $M$ is 4.
Answer: $\boxed{4}$
"
:::

:::question type="MSQ" question="Consider the homogeneous system of linear equations $A\mathbf{x} = \mathbf{0}$, where $A$ is a $3 \times 5$ matrix with $\operatorname{rank}(A) = 2$. Which of the following statements is/are necessarily TRUE?" options=["The system has a unique solution.","The system has exactly 3 linearly independent solutions.","Any solution can be written as a linear combination of 3 basis vectors.","The system has a non-trivial solution."] answer="The system has exactly 3 linearly independent solutions.,Any solution can be written as a linear combination of 3 basis vectors.,The system has a non-trivial solution." hint="Apply the Rank-Nullity theorem to find the dimension of the null space. A non-trivial solution exists if the $\operatorname{nullity}$ is greater than 0." solution="
Let's analyze each option based on the given information: $A$ is a $3 \times 5$ matrix, so $n=3$ and $m=5$. We are given $\operatorname{rank}(A) = 2$.

1. Analyze the nullity:
Using the Rank-Nullity Theorem: $\operatorname{rank}(A) + \operatorname{nullity}(A) = m$.
$2 + \operatorname{nullity}(A) = 5$
$\operatorname{nullity}(A) = 3$.
The $\operatorname{nullity}$ is the dimension of the solution space (null space).

2. Evaluate the options:

• "The system has a unique solution." This is FALSE. A unique solution (the trivial one) exists only if $\operatorname{nullity}(A) = 0$ (or $\operatorname{rank}(A)=m$). Here, $\operatorname{nullity}(A) = 3$, so there are infinite solutions.

• "The system has exactly 3 linearly independent solutions." This is TRUE. The $\operatorname{nullity}$ represents the maximum number of linearly independent solutions. A basis for the null space will contain exactly 3 vectors.

• "Any solution can be written as a linear combination of 3 basis vectors." This is TRUE. This is the definition of a basis for a vector space. Since the dimension of the solution space is 3, there exists a basis of 3 vectors, and every solution is a linear combination of these basis vectors.

• "The system has a non-trivial solution." This is TRUE. Since $\operatorname{nullity}(A) = 3 > 0$, there are infinitely many non-trivial solutions.

Result: The correct options are B, C, and D.
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:::

:::question type="MCQ" question="For the system of equations $x + 2y = 3$ and $2x + ay = 6$, the condition for having an infinite number of solutions is:" options=[" $a = 2$ "," $a = 4$ "," $a \neq 2$ "," $a \neq 4$ "] answer="$a = 4$" hint="For a non-homogeneous system to have infinite solutions, the $\operatorname{rank}$ of the coefficient matrix and the augmented matrix must be equal, and this $\operatorname{rank}$ must be less than the number of variables." solution="
Step 1: Write the augmented matrix for the system.

Step 2: Reduce the matrix to row echelon form.

Apply the row operation $R_2 \rightarrow R_2 - 2R_1$.

Step 3: Analyze the conditions for infinite solutions.
For infinite solutions, we need $\operatorname{rank}(A) = \operatorname{rank}([A|\mathbf{b}]) < m$.
Here, the number of variables $m=2$. So we need the common $\operatorname{rank}$ to be 1.
This occurs when the second row becomes all zeros.
For the coefficient part, we need $a-4 = 0$, which gives $a=4$.
For the augmented part, the last element is already 0.
So, if $a=4$, the matrix becomes:

Step 4: Verify the ranks.
If $a=4$, $\operatorname{rank}(A) = 1$ and $\operatorname{rank}([A|\mathbf{b}]) = 1$.
Since $\operatorname{rank}(A) = \operatorname{rank}([A|\mathbf{b}]) = 1 < 2$ (number of variables), the system has infinite solutions.

Result: The condition is $a=4$.
Answer: $\boxed{a=4}$
"
:::

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Summary

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What's Next?

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Part 2: Solving Linear Systems

Introduction

A system of linear equations represents a collection of linear relationships between multiple variables. In the context of engineering and computer science, such systems arise frequently in fields ranging from circuit analysis and network flow problems to machine learning and computer graphics. The ability to characterize and solve these systems is, therefore, a fundamental skill.

We shall explore the conditions under which a linear system possesses a unique solution, infinite solutions, or no solution at all. Our primary analytical tool will be the concept of matrix rank, which provides a powerful and elegant method for determining the nature of the solution set without necessarily computing the solution itself. We will also examine the standard procedure of Gaussian elimination for finding a solution when one exists. The focus will remain on the methods and criteria most pertinent to competitive examinations like GATE.

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Key Concepts

1. Consistency and Types of Solutions

A system of linear equations is termed consistent if it has at least one solution. If no solution exists, the system is inconsistent. A consistent system can have either a unique solution or infinitely many solutions.

The geometric interpretation in two dimensions provides clarity. A system of two linear equations in two variables represents two lines in a plane.

• A unique solution corresponds to the lines intersecting at a single point.


• Infinite solutions correspond to the two lines being coincident (the same line).


• No solution corresponds to the lines being parallel and distinct.

Unique Solution

No Solution

Infinite Solutions

2. Condition for Consistency using Rank

The most reliable method for determining the nature of the solution for a system $Ax=b$ involves comparing the rank of the coefficient matrix $A$ with the rank of the augmented matrix $[A|b]$.

📖 Augmented Matrix

For a system $Ax=b$, the augmented matrix, denoted $[A|b]$, is formed by appending the column vector $b$ to the coefficient matrix $A$.

📐 Rouché-Capelli Theorem (Consistency Condition)

Let $\rho(A)$ be the rank of the coefficient matrix $A$ and $\rho([A|b])$ be the rank of the augmented matrix. Let $n$ be the number of variables.

• No Solution (Inconsistent):

$$\rho(A) \neq \rho([A|b])$$

• Unique Solution (Consistent):

$$\rho(A) = \rho([A|b]) = n$$

• Infinite Solutions (Consistent):

$$\rho(A) = \rho([A|b]) < n$$

When to use: This is the fundamental test to apply to any linear system to determine the nature of its solution set before attempting to solve it.

3. Homogeneous Systems

A system of linear equations is called homogeneous if all the constant terms are zero. It is of the form $Ax=0$.

A key observation is that a homogeneous system is always consistent, because $x=0$ (the zero vector) is always a solution. This is known as the trivial solution. The central question for homogeneous systems is whether a non-trivial solution (a solution where at least one variable is non-zero) exists.

❗ Condition for Non-Trivial Solutions

A homogeneous system $Ax=0$ has a non-trivial solution if and only if the rank of the coefficient matrix $A$ is less than the number of variables $n$.

$$\rho(A) < n$$

For a square matrix $A$ of size $n \times n$, this condition is equivalent to the determinant of $A$ being zero.

$$\det(A) = 0$$

4. Gaussian Elimination

Gaussian elimination is a systematic algorithm for solving systems of linear equations. The procedure transforms the augmented matrix $[A|b]$ into row echelon form using elementary row operations. From this simplified form, the solution can be found by back substitution.

Elementary Row Operations:

Swapping two rows.

Multiplying a row by a non-zero scalar.

Adding a multiple of one row to another row.

Worked Example:

Problem: Solve the following system of linear equations:

$$x + 2y + z = 8$$

$$2x + 3y + 4z = 20$$

$$4x + 3y + 2z = 16$$

Solution:

Step 1: Write the augmented matrix $[A|b]$.

$$[A|b] = \left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 2 & 3 & 4 & 20 \\ 4 & 3 & 2 & 16 \end{array}\right]$$

Step 2: Apply row operations to create zeros below the first pivot (the element in the first row, first column). Perform $R_2 \rightarrow R_2 - 2R_1$ and $R_3 \rightarrow R_3 - 4R_1$.

$$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 0 & -1 & 2 & 4 \\ 0 & -5 & -2 & -16 \end{array}\right]$$

Step 3: Apply a row operation to create a zero below the second pivot. Perform $R_3 \rightarrow R_3 - 5R_2$.

$$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 0 & -1 & 2 & 4 \\ 0 & 0 & -12 & -36 \end{array}\right]$$

This is the row echelon form. The system is now:

$$x + 2y + z = 8$$

$$-y + 2z = 4$$

$$-12z = -36$$

Step 4: Use back substitution to solve for the variables.
From the last equation:

$$z = \frac{-36}{-12} = 3$$

Substitute $z=3$ into the second equation:

$$-y + 2(3) = 4 \implies -y = -2 \implies y = 2$$

Substitute $y=2$ and $z=3$ into the first equation:

$$x + 2(2) + 3 = 8 \implies x + 7 = 8 \implies x = 1$$

Answer: The unique solution is $\boxed{x=1, y=2, z=3}$.

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Problem-Solving Strategies

💡 GATE Strategy: Check Rank First

Before investing time in solving a system using Gaussian elimination, always perform a quick check on the nature of the solution.

• Form the augmented matrix $[A|b]$.


• Reduce it to row echelon form. The number of non-zero rows gives the rank.


• Compare $\rho(A)$ and $\rho([A|b])$ with the number of variables $n$. This immediately tells you whether to expect a unique solution, infinite solutions, or no solution, preventing wasted effort on inconsistent systems.

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Common Mistakes

⚠️ Avoid These Errors

• ❌ Confusing conditions for homogeneous and non-homogeneous systems. The condition $\det(A)=0$ implies infinite solutions for a homogeneous system $Ax=0$, but it could imply either infinite solutions or no solution for a non-homogeneous system $Ax=b$.

✅ Always use the rank condition: $\rho(A) = \rho([A|b]) < n$ for infinite solutions in the non-homogeneous case.

• ❌ Assuming a system with more variables than equations ($n > m$) must have infinite solutions.

✅ This is not guaranteed. The system could be inconsistent. For example, $x+y+z = 1$ and $x+y+z = 2$ has no solution. The rank condition is the only definitive test.

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Practice Questions

:::question type="MCQ" question="The system of equations $x + y + z = 6$, $x + 2y + 3z = 10$, and $x + 2y + \lambda z = \mu$ has no solution if:" options=["$\lambda = 3, \mu = 10$","$\lambda = 3, \mu \neq 10$","$\lambda \neq 3, \mu = 10$","$\lambda \neq 3, \mu \neq 10$"] answer="$\lambda = 3, \mu \neq 10$" hint="For no solution, we require $\rho(A) \neq \rho([A|b])$. Use row reduction on the augmented matrix to find the condition on $\lambda$ and $\mu$ that satisfies this." solution="
Step 1: Form the augmented matrix.

$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 10 \\ 1 & 2 & \lambda & \mu \end{array}\right]$$

Step 2: Apply row operations. $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$.

$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 1 & \lambda-1 & \mu-6 \end{array}\right]$$

Step 3: Apply $R_3 \rightarrow R_3 - R_2$.

$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & \lambda-3 & \mu-10 \end{array}\right]$$

Step 4: Analyze the condition for no solution, which is $\rho(A) \neq \rho([A|b])$.
This occurs when the last row of the coefficient matrix part is all zeros, but the corresponding element in the augmented part is non-zero.
This means we need $\lambda-3 = 0$ and $\mu-10 \neq 0$.

Result:

$$\lambda = 3 \quad \text{and} \quad \mu \neq 10$$

Answer: \boxed{\lambda = 3 \text{ and } \mu \neq 10}
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:::

:::question type="NAT" question="Consider the homogeneous system of equations: $2x - y + 3z = 0$, $3x + 2y + z = 0$, $x - 4y + kz = 0$. If the system has a non-trivial solution, what is the value of $k$?" answer="5" hint="A homogeneous system has a non-trivial solution if and only if the determinant of the coefficient matrix is zero." solution="
Step 1: Write the coefficient matrix $A$.

$$A = \begin{pmatrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & -4 & k \end{pmatrix}$$

Step 2: For a non-trivial solution to exist, we must have $\det(A) = 0$.
Calculate the determinant:

$$\det(A) = 2(2k - (-4)) - (-1)(3k - 1) + 3(-12 - 2) = 0$$

Step 3: Simplify the expression.

$$\begin{aligned}2(2k + 4) + (3k - 1) + 3(-14) & = 0 \\$$
4k + 8 + 3k - 1 - 42 & = 0 \\
7k - 35 & = 0\end{aligned}

Step 4: Solve for $k$.

$$\begin{aligned}7k & = 35 \\$$
k & = 5\end{aligned}

Result: The value of $k$ is 5.
Answer: \boxed{5}
"
:::

:::question type="MSQ" question="Which of the following statements about the system $x+y=2$ and $2x+2y=4$ is/are true?" options=["The system is inconsistent.","The system has a unique solution.","The system has infinitely many solutions.","The rank of the coefficient matrix is 1."] answer="The system has infinitely many solutions.,The rank of the coefficient matrix is 1." hint="Analyze the system using the rank method. Form the augmented matrix and find the ranks of A and [A|b]." solution="
The system is:
$x + y = 2$
$2x + 2y = 4$

The coefficient matrix is

$$A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$$

The augmented matrix is

$$[A|b] = \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 2 & 2 & 4 \end{array}\right]$$

Let's find the ranks by reducing to row echelon form.
Applying $R_2 \rightarrow R_2 - 2R_1$ to $[A|b]$ gives:

$$\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]$$

From this form, we can see:

• The rank of A, $\rho(A)$, is the number of non-zero rows in the coefficient part, which is 1. So, the option "The rank of the coefficient matrix is 1" is correct.


• The rank of [A|b], $\rho([A|b])$, is the number of non-zero rows in the entire matrix, which is also 1.


• The number of variables is $n=2$.

Since $\rho(A) = \rho([A|b]) = 1 < n=2$, the system has infinitely many solutions.
Therefore, the option "The system has infinitely many solutions" is correct.
The options "The system is inconsistent" and "The system has a unique solution" are incorrect.
Answer: \boxed{\text{The system has infinitely many solutions.,The rank of the coefficient matrix is 1.}}
"
:::

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Summary

❗ Key Takeaways for GATE

• Consistency is Key: The first step is always to determine if a system is consistent. Use the rank condition: the system $Ax=b$ is consistent if and only if $\operatorname{rank}(A) = \operatorname{rank}([A|b])$.


• Nature of Solution: If consistent, the type of solution depends on the rank relative to the number of variables, $n$.

- Unique solution if $\operatorname{rank}(A) = n$.
- Infinite solutions if $\operatorname{rank}(A) < n$.

• Homogeneous Systems ($Ax=0$): These are always consistent. They have a non-trivial (non-zero) solution if and only if $\operatorname{rank}(A) < n$. For a square matrix, this is equivalent to $\det(A) = 0$.

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What's Next?

💡 Continue Learning

This topic serves as a foundation for more advanced concepts in Linear Algebra:

• Eigenvalues and Eigenvectors: The calculation of eigenvectors involves solving the homogeneous system $(A - \lambda I)x = 0$. Understanding the condition for non-trivial solutions is critical here.


• Vector Spaces: The set of all solutions to a homogeneous system $Ax=0$ forms a vector space called the null space of A. The number of free variables in the solution ($n - \rho(A)$) gives the dimension of this null space.

Master these connections for a comprehensive understanding of Linear Algebra for GATE!

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Chapter Summary

📖 System of Linear Equations - Key Takeaways

In our study of linear systems, we have developed a systematic framework for analyzing and solving them. The following points encapsulate the essential concepts that are critical for the GATE examination.

• Matrix Representation: Any system of $m$ linear equations in $n$ variables can be concisely represented in the matrix form $AX = B$, where $A$ is the $m \times n$ coefficient matrix, $X$ is the $n \times 1$ column vector of variables, and $B$ is the $m \times 1$ column vector of constants.

• The Concept of Consistency: A system of linear equations is termed consistent if it possesses at least one solution. Otherwise, it is inconsistent. The consistency of a system is fundamentally determined by the relationship between the rank of the coefficient matrix $A$ and the augmented matrix $[A|B]$.

• The Rank Condition for Solutions (Rouché-Capelli Theorem): We have established the definitive criteria for the nature of the solution set:

- No Solution: If $\operatorname{rank}(A) \neq \operatorname{rank}([A|B])$, the system is inconsistent.
- Unique Solution: If $\operatorname{rank}(A) = \operatorname{rank}([A|B]) = n$ (where $n$ is the number of variables), the system is consistent and has exactly one solution.
- Infinitely Many Solutions: If $\operatorname{rank}(A) = \operatorname{rank}([A|B]) = r < n$, the system is consistent and has an infinite number of solutions. The number of free variables (or independent parameters) in the solution is given by $n-r$.

• Homogeneous Systems: For a homogeneous system, $AX = 0$, the vector $B$ is the zero vector. Such systems are always consistent, as they always admit the trivial solution ($X=0$). A non-trivial solution exists if and only if $\operatorname{rank}(A) < n$. For a square matrix $A$, this condition is equivalent to $\det(A) = 0$.

• Gaussian Elimination: The primary computational tool for analyzing linear systems is Gaussian elimination. By reducing the augmented matrix to its row echelon form, we can efficiently determine the ranks of $A$ and $[A|B]$ and subsequently find the solution(s) through back substitution.

• Structure of Solutions: For a consistent non-homogeneous system $AX=B$, the general solution can be expressed as $X = X_p + X_h$, where $X_p$ is any particular solution to $AX=B$, and $X_h$ is the general solution to the corresponding homogeneous system $AX=0$.

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Chapter Review Questions

:::question type="MCQ" question="Consider the system of linear equations:
$x + y + z = 3$
$x + 2y + 2z = 5$
$x + 2y + \lambda z = \mu$
For which of the following values of $\lambda$ and $\mu$ does the system have no solution?" options=["$\lambda = 2, \mu = 5$","$\lambda \neq 2, \mu = 5$","$\lambda = 2, \mu \neq 5$","$\lambda \neq 2, \mu \neq 5$"] answer="C" hint="For a system to have no solution, the rank of the coefficient matrix must be less than the rank of the augmented matrix. Use Gaussian elimination to find the condition on $\lambda$ and $\mu$ that satisfies this." solution="We begin by writing the augmented matrix $[A|B]$ for the given system:

$$[A|B] = \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 1 & 2 & 2 & 5 \\ 1 & 2 & \lambda & \mu \end{array} \right]$$

Now, we apply Gaussian elimination to reduce the matrix to its row echelon form.
Applying the row operations $R_2 \leftarrow R_2 - R_1$ and $R_3 \leftarrow R_3 - R_1$:

$$\sim \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & \lambda-1 & \mu-3 \end{array} \right]$$

Next, we apply the operation $R_3 \leftarrow R_3 - R_2$:

$$\sim \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & \lambda-2 & \mu-5 \end{array} \right]$$

For the system to have no solution, we require $\operatorname{rank}(A) < \operatorname{rank}([A|B])$.
From the echelon form, $\operatorname{rank}(A)$ will be 2 if the last row of the coefficient part is all zeros. This occurs when:

$$\lambda - 2 = 0 \implies \lambda = 2$$

At the same time, for $\operatorname{rank}([A|B])$ to be 3, the last element in the augmented part of the third row must be non-zero. This occurs when:

$$\mu - 5 \neq 0 \implies \mu \neq 5$$

Therefore, the system has no solution if $\lambda = 2$ and $\mu \neq 5$. This corresponds to option C.
Answer: \boxed{C}
"
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:::question type="NAT" question="The system of equations
$x + y + z = 6$
$x + 4y + 6z = 20$
$x + 4y + \lambda z = \mu$
has infinitely many solutions. The value of $\lambda + \mu$ is:" answer="26" hint="For a system to have infinitely many solutions, the rank of the coefficient matrix and the augmented matrix must be equal, and this rank must be less than the number of variables. Use Gaussian elimination to find the conditions on $\lambda$ and $\mu$." solution="We write the augmented matrix $[A|B]$ and reduce it to row echelon form.

$$[A|B] = \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 1 & 4 & 6 & 20 \\ 1 & 4 & \lambda & \mu \end{array} \right]$$

Perform the row operation $R_2 \leftarrow R_2 - R_1$ and $R_3 \leftarrow R_3 - R_1$:

$$\sim \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 3 & 5 & 14 \\ 0 & 3 & \lambda-1 & \mu-6 \end{array} \right]$$

Next, perform the row operation $R_3 \leftarrow R_3 - R_2$:

$$\sim \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 3 & 5 & 14 \\ 0 & 0 & (\lambda-1)-5 & (\mu-6)-14 \end{array} \right] \\$$
\sim \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 3 & 5 & 14 \\ 0 & 0 & \lambda-6 & \mu-20 \end{array} \right]

For the system to have infinitely many solutions, we must have $\operatorname{rank}(A) = \operatorname{rank}([A|B]) < 3$. This condition is satisfied if the last row of the augmented matrix is a zero row.
This implies:

$$\begin{aligned}\lambda - 6 & = 0 \implies \lambda = 6 \\$$
\mu - 20 & = 0 \implies \mu = 20\end{aligned}

The question asks for the value of $\lambda + \mu$.

$$\lambda + \mu = 6 + 20 = 26$$

Thus, the required value is 26.
Answer: \boxed{26}
"
:::

:::question type="MSQ" question="Consider the system of linear equations $AX=B$, where

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 2 & 5 & 7 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix}$$

Which of the following statements is/are correct?" options=["The system is consistent.","The determinant of matrix $A$ is zero.","The system has a unique solution.","The number of free variables is 1."] answer="A,B,D" hint="Analyze the system by finding the rank of the coefficient matrix $A$ and the augmented matrix $[A|B]$. Remember that the number of free variables is given by $n - \operatorname{rank}(A)$." solution="We first analyze the coefficient matrix $A$ and the augmented matrix $[A|B]$.

$$[A|B] = \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 1 \\ 1 & 3 & 4 & 3 \\ 2 & 5 & 7 & 4 \end{array} \right]$$

Let's find the rank by row reduction. Apply $R_2 \leftarrow R_2 - R_1$ and $R_3 \leftarrow R_3 - 2R_1$:

$$\sim \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & 2 \end{array} \right]$$

Now, apply $R_3 \leftarrow R_3 - R_2$:

$$\sim \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

From the row echelon form, we can draw our conclusions.
The number of non-zero rows in the echelon form of $A$ is 2, so $\operatorname{rank}(A) = 2$.
The number of non-zero rows in the echelon form of $[A|B]$ is also 2, so $\operatorname{rank}([A|B]) = 2$.

Let's evaluate each option:
(A) The system is consistent. Since $\operatorname{rank}(A) = \operatorname{rank}([A|B]) = 2$, the system is consistent. This statement is correct.

(B) The determinant of matrix $A$ is zero. Since $\operatorname{rank}(A) = 2$, which is less than the order of the matrix (3), the matrix $A$ is singular. Therefore, $\det(A) = 0$. This statement is correct. We can also verify this directly from the row operations: $R_3$ is the sum of $R_1$ and $R_2$ ($R_1+R_2=R_3$ is not true, but $R_3 = R_1 + R_2'$ where $R_2'$ is the original second row. Let's check original rows: $R_1+R_2 = [2, 5, 7]$. This is exactly $R_3$. Hence rows are linearly dependent and $\det(A)=0$).

(C) The system has a unique solution. A unique solution exists if $\operatorname{rank}(A) = n$, where $n$ is the number of variables. Here, $\operatorname{rank}(A) = 2$ and $n=3$. Since $\operatorname{rank}(A) < n$, the system does not have a unique solution. This statement is incorrect.

(D) The number of free variables is 1. The number of free variables is given by $n - \operatorname{rank}(A)$. Here, $n=3$ and $\operatorname{rank}(A)=2$. So, the number of free variables is $3 - 2 = 1$. The system has infinitely many solutions. This statement is correct.

Therefore, the correct statements are A, B, and D.
Answer: \boxed{A, B, D}
"
:::

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What's Next?

💡 Continue Your GATE Journey

Having completed our study of the System of Linear Equations, you have established a firm foundation for several advanced and interconnected chapters in Engineering Mathematics. The principles of consistency, rank, and solution spaces are not isolated concepts but rather the bedrock upon which more abstract structures are built.

Key connections to leverage in your preparation:

• Relation to Previous Chapters: This chapter is a direct application of the concepts from Matrices and Determinants. Your ability to calculate the rank of a matrix and evaluate determinants is the primary tool used to classify and solve linear systems.

• Foundation for Future Chapters:

- Vector Spaces: The concepts explored here are central to the study of vector spaces. The set of all solutions to a homogeneous system $AX=0$ forms a vector space known as the null space or kernel of matrix $A$. The number of free variables corresponds to the dimension of this null space. - Eigenvalues and Eigenvectors: One of the most important problems in linear algebra is finding the eigenvalues and eigenvectors of a matrix, which involves solving the characteristic equation $(A - \lambda I)X = 0$. This is a direct application of finding non-trivial solutions to a homogeneous system of linear equations, a core skill you have just mastered.