I have no clue where to go with this matrix:

1 0 k = 1

k 1 0 = 1

0 k 1 = 1

The problem says to find all values of k, if any, for which the system of linear equations has no solution.

x + kz = 1

kx + y = 1

ky + z = 1

Printable View

- Feb 5th 2013, 05:46 PMwidenerl194Linear Algebra matrix with variables
I have no clue where to go with this matrix:

1 0 k = 1

k 1 0 = 1

0 k 1 = 1

The problem says to find all values of k, if any, for which the system of linear equations has no solution.

x + kz = 1

kx + y = 1

ky + z = 1 - Feb 5th 2013, 06:41 PMjakncokeRe: Linear Algebra matrix with variables
take the determinant of $\displaystyle A = \begin{bmatrix} 1 & 0 & k \\ k & 1 & 0 \\ 0 & k & 1 \end{bmatrix} $ to get $\displaystyle Det(A) = k^3 + 1 $, we want the determinant equal to zero. (Why? Can you anwer this?) so $\displaystyle k = -1 $. so the matrix $\displaystyle Ax = b $ either has zero solutions or infinite solutions. Now how do we know which? Well note that $\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} $ are linearly. independent (the 3rd vector is not independent) so clearly if $\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ has a solution at all. $\displaystyle x = \begin{bmatrix} x_1 \\ -1 \end{bmatrix} $ which means $\displaystyle x_2 = -2 $, but clearly $\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -2 \\ -1 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} \not =\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $ so. for k = -1, no soln. exist.

- Feb 6th 2013, 08:20 AMDevenoRe: Linear Algebra matrix with variables
if k = -1, our system becomes:

x - z = 1 --> x > z.

y - x = 1 --> y > x > z.

z - y = 1 --> z > y > x > z, so z > z, which is impossible.