Determining Λ Values For No Solution In A System Of Equations

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In linear algebra, understanding the conditions under which a system of linear equations has no solution is crucial. This article delves into how to determine the values of $\\lambda$ for which a given system of equations, represented in matrix form as $AX = b$, has no solution. We will explore the underlying principles, the methods to apply, and provide a detailed walkthrough to solve such problems. This topic is fundamental in various fields, including engineering, physics, computer science, and economics, where systems of equations are frequently used to model real-world phenomena. Let's dive deep into the process of identifying when a system lacks a solution and what that implies about the matrices involved.

To determine the values of λ for which a system of linear equations has no solution, we must first understand the structure of the system. The given system is represented in matrix form as $AX = b$, where $A$ is the coefficient matrix, $X$ is the column vector of unknowns, and $b$ is the column vector of constants. In this specific case, we have:

A=[12−3 3−15 41−2+lambda2]A = \begin{bmatrix} 1 & 2 & -3 \\\ 3 & -1 & 5 \\\ 4 & 1 & -2 + \\\\lambda^2 \end{bmatrix}

The vector $X$ represents the unknowns (typically $x$, $y$, and $z$), and $b$ is the constant vector. For the system to have no solution, the equations must be inconsistent. This inconsistency arises when the determinant of matrix $A$ is zero, and the augmented matrix $[A | b]$ has a different rank than the matrix $A$. In simpler terms, this means that the rows of matrix $A$ are linearly dependent, leading to a situation where no combination of variables can satisfy all equations simultaneously. The determinant plays a crucial role because a non-zero determinant indicates that the matrix is invertible, implying a unique solution exists for the system. When the determinant is zero, the matrix is singular, which means it does not have an inverse, and the system may have either infinitely many solutions or no solution at all. To differentiate between these two cases, we analyze the rank of the augmented matrix. If the rank of the augmented matrix is higher than the rank of matrix $A$, the system is inconsistent and has no solution. This condition signifies that the addition of the constant vector $b$ introduces a new linearly independent row, making it impossible to find a solution that satisfies all equations.

The determinant of matrix A is a scalar value that provides crucial information about the matrix's properties. A zero determinant indicates that the matrix is singular, meaning it is not invertible and the system of equations may have no unique solution. To find the determinant of the given 3x3 matrix, we use the formula:

det(A)=a(ei−fh)−b(di−fg)+c(dh−eg)det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)

where A is defined as:

A=[abc def ghi]A = \begin{bmatrix} a & b & c \\\ d & e & f \\\ g & h & i \end{bmatrix}

Applying this to our matrix:

A=[12−3 3−15 41−2+lambda2]A = \begin{bmatrix} 1 & 2 & -3 \\\ 3 & -1 & 5 \\\ 4 & 1 & -2 + \\\\lambda^2 \end{bmatrix}

We calculate the determinant as follows:

det(A)=1[(−1)(−2+lambda2)−(5)(1)]−2[(3)(−2+lambda2)−(5)(4)]+(−3)[(3)(1)−(−1)(4)]det(A) = 1[(-1)(-2 + \\\\lambda^2) - (5)(1)] - 2[(3)(-2 + \\\\lambda^2) - (5)(4)] + (-3)[(3)(1) - (-1)(4)]

det(A)=1[2−lambda2−5]−2[−6+lambda2−20]−3[3+4]det(A) = 1[2 - \\\\lambda^2 - 5] - 2[-6 + \\\\lambda^2 - 20] - 3[3 + 4]

det(A)=1[−3−lambda2]−2[lambda2−26]−3[7]det(A) = 1[-3 - \\\\lambda^2] - 2[\\\\lambda^2 - 26] - 3[7]

det(A)=−3−lambda2−2lambda2+52−21det(A) = -3 - \\\\lambda^2 - 2\\\\lambda^2 + 52 - 21

det(A)=−3lambda2+28det(A) = -3\\\\lambda^2 + 28

Setting the determinant to zero to find the values of $\\\lambda$ for which the matrix is singular:

−3lambda2+28=0-3\\\\lambda^2 + 28 = 0

3lambda2=283\\\\lambda^2 = 28

lambda2=frac283{\\\\lambda}^2 = \\\\frac{28}{3}

λ=pmsqrtfrac283=pmfrac2sqrt213\\\\\lambda = \\\\pm \\\\sqrt{\\\\frac{28}{3}} = \\\\pm \\\\frac{2\\\\sqrt{21}}{3}

Therefore, the determinant of matrix A is zero when $\\\lambda = \\pm \\frac{2\\sqrt{21}}{3}$, which are the critical values we need to further investigate for the system's solvability.

To analyze whether the system has no solution, we must consider the augmented matrix $[A | b]$ and compare its rank with the rank of matrix $A$. The augmented matrix is formed by appending the column vector $b$ to matrix $A$. In our case, we have not explicitly defined the vector $b$, so we will proceed with a general analysis. The system has no solution if the rank of $[A | b]$ is greater than the rank of $A$. This condition implies that the addition of the column vector $b$ introduces a linear dependency that makes the system inconsistent.

Given the matrix:

A=[12−3 3−15 41−2+lambda2]A = \begin{bmatrix} 1 & 2 & -3 \\\ 3 & -1 & 5 \\\ 4 & 1 & -2 + \\\\lambda^2 \end{bmatrix}

We found that $det(A) = -3\\lambda^2 + 28$, and the determinant is zero when $\\\lambda = \\pm \\frac{2\\sqrt{21}}{3}$. At these values, matrix $A$ is singular, and we need to examine the rank of the augmented matrix to determine if the system has no solution.

Let's assume a general form for the vector $b$:

b=[b1 b2 b3]b = \begin{bmatrix} b_1 \\\ b_2 \\\ b_3 \end{bmatrix}

The augmented matrix $[A | b]$ is:

[A∣b]=[12−3∣b1 3−15∣b2 41−2+lambda2∣b3][A | b] = \begin{bmatrix} 1 & 2 & -3 & | & b_1 \\\ 3 & -1 & 5 & | & b_2 \\\ 4 & 1 & -2 + \\\\lambda^2 & | & b_3 \end{bmatrix}

For the system to have no solution when $det(A) = 0$, the rank of $[A | b]$ must be 3, while the rank of $A$ is less than 3 (which is 2 or 1). This means that the columns of $A$ are linearly dependent, but the column vector $b$ introduces a new linearly independent column, making the system inconsistent.

To check this, we can perform row operations on the augmented matrix to see if we can obtain a row of the form $[0 \ 0 \ 0 | c]$ where $c \\neq 0$. This would indicate an inconsistency in the system. However, without specific values for $b_1$, $b_2$, and $b_3$, we cannot perform these operations definitively. In general, we can say that if the determinant of any 3x3 submatrix of $[A | b]$ that includes $b$ is non-zero, then the rank of $[A | b]$ is 3, and the system has no solution.

The conditions for a system of linear equations to have no solution are crucial in understanding the behavior of the system. For the system $AX = b$ to have no solution, two primary conditions must be met:

  1. The determinant of the coefficient matrix $A$ must be zero. This indicates that the matrix is singular and does not have an inverse, which is a prerequisite for the system to have either no solution or infinitely many solutions.
  2. The rank of the augmented matrix $[A | b]$ must be greater than the rank of the coefficient matrix $A$. This condition ensures that the system is inconsistent. In simpler terms, the vector $b$ introduces a linear dependency that cannot be satisfied by the solutions of the equations represented by matrix $A$.

In our specific case, we found that the determinant of matrix $A$ is given by:

det(A)=−3lambda2+28det(A) = -3\\\\lambda^2 + 28

Setting this to zero, we obtained the values of $\\\lambda$ for which the matrix $A$ is singular:

λ=pmfrac2sqrt213\\\\\lambda = \\\\pm \\\\frac{2\\\\sqrt{21}}{3}

Now, we need to verify the second condition. We consider the augmented matrix:

[A∣b]=[12−3∣b1 3−15∣b2 41−2+lambda2∣b3][A | b] = \begin{bmatrix} 1 & 2 & -3 & | & b_1 \\\ 3 & -1 & 5 & | & b_2 \\\ 4 & 1 & -2 + \\\\lambda^2 & | & b_3 \end{bmatrix}

For the system to have no solution, the rank of this augmented matrix must be 3, while the rank of $A$ is less than 3. This implies that at least one determinant formed by a 3x3 submatrix of $[A | b]$ that includes the column $b$ must be non-zero. Without specific values for $b_1$, $b_2$, and $b_3$, we cannot perform a definitive calculation. However, we can express the condition generally: the system has no solution if the determinant of at least one 3x3 submatrix including $b$ is non-zero when $\\\lambda = \\pm \\frac{2\\sqrt{21}}{3}$.

To provide a step-by-step solution for determining the values of λ, we will recap the process and outline the specific steps to follow. Given the system $AX = b$, where

A=[12−3 3−15 41−2+lambda2]A = \begin{bmatrix} 1 & 2 & -3 \\\ 3 & -1 & 5 \\\ 4 & 1 & -2 + \\\\lambda^2 \end{bmatrix}

Step 1: Calculate the Determinant of Matrix A

We have already computed the determinant of matrix $A$:

det(A)=−3lambda2+28det(A) = -3\\\\lambda^2 + 28

Step 2: Set the Determinant to Zero and Solve for λ

To find the values of $\\\lambda$ for which the system might have no solution, we set the determinant equal to zero:

−3lambda2+28=0-3\\\\lambda^2 + 28 = 0

Solving for $\\\lambda$:

3lambda2=283\\\\lambda^2 = 28

lambda2=frac283{\\\\lambda}^2 = \\\\frac{28}{3}

λ=pmsqrtfrac283=pmfrac2sqrt213\\\\\lambda = \\\\pm \\\\sqrt{\\\\frac{28}{3}} = \\\\pm \\\\frac{2\\\\sqrt{21}}{3}

Step 3: Analyze the Augmented Matrix [A | b]

Construct the augmented matrix with a general column vector $b$:

[A∣b]=[12−3∣b1 3−15∣b2 41−2+lambda2∣b3][A | b] = \begin{bmatrix} 1 & 2 & -3 & | & b_1 \\\ 3 & -1 & 5 & | & b_2 \\\ 4 & 1 & -2 + \\\\lambda^2 & | & b_3 \end{bmatrix}

Step 4: Determine Conditions for No Solution

For the system to have no solution, the rank of the augmented matrix must be greater than the rank of matrix $A$ when $det(A) = 0$. This requires that at least one 3x3 submatrix of $[A | b]$ including $b$ has a non-zero determinant.

Step 5: Express the Condition for Inconsistency

The system has no solution if, for $\\\lambda = \\pm \\frac{2\\sqrt{21}}{3}$, at least one of the following determinants is non-zero:

  1. Determinant of the submatrix formed by columns 1, 2, and $b$
  2. Determinant of the submatrix formed by columns 1, 3, and $b$
  3. Determinant of the submatrix formed by columns 2, 3, and $b$

These conditions depend on the specific values of $b_1$, $b_2$, and $b_3$. If any of these determinants are non-zero, the system has no solution for the calculated values of $\\\lambda$. This step-by-step approach provides a comprehensive method to determine the values of $\\\lambda$ for which the system $AX = b$ has no solution, emphasizing the critical role of the determinant and the ranks of the coefficient and augmented matrices.

In conclusion, determining the values of $\\\lambda$ for which a system of linear equations has no solution involves a systematic approach. We start by calculating the determinant of the coefficient matrix $A$ and finding the values of $\\\lambda$ that make the determinant zero. These values indicate the potential for the system to have no solution or infinitely many solutions. The next crucial step is to analyze the augmented matrix $[A | b]$ and compare its rank with the rank of matrix $A$. If the rank of the augmented matrix is greater than the rank of the coefficient matrix when the determinant of $A$ is zero, then the system has no solution. This condition signifies that the addition of the column vector $b$ introduces a linear dependency that the system cannot resolve. The specific conditions under which this occurs depend on the values of the elements in the vector $b$. By following a step-by-step process, we can effectively identify the values of $\\\lambda$ that lead to an inconsistent system, providing a solid understanding of the system's solvability. This knowledge is vital in various applications where linear systems are used to model and solve real-world problems. The interplay between the determinant, matrix rank, and the augmented matrix provides a comprehensive framework for analyzing the solvability of linear systems.