Tafuta zao la matriki A na B (AB).

Tumepewa matriki $A = \begin{pmatrix} k & 4 \\ 3 & 2 \end{pmatrix}$ na $B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

$$AB = \begin{pmatrix} k & 4 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$$ $$AB = \begin{pmatrix} (k)(1) + (4)(3) & (k)(2) + (4)(4) \\ (3)(1) + (2)(3) & (3)(2) + (2)(4) \end{pmatrix}$$ $$AB = \begin{pmatrix} k + 12 & 2k + 16 \\ 3 + 6 & 6 + 8 \end{pmatrix}$$ $$AB = \begin{pmatrix} k + 12 & 2k + 16 \\ 9 & 14 \end{pmatrix}$$

Kwa matriki $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, kigezo ni $ad - bc$.

$$\det(AB) = (k + 12)(14) - (2k + 16)(9)$$ $$\det(AB) = 14k + 168 - (18k + 144)$$ $$\det(AB) = 14k + 168 - 18k - 144$$ $$\det(AB) = -4k + 24$$

Tumepewa kuwa $\det(AB) = 4$.

$$-4k + 24 = 4$$

Toa 24 pande zote mbili:

$$-4k = 4 - 24$$ $$-4k = -20$$

Gawanya kwa -4:

$$k = \frac{-20}{-4}$$ $$k = 5$$

Thamani ya $k$ ni 5.

Thamani ya $k$ ni $\boxed{\text{5}}$.