Question

The system of equations

$kx + y + z = 1$ $x + ky + z = k$ $x + y + zk = k^2$

has no solution, if k is equal to

• A. 1
• B. -2
• C. 2
• D. -1

Answer 1

Answer: (B)

-2

Answer 2

The given system of linear equations can be represented in matrix form $AX = B$, where: $A = \begin{pmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, $B = \begin{pmatrix} 1 \\ k \\ k^2 \end{pmatrix}$.

A system of linear equations $AX = B$ has no solution if $\det(A) = 0$ and the rank of the augmented matrix $[A|B]$ is greater than the rank of the coefficient matrix $A$.

First, we calculate the determinant of the coefficient matrix $A$: $\det(A) = k(k^2 - 1) - 1(k - 1) + 1(1 - k)$ $\det(A) = k(k-1)(k+1) - (k-1) - (k-1)$ $\det(A) = (k-1) [k(k+1) - 1 - 1]$ $\det(A) = (k-1) [k^2 + k - 2]$ $\det(A) = (k-1) (k+2)(k-1)$ $\det(A) = (k-1)^2 (k+2)$

For the system to have no solution or infinitely many solutions, $\det(A)$ must be zero. $\det(A) = 0 \implies (k-1)^2 (k+2) = 0$ This gives $k=1$ or $k=-2$.

Case 1: $k=1$ If $k=1$, the system becomes: $x + y + z = 1$ $x + y + z = 1$ $x + y + z = 1$ This system has infinitely many solutions.

Case 2: $k=-2$ If $k=-2$, the system becomes: $-2x + y + z = 1$ $x - 2y + z = -2$ $x + y - 2z = 4$

The augmented matrix is: $[A|B] = \begin{pmatrix} -2 & 1 & 1 & | & 1 \\ 1 & -2 & 1 & | & -2 \\ 1 & 1 & -2 & | & 4 \end{pmatrix}$

Performing row operations: $\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ 0 & -3 & 3 & | & -3 \\ 0 & 3 & -3 & | & 6 \end{pmatrix}$ (after swapping $R_1, R_2$ and applying $R_2 \to R_2 + 2R_1$, $R_3 \to R_3 - R_1$) $\begin{pmatrix} 1 & -2 & 1 & | & -2 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & 0 & | & 3 \end{pmatrix}$ (after $R_2 \to R_2 / (-3)$ and $R_3 \to R_3 - 3R_2$)

The rank of $A$ is 2, and the rank of $[A|B]$ is 3. Since $\text{rank}(A) \neq \text{rank}([A|B])$, the system has no solution for $k=-2$.