Determine the values of k such that the system of linear equ

Determine the values of k such that the system of linear equations does not have a unique solution. x + y + kz = 2 x + ky + z = 5 kx + y + z = 3

Solution

We are given the following three linear equations:

x + y + kz = 3 .....(1)

2x + ky +z = 3 .....(2)

5kx + y + z = 3.....(3)

On subtracting the 1st equatuion from the 3rd equation, we get ( 5k-1)x + (1- k)z = 0 ....(4)

On mutiplying both the sides of the 3rd equation by k, we get 5k²x + ky + kz = 3k...(5)

Now, on subtracting the 2nd equation from the 5th equation, we get (5k² - 2)x + (k - 1)z = 3(k - 1)....(6)

Now, on adding the 4th and the 6th equations, we get ( 5k² -2 + 5k - 1)x = 3(k -1) or, x = 3(k -1) / ( 5k² +5k - 3)

Since division by 0 is undefined, the given system of equations will not have a unique solution when   ( 5k² +5k - 3) = 0 i.e. when k = [-5 ± (25+60)]/10 or k = [-5 ± 85]/10.