Key Concept: Conditions for Solutions of a System of Linear Equations

For a system of linear equations given in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants:

1. Unique Solution: The system has a unique solution if and only if the determinant of the coefficient matrix, $D = \det(A)$, is non-zero ($D \neq 0$). In this case, Cramer's Rule can be directly applied.
2. No Solution (Inconsistent System): The system has no solution if and only if $D = 0$ AND at least one of the determinants $D_x, D_y, D_z$ (obtained by replacing the respective column in $A$ with $B$) is non-zero. Equivalently, using rank, if $rank(A) \neq rank([A|B])$.
3. Infinitely Many Solutions (Consistent and Dependent System): The system has infinitely many solutions if and only if $D = 0$ AND $D_x = D_y = D_z = 0$. Equivalently, using rank, if $rank(A) = rank([A|B]) < n$, where $n$ is the number of variables.

Our goal is to find the value of $k$ for which the given system has no solution. This means we first need to find the values of $k$ for which $D=0$, and then test those values to see which one leads to an inconsistent system.

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Step-by-Step Solution

Step 1: Formulate the Coefficient Matrix and Calculate its Determinant

The given system of equations is: $kx + y + z = 1$ $x + ky + z = k$ $x + y + zk = k^2$

We can write this system in matrix form $AX=B$, where: $A = \begin{pmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ k \\ k^2 \end{pmatrix}$

First, we calculate the determinant of the coefficient matrix $A$, denoted as $D$. $D = \det(A) = \left| \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right|$

To calculate the determinant, we expand along the first row: $D = k \left| \begin{matrix} k & 1 \\ 1 & k \end{matrix} \right| - 1 \left| \begin{matrix} 1 & 1 \\ 1 & k \end{matrix} \right| + 1 \left| \begin{matrix} 1 & k \\ 1 & 1 \end{matrix} \right|$ $D = k(k \cdot k - 1 \cdot 1) - 1(1 \cdot k - 1 \cdot 1) + 1(1 \cdot 1 - k \cdot 1)$ $D = k(k^2 - 1) - (k - 1) + (1 - k)$

Now, we simplify and factor the expression: $D = k(k^2 - 1) - (k - 1) - (k - 1)$ $D = k(k-1)(k+1) - 2(k - 1)$ We can factor out $(k-1)$: $D = (k-1) [k(k+1) - 2]$ $D = (k-1) [k^2 + k - 2]$ The quadratic term $k^2 + k - 2$ can be factored as $(k+2)(k-1)$. $D = (k-1)(k-1)(k+2)$ $D = (k-1)^2(k+2)$

For the system to have no unique solution (i.e., either no solution or infinitely many solutions), we must have $D=0$. Setting $D=0$: $(k-1)^2(k+2) = 0$ This gives us two critical values for $k$: $k = 1 \quad \text{or} \quad k = -2$

Step 2: Analyze Each Case where $D=0$

We must now examine these values of $k$ to determine which one leads to no solution.

Case A: When $k = 1$

Substitute $k=1$ into the original system of equations: $(1)x + y + z = 1 \implies x + y + z = 1$ $x + (1)y + z = 1 \implies x + y + z = 1$ $x + y + z(1) = 1^2 \implies x + y + z = 1$

In this case, all three equations become identical ($x+y+z=1$). This means the system represents a single plane in 3D space. Any point $(x, y, z)$ satisfying this equation is a solution. Since there are infinitely many such points, the system has infinitely many solutions for $k=1$. Therefore, $k=1$ is not the answer for "no solution".

Case B: When $k = -2$

Substitute $k=-2$ into the original system of equations: $(-2)x + y + z = 1 \implies -2x + y + z = 1$ $x + (-2)y + z = -2 \implies x - 2y + z = -2$ $x + y + z(-2) = (-2)^2 \implies x + y - 2z = 4$

Now we have the system:

1. $-2x + y + z = 1$
2. $x - 2y + z = -2$
3. $x + y - 2z = 4$

To determine if this system has no solution or infinitely many solutions, we can use the concept of ranks of matrices or check the consistency by elimination. Let's use the augmented matrix method, which is robust: The augmented matrix $[A|B]$ for this system is: $\left[ \begin{array}{ccc|c} -2 & 1 & 1 & 1 \\ 1 & -2 & 1 & -2 \\ 1 & 1 & -2 & 4 \end{array} \right]$

We perform row operations to bring the matrix to row echelon form:

1. Swap $R_1$ and $R_2$ to get a leading 1: $\left[ \begin{array}{ccc|c} 1 & -2 & 1 & -2 \\ -2 & 1 & 1 & 1 \\ 1 & 1 & -2 & 4 \end{array} \right]$
2. Perform $R_2 \to R_2 + 2R_1$ and $R_3 \to R_3 - R_1$ to create zeros below the leading 1 in the first column: $\left[ \begin{array}{ccc|c} 1 & -2 & 1 & -2 \\ 0 & -3 & 3 & -3 \\ 0 & 3 & -3 & 6 \end{array} \right]$
3. Divide $R_2$ by $-3$ to get a leading 1: $\left[ \begin{array}{ccc|c} 1 & -2 & 1 & -2 \\ 0 & 1 & -1 & 1 \\ 0 & 3 & -3 & 6 \end{array} \right]$
4. Perform $R_3 \to R_3 - 3R_2$ to create a zero below the leading 1 in the second column: $\left[ \begin{array}{ccc|c} 1 & -2 & 1 & -2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 3 \end{array} \right]$

Now, let's interpret the final row echelon form: The last row of the augmented matrix corresponds to the equation $0x + 0y + 0z = 3$, which simplifies to $0 = 3$. This is a contradiction.

A contradiction implies that the system of equations has no solution. In terms of rank, the rank of the coefficient matrix $A$ (ignoring the last column) is $rank(A)=2$ (as there are two non-zero rows: $\begin{pmatrix} 1 & -2 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 & -1 \end{pmatrix}$). The rank of the augmented matrix $[A|B]$ is $rank([A|B])=3$ (as all three rows are non-zero: $\begin{pmatrix} 1 & -2 & 1 & -2 \end{pmatrix}$, $\begin{pmatrix} 0 & 1 & -1 & 1 \end{pmatrix}$, and $\begin{pmatrix} 0 & 0 & 0 & 3 \end{pmatrix}$). Since $rank(A) \neq rank([A|B])$, the system is inconsistent and has no solution.