Question

Determmine value of k such that

Kx+y+z=1

X+ky+z=1

X+y+kz=1 has a) no solution b) unique solution c) more than one solution

Answer 1

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

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

$$k^3-k-k+1+1-k=k^3-3k+2$$

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

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

Nonhomogeneous system of linear equations has a unique non-trivial solution if and only if

$$\det A\not=0=>(k-1)^2(k+2)\not=0$$

The system has a unique non-trivial solution if $k\in \R, k\not=1, k\not=-2.$

If $k=1,$ we have

$$\begin{matrix} x+y+z=1\\ x+y+z=1\\ x+y+z=1\\ \end{matrix}$$

The system has an infinite number of solutions if $k=1.$

If $k=-2,$ we have

$$\begin{matrix} -2x+y+z=1\\ x-2y+z=1\\ x+y-2z=1\\ \end{matrix}$$

$$\begin{matrix} -3x+3y=0\\ 3y-3z=0\\ x+y-2z=1\\ \end{matrix}$$

$$\begin{matrix} x=y\\ y=z\\ y+y-2y=1\\ \end{matrix}$$

The system has no solution if $k=-2.$