# Thread: Linear Algebra matrix with variables

1. ## Linear 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

2. ## Re: Linear Algebra matrix with variables

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

3. ## Re: 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.