# Solution of system of linear equations

• Nov 22nd 2009, 11:24 PM
Solution of system of linear equations
1.Using either Gaussian or Gauss-Jordan elimination, find the value(s) of k, if any, for which the following system will have (i) no solution (ii) a unique solution, and (iii) infinitely many solutions

x + y + kz = 1
x+ky + z = 1
kx +y + z = -2

For the value of k for which the system has a unique solution, give unique values of x, y, and z.

Please explain to me how to do it. thanks! (Wink)
• Nov 23rd 2009, 03:58 AM
HallsofIvy
Quote:

1.Using either Gaussian or Gauss-Jordan elimination, find the value(s) of k, if any, for which the following system will have (i) no solution (ii) a unique solution, and (iii) infinitely many solutions

x + y + kz = 1
x+ky + z = 1
kx +y + z = -2

For the value of k for which the system has a unique solution, give unique values of x, y, and z.

Please explain to me how to do it. thanks! (Wink)

Just do exactly what you are told to do. Use "Gauss Jordan elimination" on the augmented matrix $\begin{bmatrix}1 & 1 & k & 1 \\ 1 & k & 1 & 1 \\ k & 1 & 1 & -2$. The last row will have "0" in the first two places, expressions in k in the third and fourth places.

Values of k that make the expression in the third place non-zero give a unique solution since you can just divide through by it to get the solution.

Values of k that make the expression in the third place zero and in the fourth place non-zero give no solution since that corresponds to 0z= non-zero which is not true for any z.

Values of k that make the expressions in both third and fourth places 0 give infinitely many solutions since that corresponds to 0z= 0 which is true for all z.
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