Question

For what real values of k does the following system of equations have a non-zero solution? kx+y+z=0 x+ky+z=0 x+y+kz=0

Answer 1

The real values of k are 1 and -2.

Answer 2

The given system of linear equations is:

$kx + y + z = 0$

$x + ky + z = 0$

$x + y + kz = 0$

This is a homogeneous system of linear equations. A homogeneous system has a non-zero solution if and only if the determinant of the coefficient matrix is zero.

The coefficient matrix is:

$A = \begin{pmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{pmatrix}$

The determinant of the matrix $A$ is calculated as:

$\det(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix}$

$\det(A) = k(k \cdot k - 1 \cdot 1) - 1(1 \cdot k - 1 \cdot 1) + 1(1 \cdot 1 - k \cdot 1)$

$\det(A) = k(k^2 - 1) - (k - 1) + (1 - k)$

$\det(A) = k^3 - k - k + 1 + 1 - k$

$\det(A) = k^3 - 3k + 2$

For the system to have a non-zero solution, the determinant must be zero:

$\det(A) = k^3 - 3k + 2 = 0$

We need to find the real roots of this cubic equation. We can test integer divisors of the constant term (2), which are $\pm 1, \pm 2$.

For $k=1$: $1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$. So, $k=1$ is a root.

This means $(k-1)$ is a factor of the polynomial $k^3 - 3k + 2$. We can perform polynomial division or synthetic division to find the other factor(s).

Using synthetic division with root 1:

1 | 1  0  -3   2
  |    1   1  -2
  ----------------
    1  1  -2   0

The quotient is $k^2 + k - 2$.

So, $k^3 - 3k + 2 = (k-1)(k^2 + k - 2)$.

Now, we factor the quadratic $k^2 + k - 2$. We look for two numbers that multiply to -2 and add to 1. These are 2 and -1.

$k^2 + k - 2 = (k+2)(k-1)$.

Substituting this back into the equation:

$(k-1)(k-1)(k+2) = 0$

$(k-1)^2(k+2) = 0$

The real values of $k$ for which the determinant is zero are the roots of this equation:

$(k-1)^2 = 0 \implies k-1 = 0 \implies k = 1$

$k+2 = 0 \implies k = -2$

Thus, the real values of $k$ for which the system of equations has a non-zero solution are $k=1$ and $k=-2$.