Determine the value of k such that the linear system kx y

Determine the value of k such that the linear system kx + y + z = 0 x + ky + z = 0 x + y + kz = 0 has a) A unique solution. b) A one-parameter family of solutions. c) A two-parameter family of solutions.

Solution

The coefficient matrix of the given linear system of equations is A (say) =

k

1

1

1

k

1

1

1

k

To solve the given linear system of equations, we will reduce A to its RREF as under:

Then the RREF of A is I₃=

1

0

0

0

1

0

0

0

1

Then the given linear system of equations has a unique trivial solution x = y = z = 0. However, we may observe that the 3^rd row operation is possible only if k-1 0 i.e. k 1. Also, the 7^th and the 8^th row operations are possible only if k²+k-20 i.e. ( k-1)(k+2) 0 i.e. k 1 and k -2.

(a). Thus, the given linear system of equations has a unique trivial solution x = y = z = 0 when k 1 and k -2.

(c). When k = 1, the matrix A changes to

1

1

1

1

1

1

1

1

1

In this case, the RREF of A is

1

1

1

0

0

0

0

0

0

Therefore, the the given linear system of equations is equivalent to x+y +z = 0 or, x = -y-z so that (x,y,z) = (-y-z,y,z) = y(-1,1,0)+z(-1,0,1). Thus, when k =1, the given linear system of equations has a 2 parameter family of solutions.

(b). When k = -2, A changes to

-2

1

1

1

-2

1

1

1

-2

In this case, the RREF of A is

1

0

-1

0

1

-1

0

0

0

Therefore, the the given linear system of equations is equivalent to x-z=0 or, x = z and y –z = 0 or, x = z. Then, (x,y,z) = (z,z,z) = z(1,1,1). Thus, when k =-2, the given linear system of equations has a 1 parameter family of solutions.

k	1	1
1	k	1
1	1	k