Understanding Systems of Linear Equations and No Solution

A system of linear equations can have a unique solution, infinitely many solutions, or no solution. We are given the following system of three linear equations with three variables x, y, and z, and a parameter k:

• kx + y + z = 1
• x + ky + z = k
• x + y + kz = k2

To determine when this system has no solution, we can use the concept of determinants and matrix rank. The system can be written in matrix form as Ax = b, where:

For a system Ax = b:

• If det(A) is non-zero, the matrix A is invertible, and the system has a unique solution given by x = A^-1b.
• If det(A) is zero, the system has either no solution or infinitely many solutions. This occurs when the rank of the coefficient matrix A is less than the number of variables. To distinguish between no solution and infinite solutions when det(A) = 0, we examine the augmented matrix [A | b]. The system has no solution if and only if the rank of A is less than the rank of the augmented matrix [A | b]. The system has infinitely many solutions if the rank of A is equal to the rank of [A | b], and this rank is less than the number of variables.

Calculating the Determinant for Finding k

First, let's calculate the determinant of the coefficient matrix A:

\(det(A) = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}\)

Expanding along the first row:

\(det(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix}\)

\(det(A) = k(k \cdot k - 1 \cdot 1) - 1(1 \cdot k - 1 \cdot 1) + 1(1 \cdot 1 - k \cdot 1)\)

\(det(A) = k(k^2 - 1) - (k - 1) + (1 - k)\)

\(det(A) = k^3 - k - k + 1 + 1 - k\)

\(det(A) = k^3 - 3k + 2\)

Now, we find the values of k for which \(det(A) = 0\):

\(k^3 - 3k + 2 = 0\)

We can notice that \(k=1\) is a root, since \(1^3 - 3(1) + 2 = 1 - 3 + 2 = 0\). This means \((k-1)\) is a factor. We can perform polynomial division or synthetic division:

\((k^3 - 3k + 2) \div (k-1) = k^2 + k - 2\)

Factoring the quadratic term:

\(k^2 + k - 2 = (k+2)(k-1)\)

So, \(det(A) = (k-1)(k^2 + k - 2) = (k-1)(k+2)(k-1) = (k-1)^2(k+2)\).

The determinant is zero when \((k-1)^2(k+2) = 0\), which means \(k = 1\) or \(k = -2\). These are the values of k for which the system might have no solution or infinitely many solutions.

Analyzing Solutions for Different Values of k

Case 1: k = 1

If \(k = 1\), the system becomes:

• 1x + y + z = 1 → x + y + z = 1
• x + 1y + z = 1 → x + y + z = 1
• x + y + 1z = 12 → x + y + z = 1

All three equations are identical. This means the equations are linearly dependent, and the system is consistent. It has infinitely many solutions (e.g., any (x, y, z) such that x + y + z = 1).

Case 2: k = -2

If \(k = -2\), the system becomes:

• -2x + y + z = 1
• x - 2y + z = -2
• x + y - 2z = (-2)2 = 4

The augmented matrix is:

Let's perform row operations to find the rank of A and [A | b].

• Swap R1 and R2: \(\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ -2 & 1 & 1 & | & 1 \\ 1 & 1 & -2 & | & 4 \end{pmatrix}\)
• R2 → R2 + 2*R1, R3 → R3 - R1: \(\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ 0 & -3 & 3 & | & -3 \\ 0 & 3 & -3 & | & 6 \end{pmatrix}\)
• R2 → (-1/3)*R2: \(\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ 0 & 1 & -1 & | & 1 \\ 0 & 3 & -3 & | & 6 \end{pmatrix}\)
• R3 → R3 - 3*R2: \(\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & 0 & | & 3 \end{pmatrix}\)

The last row corresponds to the equation 0x + 0y + 0z = 3, which simplifies to 0 = 3. This is a contradiction, indicating that the system is inconsistent and has no solution for \(k = -2\).

Evaluating the Given Options for k

The options provided are 0, 1, and 2. Let's analyze each option:

Option: k = 0

If \(k = 0\), the determinant is \(det(A) = (0-1)^2(0+2) = (-1)^2(2) = 1 \cdot 2 = 2\). Since \(det(A) \neq 0\), the system has a unique solution when \(k=0\).

Option: k = 1

If \(k = 1\), as shown in Case 1, \(det(A) = 0\), and the system has infinitely many solutions.

Option: k = 2

If \(k = 2\), the determinant is \(det(A) = (2-1)^2(2+2) = (1)^2(4) = 1 \cdot 4 = 4\). Since \(det(A) \neq 0\), the matrix A is invertible, and the system has a unique solution when \(k=2\).

Conclusion Based on Analysis

Based on our mathematical analysis using determinants and row operations:

• For \(k = -2\), the system has no solution.
• For \(k = 1\), the system has infinitely many solutions.
• For \(k \neq 1\) and \(k \neq -2\), the system has a unique solution.

Among the given options (0, 1, 2), our analysis shows that for \(k=0\) and \(k=2\), the system has a unique solution, and for \(k=1\), it has infinitely many solutions. Therefore, based on the standard mathematical methods, none of the given options for k result in the system having no solution. The value of k that leads to no solution is -2.

Revision Table: System Solution Types based on k

Additional Information: Rank and Consistency

The Rouché-Capelli theorem or the Rank theorem provides a complete picture of the solutions to a system of linear equations Ax = b:

• The system is consistent (has at least one solution) if and only if the rank of the coefficient matrix A is equal to the rank of the augmented matrix [A | b].
• If the system is consistent and the rank of A equals the number of variables, there is a unique solution.
• If the system is consistent and the rank of A is less than the number of variables, there are infinitely many solutions.
• The system is inconsistent (has no solution) if and only if the rank of A is less than the rank of [A | b].

In our analysis for \(k = -2\), the rank of A (matrix without the last column) after row operations is 2 (two non-zero rows: \(\begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}\)), while the rank of the augmented matrix [A | b] is 3 (three non-zero rows: \(\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & 0 & | & 3 \end{pmatrix}\)). Since rank(A) < rank([A | b]), the system has no solution, which confirms our finding for \(k=-2\).

For \(k=1\), the augmented matrix reduces to a single non-zero row, meaning rank(A) = rank([A | b]) = 1, which is less than the number of variables (3). This indicates infinitely many solutions.

For \(k=0\) and \(k=2\) (and any \(k \neq 1, -2\)), \(det(A) \neq 0\), which implies rank(A) = 3. Since the maximum rank of the augmented matrix is also 3, rank(A) = rank([A | b]) = 3. This equals the number of variables, resulting in a unique solution.