Systems of Linear Equations

Question 1 & 2: Solving a System of Linear Equations

Problem Consider a system of linear equations (System 1): $-2x_1 + 3x_2 + x_3 = 1$ $-x_1 + x_3 = 0$ $2x_2 = 5$

1) If the matrix representation of system (1) is $Ax = b$, where $x = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}$, then identify A and b. 2) System (1) has:

• a unique solution.
• no solution.
• infinitely many solutions.
• None of the above.

Answers

1) The correct representation is the last option: $A = \begin{bmatrix} -2 & 3 & 1 \ -1 & 0 & 1 \ 0 & 2 & 0 \end{bmatrix}$, and $b = \begin{bmatrix} 1 \ 0 \ 5 \end{bmatrix}$

2) The system has a unique solution.

Concepts Explained 💡

• Matrix Representation: A system of linear equations can be written as $Ax=b$. The coefficient matrix A contains the coefficients of the variables, and the constant vector b contains the constants on the right side. Be careful with missing terms; their coefficient is 0.
• Solving a System: You can solve a system by substitution or elimination. If you find a single, specific value for each variable, the system has a unique solution.

Step-by-Step Solution

1. Matrix Representation (Q1):

   • The coefficients of the variables in the three equations form the rows of matrix A.
     • Row 1: [-2, 3, 1]
     • Row 2: [-1, 0, 1] (since $x_2$ is missing, its coefficient is 0)
     • Row 3: [0, 2, 0] (since $x_1$ and $x_3$ are missing, their coefficients are 0)
   • The constants on the right side form the vector b: [1, 0, 5]ᵀ.
   • This matches the last option provided.

2. Solving the System (Q2):

   • From the 3rd equation: $2x_2 = 5 \implies x_2 = \frac{5}{2}$.
   • From the 2nd equation: $-x_1 + x_3 = 0 \implies x_1 = x_3$.
   • Substitute these results into the 1st equation: $-2x_1 + 3(\frac{5}{2}) + x_3 = 1$
   • Now substitute $x_1$ for $x_3$: $-2x_1 + \frac{15}{2} + x_1 = 1$ $-x_1 = 1 - \frac{15}{2} = \frac{2}{2} - \frac{15}{2} = -\frac{13}{2}$ $x_1 = \frac{13}{2}$
   • Since $x_3 = x_1$, we have $x_3 = \frac{13}{2}$.
   • We found a single value for each variable: $(x_1, x_2, x_3) = (\frac{13}{2}, \frac{5}{2}, \frac{13}{2})$. This is a unique solution.

Question 3: Properties of Linear Systems

Problem Choose the set of correct options.

Options

• Every system of linear equations has either a unique solution, no solution or infinitely many solutions.
• If each equation of a system of linear equations is multiplied by a non-zero constant $c$, then the solution of the new system of equations is $c$ times the solution of the old system of equations.
• If $Ax = b$ is a system of linear equations which has a solution, then the system of linear equations $cAx = b$, where $c \neq 0$, will also have a solution.
• If $Ax = b$ is a system of linear equations which has a solution, then the system $\frac{1}{c}Ax = b$, where $c \neq 0$, will also have a solution.

Correct Options

• Every system of linear equations has either a unique solution, no solution or infinitely many solutions.
• If $Ax = b$ is a system of linear equations which has a solution, then the system of linear equations $cAx = b$, where $c \neq 0$, will also have a solution.
• If $Ax = b$ is a system of linear equations which has a solution, then the system $\frac{1}{c}Ax = b$, where $c \neq 0$, will also have a solution.

Concepts Explained 💡

• Solution Types: A key theorem in linear algebra states that any system of linear equations will have exactly one of three outcomes: one unique solution, no solution at all, or infinitely many solutions.
• Column Space: A system $Ax=b$ has a solution if and only if the vector $b$ is in the column space of $A$ (i.e., $b$ can be written as a linear combination of the columns of $A$). A column space is a vector space, which means it is closed under scalar multiplication.

Analysis of Options

• Statement 1: True. This is a fundamental property of linear systems.
• Statement 2: False. If $x_0$ is a solution to $Ax=b$, then $A x_0 = b$. If we multiply each equation by $c$, the new system is $(cA)x = cb$. Substituting $x_0$ gives $(cA)x_0 = c(Ax_0) = cb$. So, $x_0$ is also the solution to the new system, not $cx_0$. The solution set remains unchanged.
• Statement 3: True. The new system can be written as $Ax = \frac{1}{c}b$. Since the original system $Ax=b$ has a solution, the vector $b$ must be in the column space of $A$. Because a column space is a vector space, it is closed under scalar multiplication, meaning any scalar multiple of $b$ (like $\frac{1}{c}b$) is also in the column space of $A$. Therefore, the new system must also have a solution.
• Statement 4: True. This follows the same logic as statement 3. The new system can be written as $Ax = cb$. Since $b$ is in the column space of $A$, $cb$ must also be in the column space of $A$. Therefore, a solution must exist.

Question 4: Geometric Interpretation (Planes)

Problem The Plane 1 and Plane 2 in Figure M2W1AQ3, correspond to two different linear equations, which form a system of linear equations. The above system of linear equations has… (The figure shows two distinct, parallel planes).

Correct Option

• no solution.

Concepts Explained 💡

• Geometric Interpretation: A linear equation with three variables (e.g., $ax + by + cz = d$) represents a plane in 3D space.
• Solution of a System: A solution to a system of equations is a point $(x, y, z)$ that satisfies all equations simultaneously. Geometrically, this means the point must lie on all the planes represented by the equations.
• Parallel Planes: If two planes are parallel and distinct, they never intersect.

Explanation

The figure shows two parallel planes that do not touch. Since there is no point of intersection, there is no point $(x, y, z)$ that lies on both planes. Therefore, the corresponding system of linear equations has no solution.

Question 5: Geometric Interpretation (Lines)

Problem Consider the geometric representations (Figures (a), (b), and (c)) of three systems of linear equations. Choose the set of correct options.

Correct Options

• Figure (a) represents a system of linear equations which has no solution.
• Figure (b) represents a system of linear equations which has a unique solution.
• Figure (c) represents a system of linear equations which has infinitely many solutions.

Concepts Explained 💡

• Geometric Interpretation: A linear equation with two variables (e.g., $ax + by = c$) represents a line in 2D space.
• Solution Types (Lines):
  • No Solution: The lines are parallel and distinct; they never intersect.
  • Unique Solution: The lines intersect at exactly one point.
  • Infinitely Many Solutions: The lines are coincident (they are the same line), so every point on the line is an intersection point.

Analysis of Figures

• Figure (a): Shows two distinct parallel lines. They will never intersect, so there is no solution.
• Figure (b): Shows three distinct lines all intersecting at a single, common point. This represents a unique solution.
• Figure (c): Shows multiple lines that are all lying on top of each other (coincident). They share all their points, so there are infinitely many solutions.

Question 6: Analyzing a System with Parameters

Problem Consider a system of equations: $2x_1 + 3x_2 = 6$ $-2x_1 + kx_2 = d$ $4x_1 + 6x_2 = 12$ Choose the set of correct options.

Correct Options

• $Ax = b$ represents the above system, where $x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, A = \begin{bmatrix} 2 & 3 \ -2 & k \ 4 & 6 \end{bmatrix}$, and $b = \begin{bmatrix} 6 \ d \ 12 \end{bmatrix}$.
• The system has no solution if $k = -3, d = 0$.
• The system has a unique solution if $k = 3, d = 0$.
• The system has infinitely many solutions if $k = -3, d = -6$.

Concepts Explained 💡

For a system of linear equations, the relationship between equations determines the nature of the solution.

• Redundant Equation: If one equation is a multiple of another, it provides no new information and can be ignored.
• Conditions for Solutions (2x2 system):
  • Unique Solution: The lines have different slopes. For $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$, this means $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
  • No Solution / Infinite Solutions: The lines have the same slope ($\frac{a_1}{a_2} = \frac{b_1}{b_2}$). It’s no solution if the intercepts are different ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$) and infinite solutions if the intercepts are the same ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$).

Step-by-Step Solution

1. Simplify the System: Notice that the third equation ($4x_1 + 6x_2 = 12$) is exactly 2 times the first equation ($2x_1 + 3x_2 = 6$). This means the third equation is redundant, and we only need to analyze the first two:
   • Eq 1: 2x₁ + 3x₂ = 6
   • Eq 2: -2x₁ + kx₂ = d
2. Solve by Elimination: Add the two equations to eliminate $x_1$:
   • $(2 - 2)x_1 + (3 + k)x_2 = 6 + d$
   • $(3 + k)x_2 = 6 + d$
3. Analyze the Result:
   • Unique Solution: If the coefficient of $x_2$ is non-zero (i.e., $3+k \neq 0$ or $k \neq -3$), we can solve for a unique value of $x_2$. The option k=3, d=0 satisfies $k \neq -3$, so it has a unique solution.
   • No or Infinite Solutions: If the coefficient of $x_2$ is zero (i.e., $3+k = 0$ or $k = -3$), the equation becomes $0 = 6+d$.
     • No Solution: If $6+d \neq 0$ (i.e., $d \neq -6$), the equation is $0 = (\text{non-zero})$, which is a contradiction. The option k=-3, d=0 satisfies this, leading to no solution.
     • Infinitely Many Solutions: If $6+d = 0$ (i.e., $d = -6$), the equation is $0 = 0$, which is always true. This means the two original equations were dependent, leading to infinitely many solutions. The option k=-3, d=-6 satisfies this.

Question 7 & 8: Properties of Solutions

Question 7

Let $x_1$ and $x_2$ be solutions of the system of linear equations $Ax = b$. Which of the following options are correct?

Question 8

Let $v$ be a solution of the systems of linear equations $A_1x = b$ and $A_2x = b$. Which of the following options are correct?

Correct Options

• For Q7:
  • $x_1 + x_2$ is a solution of the system of linear equations $Ax = 2b$.
  • $x_1 - x_2$ is a solution of the system of linear equations $Ax = 0$.
• For Q8:
  • $v$ is a solution of the system of linear equations $(A_1 + A_2)x = 2b$.
  • $v$ is a solution of the system of linear equations $(A_1 - A_2)x = 0$.

Concepts Explained 💡

These questions test the linearity properties of matrix-vector multiplication. For any matrices A, B and vectors x, y:

• $A(x+y) = Ax + Ay$ (Distributive property)
• $(A+B)x = Ax + Bx$ (Distributive property)

Step-by-Step Derivations

For Question 7:

• We are given Ax₁ = b and Ax₂ = b.
• Test Sum: $A(x_1 + x_2) = Ax_1 + Ax_2 = b + b = 2b$. Thus, $x_1 + x_2$ is a solution to $Ax=2b$.
• Test Difference: $A(x_1 - x_2) = Ax_1 - Ax_2 = b - b = 0$. Thus, $x_1 - x_2$ is a solution to $Ax=0$.

For Question 8:

• We are given A₁v = b and A₂v = b.
• Test Sum: $(A_1 + A_2)v = A_1v + A_2v = b + b = 2b$. Thus, $v$ is a solution to $(A_1+A_2)x=2b$.
• Test Difference: $(A_1 - A_2)v = A_1v - A_2v = b - b = 0$. Thus, $v$ is a solution to $(A_1-A_2)x=0$.

Question 9: Condition for a Unique Solution

Problem Consider a system of equations: $x_1 - 3x_2 = 4$ $3x_1 + kx_2 = -12$ Where $k \in \mathbb{R}$. If the given system has a unique solution, then $k$ should not be equal to

Answer: -9

Concepts Explained 💡

A $2 \times 2$ system of linear equations has a unique solution if and only if the determinant of its coefficient matrix is non-zero. If the determinant is zero, the system has either no solution or infinitely many solutions.

Step-by-Step Solution

1. Write the coefficient matrix A: $A = \begin{bmatrix} 1 & -3 \ 3 & k \end{bmatrix}$
2. Calculate the determinant of A: $det(A) = (1)(k) - (-3)(3) = k + 9$
3. Set the condition for a unique solution: For a unique solution, we must have $det(A) \neq 0$. $k + 9 \neq 0 \implies k \neq -9$
4. Interpret the question: The question asks for the value that $k$ should not be equal to if the system is to have a unique solution. Based on our condition, the value that prevents a unique solution is -9.

Question 10: Solving Coupled Systems

Problem Consider two system of linear equations: System 1: $x_1 + 2x_2 = -5$ $0x_1 - x_2 = 5$ System 2: $-2x_3 + x_4 = -5$ $3x_3 + x_4 = 5$ Suppose there is another system of linear equation given by $(1-2)(x_1+x_3) + (2+1)(x_2+x_4) = m$ $(0+3)(x_1+x_3) + (-1+1)(x_2+x_4) = n$ for some real values of $m$ and $n$. Find the value of $n - m$.

Answer: 46

Concepts Explained 💡

This problem requires solving two independent systems of linear equations and then substituting their solutions into a third system to find the unknown parameters.

Step-by-Step Solution

1. Solve System 1:

   • From the second equation: $-x_2 = 5 \implies x_2 = -5$.
   • Substitute into the first equation: $x_1 + 2(-5) = -5 \implies x_1 - 10 = -5 \implies x_1 = 5$.
   • Solution 1 is (x₁, x₂) = (5, -5).

2. Solve System 2:

   • We have:
     • $-2x_3 + x_4 = -5$
     • $3x_3 + x_4 = 5$
   • Subtract the first equation from the second to eliminate $x_4$:
     • $(3x_3 - (-2x_3)) + (x_4 - x_4) = 5 - (-5)$
     • $5x_3 = 10 \implies x_3 = 2$.
   • Substitute $x_3=2$ into the second equation: $3(2) + x_4 = 5 \implies 6 + x_4 = 5 \implies x_4 = -1$.
   • Solution 2 is (x₃, x₄) = (2, -1).

3. Set up the Third System:

   • Let $y_1 = x_1 + x_3$ and $y_2 = x_2 + x_4$.
   • Calculate $y_1$ and $y_2$:
     • $y_1 = 5 + 2 = 7$
     • $y_2 = -5 + (-1) = -6$
   • Now substitute these into the third system after simplifying its coefficients:
     • $(-1)y_1 + (3)y_2 = m \implies -y_1 + 3y_2 = m$
     • $(3)y_1 + (0)y_2 = n \implies 3y_1 = n$

4. Solve for m and n:

   • From the second new equation: $n = 3y_1 = 3(7) = 21$.
   • From the first new equation: $m = -y_1 + 3y_2 = -(7) + 3(-6) = -7 - 18 = -25$.

5. Calculate n - m:

   • $n - m = 21 - (-25) = 21 + 25 = 46$.