Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if

The correct option is |k| ≠ 2.

To understand why this is correct, we can analyze the system of linear equations:
1) $kx+y=3$
2) $4x+ky=4$

A system of two linear equations in the form:
$a_{1}x+b_{1}y=c_1$
$a_{2}x+b_{2}y=c_2$
has a unique solution if and only if the determinant of the coefficient matrix is non-zero. That is:

$\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|\ne 0$

This determinant condition simplifies to:
$a_1{b}_2-a_2{b}_1\ne 0$

Substituting the coefficients from our equations ($a_1=k$, $b_1=1$, $a_2=4$, and $b_2=k$) into the formula, we obtain:

$\left(k\right)\left(k\right)-\left(4\right)\left(1\right)\ne 0$

Solving for $k$:

$k^2-4\ne 0$

$k^2\ne 4$

Taking the square root of both sides gives the condition for a unique solution:

$\left|k\right|\ne 2$

Therefore, the system has a unique solution if and only if $\left|k\right|\ne 2$ (which means $k\ne 2$ and $k\ne -2$).