Systems of equations

This chapter rigorously examines methods for solving systems of linear equations, a fundamental concept in linear algebra with broad applications across mathematics and science. Mastery of these techniques, including matrix representation, inverse matrix solutions, and consistency analysis, is crucial for success in the CMI BS Hons examination, particularly in problems involving vectors, matrices, and 3D geometry.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Matrix form of linear system | | 2 | Inverse matrix method | | 3 | Consistency of system |

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We begin with Matrix form of linear system.

Part 1: Matrix form of linear system

Matrix Form of Linear System

Overview

Writing a system of linear equations in matrix form is the natural first step before using elimination, determinants, inverse matrices, or rank ideas. The form $\qquad AX=B$ organizes the system clearly and makes its structure visible. In exam problems, matrix form is not just notation — it often tells you immediately what the coefficient matrix is, whether the system is homogeneous, and what algebraic tools are available. ---

Learning Objectives

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Core Idea

For example, the system $\qquad x+2y=3$ $\qquad 3x-y=5$ can be written as $\qquad \begin{pmatrix}1 & 2\\3 & -1\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}3\\5\end{pmatrix}$ ::: ---

Components of the Matrix Form

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Dimensions Matter

This is one of the first things to check in matrix form problems. ::: ---

Augmented Matrix

For the system $\qquad x+2y=3,\quad 3x-y=5$ the augmented matrix is $\qquad \left[\begin{array}{cc|c}1 & 2 & 3\\3 & -1 & 5\end{array}\right]$ ::: This is especially useful in elimination and row reduction. ::: ---

Homogeneous and Non-Homogeneous Systems

A non-homogeneous system has the form $\qquad AX=B$ with $B\ne 0$. ::: ---

Minimal Worked Examples

Example 1 Write the system $\qquad x+2y-z=1$ $\qquad 2x-y+3z=4$ $\qquad 3x+y+2z=5$ in matrix form. We write $\qquad A=\begin{pmatrix}1&2&-1\\2&-1&3\\3&1&2\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}1\\4\\5\end{pmatrix}$ Hence $\qquad AX=B$ --- Example 2 The system $\qquad x+y=0$ $\qquad 2x-y=0$ is homogeneous, since the constant matrix is $\qquad \begin{pmatrix}0\\0\end{pmatrix}$ ::: ---

Why Matrix Form Is Useful

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The coefficient matrix of the system $x+2y=3,\ 3x-y=5$ is" options=["$\begin{pmatrix}1 & 2\\3 & -1\end{pmatrix}$","$\begin{pmatrix}1 & 2 & 3\\3 & -1 & 5\end{pmatrix}$","$\begin{pmatrix}x\y\end{pmatrix}$","$\begin{pmatrix}3\\5\end{pmatrix}$"] answer="A" hint="Take only the coefficients of the variables." solution="The coefficient matrix is formed from the coefficients of $x$ and $y$ only, so it is $\qquad \begin{pmatrix}1 & 2\\3 & -1\end{pmatrix}$. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="A linear system has $3$ equations and $2$ unknowns. How many rows does its coefficient matrix have?" answer="3" hint="Rows correspond to equations." solution="The coefficient matrix has one row for each equation. Since there are $3$ equations, it has $\boxed{3}$ rows." ::: :::question type="MSQ" question="Which of the following are true?" options=["In $AX=B$, $A$ is the coefficient matrix","In $AX=B$, $X$ is the column matrix of variables","If $B=0$, the system is homogeneous","A non-square system can never be written in matrix form"] answer="A,B,C" hint="Think about the roles of $A,X,B$." solution="1. True. 2. True. 3. True. 4. False. Rectangular coefficient matrices are also allowed in matrix form. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Write the system $x+2y-z=1,\ 2x-y+3z=4,\ 3x+y+2z=5$ in the form $AX=B$, and also write its augmented matrix." answer="$A=\begin{pmatrix}1 & 2 & -1\\2 & -1 & 3\\3 & 1 & 2\end{pmatrix},\ X=\begin{pmatrix}x\y\z\end{pmatrix},\ B=\begin{pmatrix}1\\4\\5\end{pmatrix}$" hint="Take coefficients row by row." solution="The coefficient matrix is $\qquad A=\begin{pmatrix} 1&2&-1\\ 2&-1&3\\ 3&1&2 \end{pmatrix}$ The variable matrix is $\qquad X=\begin{pmatrix}x\y\z\end{pmatrix}$ The constant matrix is $\qquad B=\begin{pmatrix}1\\4\\5\end{pmatrix}$ So the system is $\qquad AX=B$ Its augmented matrix is $\qquad \left[\begin{array}{ccc|c} 1&2&-1&1\\ 2&-1&3&4\\ 3&1&2&5 \end{array}\right]$ Hence the required matrices are written correctly." ::: ---

Summary

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Part 2: Inverse matrix method

Inverse Matrix Method

Overview

The inverse matrix method solves a linear system by writing it in matrix form and then multiplying by the inverse of the coefficient matrix. It is one of the cleanest algebraic methods for systems of equations, but it works only when the coefficient matrix is invertible. In exam problems, the important skills are checking invertibility, finding the inverse correctly, and interpreting what invertibility means for the system. ---

Learning Objectives

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Core Idea

This is the inverse matrix method. ::: ---

When Does It Work?

So for a $2\times 2$ system, the method works exactly when the determinant is nonzero. ::: ---

Inverse of a $2\times 2$ Matrix

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What Invertibility Means for the System

Also, the homogeneous system $\qquad AX=0$ has only the trivial solution $\qquad X=0$ ::: ---

Minimal Worked Examples

Example 1 Solve $\qquad 2x+y=5$ $\qquad x+2y=4$ Write $\qquad A=\begin{pmatrix}2&1\\1&2\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}5\\4\end{pmatrix}$ Now $\qquad \det(A)=4-1=3\ne 0$ So $\qquad A^{-1}=\dfrac{1}{3}\begin{pmatrix}2 & -1\\-1 & 2\end{pmatrix}$ Hence $\qquad X=A^{-1}B =\dfrac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\begin{pmatrix}5\\4\end{pmatrix} =\dfrac{1}{3}\begin{pmatrix}6\\3\end{pmatrix} =\begin{pmatrix}2\\1\end{pmatrix}$ So $\qquad x=2,\ y=1$ --- Example 2 For $\qquad A=\begin{pmatrix}1 & 1\\2 & k\end{pmatrix}$, the inverse exists only when $\qquad \det(A)=k-2\ne 0$ So the inverse matrix method is applicable exactly when $\qquad k\ne 2$ ::: ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The inverse matrix method for solving $AX=B$ is applicable when" options=["$A$ is any matrix","$A$ is square and $\det(A)\ne 0$","$B=0$ only","$A$ is triangular"] answer="B" hint="Think about when $A^{-1}$ exists." solution="The inverse matrix method requires the inverse of $A$ to exist. This happens when $A$ is square and $\det(A)\ne 0$. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="For the system $2x+y=5,\ x+2y=4$, find $x$." answer="2" hint="Use the inverse method or solve directly after forming the matrix." solution="The coefficient matrix is $\qquad \begin{pmatrix}2 & 1\\1 & 2\end{pmatrix}$ with determinant $3\ne 0$, so the inverse method applies. Solving gives $\qquad x=2,\ y=1$. Hence the required value is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following are true?" options=["If $A$ is invertible, then $A^{-1}$ is unique","If $\det(A)=0$, then $A$ is invertible","If $A$ is invertible, then $AX=0$ has only the trivial solution","If $A$ is invertible and $AB=AC$, then $B=C$"] answer="A,C,D" hint="Use standard properties of invertible matrices." solution="1. True. 2. False. If $\det(A)=0$, then $A$ is singular. 3. True. 4. True, because multiplying by $A^{-1}$ gives $B=C$. Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Solve the system $2x+y=5,\ x+2y=4$ using the inverse matrix method." answer="$x=2,\ y=1$" hint="Form $AX=B$ and compute $A^{-1}$." solution="Write $\qquad A=\begin{pmatrix}2&1\\1&2\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}5\\4\end{pmatrix}$ Now $\qquad \det(A)=2\cdot 2-1\cdot 1=3\ne 0$ Hence $\qquad A^{-1}=\dfrac{1}{3}\begin{pmatrix}2 & -1\\-1 & 2\end{pmatrix}$ Therefore $\qquad X=A^{-1}B =\dfrac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\begin{pmatrix}5\\4\end{pmatrix} =\dfrac{1}{3}\begin{pmatrix}6\\3\end{pmatrix} =\begin{pmatrix}2\\1\end{pmatrix}$ So the solution is $\qquad \boxed{x=2,\ y=1}$." ::: ---

Summary

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Part 3: Consistency of system

Consistency of System

Overview

A system of linear equations is called consistent if it has at least one solution and inconsistent if it has no solution. For a $2\times 2$ system, consistency is governed by one central quantity: the determinant of the coefficient matrix. In CMI-style questions, this topic is tested not just by solving equations, but by deciding whether the system has exactly one solution, no solution, or infinitely many solutions, often in terms of parameters. ---

Learning Objectives

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Core Setup

This determinant is the main quantity that controls the nature of solutions. ---

Determinant Criterion

This is because the matrix is invertible, or equivalently, the two equations represent two distinct non-parallel lines. ---

The Case $\Delta = 0$

So the determinant zero case is the delicate consistency case. ---

Consistency When $\Delta = 0$

For the system $\qquad ax + by = u$ $\qquad bx + cy = v$ with $\Delta = 0$, consistency requires the two equations to represent the same line. If they represent parallel but distinct lines, there is no solution. ---

Cramer's Rule

For the special system $\qquad ax + by = \sqrt{2}$ $\qquad bx + cy = \sqrt{3}$ the unique solution is $\qquad x = \frac{c\sqrt{2} - b\sqrt{3}}{ac-b^2}$ $\qquad y = \frac{a\sqrt{3} - b\sqrt{2}}{ac-b^2}$ ::: ---

Geometric Interpretation

This geometric picture is often the fastest way to understand the answer. ---

Special Structure of the Symmetric System

Many parameter questions reduce immediately to checking whether $ac-b^2$ vanishes. ---

Minimal Worked Examples

Example 1 Determine the nature of solutions of $\qquad x + 2y = 1$ $\qquad 2x + 5y = 4$ Here $\qquad \Delta = 1\cdot 5 - 2^2 = 5-4 = 1 \ne 0$ So the system has exactly one solution. --- Example 2 Determine the nature of solutions of $\qquad x + 2y = 1$ $\qquad 2x + 4y = 3$ Here $\qquad \Delta = 1\cdot 4 - 2^2 = 0$ The coefficient rows are proportional, since the second left-hand side is twice the first. But the right-hand side should also be doubled, and $2\cdot 1 = 2 \ne 3$. Hence the system has no solution. --- Example 3 Determine the nature of solutions of $\qquad 2x + \sqrt{6}\,y = \sqrt{2}$ $\qquad \sqrt{6}\,x + 3y = \sqrt{3}$ Now $\qquad \Delta = 2\cdot 3 - (\sqrt{6})^2 = 6-6 = 0$ Also the second equation is obtained by multiplying the first by $\qquad \sqrt{\frac{3}{2}}$ because $\qquad \sqrt{\frac{3}{2}}\cdot 2 = \sqrt{6}, \qquad \sqrt{\frac{3}{2}}\cdot \sqrt{6} = 3, \qquad \sqrt{\frac{3}{2}}\cdot \sqrt{2} = \sqrt{3}$ So both equations represent the same line. Hence the system has infinitely many solutions. ---

AM-GM Insight for the Special Topic

This is a useful structural fact in problems involving determinant zero. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The system $\; ax+by=u,\; bx+cy=v \;$ has exactly one solution when" options=["$ac-b^2=0$","$ac-b^2\ne 0$","$a+b+c=0$","$u=v$"] answer="B" hint="Check the determinant of the coefficient matrix." solution="The coefficient matrix is $\qquad \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ Its determinant is $\qquad ac-b^2$ A $2\times 2$ system has exactly one solution precisely when this determinant is nonzero. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="For the system $\; x+2y=1,\; 2x+4y=3 \;$, how many solutions does the system have?" answer="0" hint="Check whether the second equation is a compatible multiple of the first." solution="The determinant is $\qquad 1\cdot 4 - 2^2 = 0$ So the system cannot have a unique solution. The second left-hand side is twice the first, but the right-hand side should then also be twice the first right-hand side. Since $2\cdot 1=2 \ne 3$, the system is inconsistent. Hence the number of solutions is $\boxed{0}$." ::: :::question type="MSQ" question="Which of the following statements are true for the system $\; ax+by=u,\; bx+cy=v \;$?" options=["If $ac-b^2\ne 0$, the system has exactly one solution","If $ac-b^2=0$, the system can have infinitely many solutions","If $ac-b^2=0$, the system can have no solution","If $ac-b^2=0$, the system has exactly one solution"] answer="A,B,C" hint="Zero determinant is the no-solution or infinitely-many-solutions case." solution="1. True. Nonzero determinant gives a unique solution.

True. If the equations are proportional including the right-hand side, the system has infinitely many solutions.

True. If the equations are proportional on the left but not on the right, the system has no solution.

False. Zero determinant rules out a unique solution.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Find all values of $k$ for which the system $\; (k-1)x+y=1,\; 2x+(k+1)y=3 \;$ has exactly one solution, no solution, or infinitely many solutions." answer="Exactly one solution for $k\ne \pm\sqrt{3}$, no solution for $k=\pm\sqrt{3}$, infinitely many solutions for no value of $k$" hint="Use the determinant first." solution="The coefficient matrix is $\qquad \begin{pmatrix} k-1 & 1 \\ 2 & k+1 \end{pmatrix}$ Its determinant is $\qquad (k-1)(k+1)-2 = k^2-3$ So the system has exactly one solution when $\qquad k^2-3 \ne 0$ that is, $\qquad k \ne \pm\sqrt{3}$ Now consider $\qquad k=\sqrt{3}$ Then the equations become $\qquad (\sqrt{3}-1)x+y=1$ $\qquad 2x+(\sqrt{3}+1)y=3$ The coefficient rows are proportional because $\qquad \frac{2}{\sqrt{3}-1} = \sqrt{3}+1 = \frac{\sqrt{3}+1}{1}$ But the right-hand side ratio is $\qquad \frac{3}{1}=3 \ne \sqrt{3}+1$ So the system has no solution. Similarly, for $\qquad k=-\sqrt{3}$ the determinant is also zero, the coefficient rows are proportional, but the right-hand side ratio again does not match, so the system has no solution. Therefore:

• exactly one solution for $\boxed{k\ne \pm\sqrt{3}}$
  • no solution for $\boxed{k=\pm\sqrt{3}}$
    • infinitely many solutions for $\boxed{\text{no value of }k}$
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    Summary

    ❗ Key Takeaways for CMI

    • For a $2\times 2$ system, the determinant is the first consistency test.

    
    

    • $\Delta \ne 0$ gives exactly one solution.

    
    

    • $\Delta = 0$ gives either no solution or infinitely many solutions.

    
    

    • In the zero-determinant case, compare the full equations, not just the coefficients.

    
    

    • For positive distinct parameters with $ac=b^2$, AM-GM often becomes useful.

    
    

    • Geometry and determinant logic should be used together.

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    Chapter Summary

    ❗ Systems of equations — Key Points

    A system of linear equations can be compactly represented in matrix form as $Ax=b$, where $A$ is the coefficient matrix, $x$ is the vector of variables, and $b$ is the constant vector.
    The solvability of a square system $Ax=b$ is largely determined by the determinant of the coefficient matrix, $\det(A)$.
    If $\det(A) \neq 0$, the matrix $A$ is non-singular and invertible. In this case, a unique solution exists, given by $x=A^{-1}b$.
    If $\det(A) = 0$, the matrix $A$ is singular. The system may then have infinitely many solutions (consistent) or no solutions (inconsistent).
    Consistency of a system $Ax=b$ is rigorously determined by comparing the rank of the coefficient matrix $A$ with the rank of the augmented matrix $[A|b]$.
    A system is consistent if and only if $\operatorname{rank}(A) = \operatorname{rank}([A|b])$. If this common rank equals the number of variables, $n$, a unique solution exists. If it is less than $n$, infinitely many solutions exist.
    * A system is inconsistent (no solutions) if $\operatorname{rank}(A) \neq \operatorname{rank}([A|b])$. Homogeneous systems ($Ax=0$) are always consistent, having at least the trivial solution $x=0$. Non-trivial solutions exist for homogeneous systems if and only if $\det(A)=0$.

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    Chapter Review Questions

    :::question type="MCQ" question="Consider the system of equations:
    

    $$\begin{aligned}x + 2y - z & = 1 \\$$
    2x - y + z & = 3 \\
    3x + y + 2z & = 4\end{aligned}
    
    Which of the following statements is true regarding the solution of this system?" options=["The system has a unique solution.","The system has infinitely many solutions.","The system is inconsistent.","The system has a trivial solution."] answer="The system has a unique solution." hint="Calculate the determinant of the coefficient matrix. If it's non-zero, a unique solution exists." solution="The coefficient matrix is $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 1 \\ 3 & 1 & 2 \end{pmatrix}$.
    The determinant of $A$ is $\det(A) = 1((-1)(2) - (1)(1)) - 2((2)(2) - (1)(3)) + (-1)((2)(1) - (-1)(3))$
    $= 1(-2-1) - 2(4-3) - 1(2+3)$
    $= -3 - 2(1) - 1(5)$
    $= -3 - 2 - 5 = -10$.
    Since $\det(A) = -10 \neq 0$, the matrix $A$ is non-singular, and thus the system has a unique solution."
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    :::question type="NAT" question="For what value of $k$ does the system of equations
    

    $$\begin{aligned}x + y + z & = 6 \\$$
    x + 2y + 3z & = 10 \\
    x + 2y + kz & = 10\end{aligned}
    
    have infinitely many solutions?" answer="3" hint="For infinitely many solutions, the determinant of the coefficient matrix must be zero, and the ranks of the coefficient and augmented matrices must be equal and less than the number of variables." solution="The coefficient matrix is $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & k \end{pmatrix}$.
    For the system to have infinitely many solutions, $\det(A)$ must be zero.
    $\det(A) = 1(2k-6) - 1(k-3) + 1(2-2) = 2k-6 - k+3 = k-3$.
    Setting $\det(A)=0$, we get $k-3=0 \implies k=3$.
    Now we check the consistency for $k=3$ using the augmented matrix:
    $[A|b] = \begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 1 & 2 & 3 & | & 10 \end{pmatrix}$
    Perform row operations:
    $R_2 \to R_2 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 1 & 2 & 3 & | & 10 \end{pmatrix}$
    $R_3 \to R_3 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 1 & 2 & | & 4 \end{pmatrix}$
    $R_3 \to R_3 - R_2$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
    Here, $\operatorname{rank}(A) = 2$ and $\operatorname{rank}([A|b]) = 2$. Since the number of variables $n=3$ and $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, the system has infinitely many solutions for $k=3$."
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    :::question type="MCQ" question="Consider the homogeneous system $Ax=0$, where $A$ is a $3 \times 3$ matrix. Which of the following conditions guarantees that the system has non-trivial solutions?" options=["$\operatorname{rank}(A) = 3$","$\det(A) \neq 0$","$A$ is invertible","$\operatorname{rank}(A) < 3$"] answer="$\operatorname{rank}(A) < 3$" hint="A homogeneous system always has the trivial solution. Non-trivial solutions exist if and only if the coefficient matrix is singular." solution="For a homogeneous system $Ax=0$, non-trivial solutions exist if and only if the matrix $A$ is singular, meaning $\det(A)=0$. This condition is equivalent to $\operatorname{rank}(A) < n$, where $n$ is the number of variables (here, $n=3$). Therefore, $\operatorname{rank}(A) < 3$ guarantees non-trivial solutions. Options A, B, and C all imply that $A$ is non-singular, which means only the trivial solution exists for $Ax=0$."
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    :::question type="MCQ" question="If $A$ is an $n \times n$ non-singular matrix, and $Ax=b$ is a system of linear equations, which of the following is true?" options=["The system has infinitely many solutions.","The system has a unique solution given by $x=A^{-1}b$.","The system is inconsistent.","The system has a trivial solution only if $b=0$."] answer="The system has a unique solution given by $x=A^{-1}b$." hint="A non-singular matrix implies its determinant is non-zero and its inverse exists." solution="A non-singular matrix $A$ means that $\det(A) \neq 0$. This implies that $A$ is invertible, and its inverse $A^{-1}$ exists. For a system $Ax=b$, if $A^{-1}$ exists, we can multiply both sides by $A^{-1}$ to get $A^{-1}(Ax) = A^{-1}b \implies Ix = A^{-1}b \implies x = A^{-1}b$. This unique solution guarantees that the system is consistent and does not have infinitely many solutions. The concept of a trivial solution (where $x=0$) applies specifically to homogeneous systems ($Ax=0$) and is only a possibility when $b=0$."
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    What's Next?

    💡 Continue Your CMI Journey

    The concepts of matrices, determinants, and rank are foundational for advanced topics in linear algebra. You will encounter these again when studying vector spaces, linear transformations, and eigenvalues and eigenvectors. Understanding how systems of equations relate to the intersection of geometric objects (lines, planes) will also be crucial in 3D Geometry.