Given that A = [8/(x+1), 2] and B = [1, 2] are singular matrices, find the value of x.

Ili kuzidisha matrisi, tunazidisha safu mlalo za matrisi ya kwanza na safu wima za matrisi ya pili.

$$AB = \begin{pmatrix} 8 & x \\ x+1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}$$ $$AB = \begin{pmatrix} (8)(1) + (x)(3) & (8)(2) + (x)(0) \\ (x+1)(1) + (2)(3) & (x+1)(2) + (2)(0) \end{pmatrix}$$ $$AB = \begin{pmatrix} 8 + 3x & 16 + 0 \\ x+1 + 6 & 2x+2 + 0 \end{pmatrix}$$ $$AB = \begin{pmatrix} 3x+8 & 16 \\ x+7 & 2x+2 \end{pmatrix}$$

Kwa matrisi ya $2 \times 2$, $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, kigeuzi ni $ad - bc$.
Hivyo, kigeuzi cha $AB$ ni:

$$\det(AB) = (3x+8)(2x+2) - (16)(x+7)$$

Panua maneno:

$$\det(AB) = (6x^2 + 6x + 16x + 16) - (16x + 112)$$ $$\det(AB) = 6x^2 + 22x + 16 - 16x - 112$$

Unganisha maneno yanayofanana:

$$\det(AB) = 6x^2 + (22x - 16x) + (16 - 112)$$ $$\det(AB) = 6x^2 + 6x - 96$$

Kwa kuwa $AB$ ni matrisi isiyo na kigeuzi, $\det(AB) = 0$.

$$6x^2 + 6x - 96 = 0$$

Gawanya mlinganyo mzima kwa 6 ili kurahisisha:

$$\frac{6x^2}{6} + \frac{6x}{6} - \frac{96}{6} = 0$$ $$x^2 + x - 16 = 0$$

Huu ni mlinganyo wa quadratic. Tutatumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Hapa, $a=1$, $b=1$, $c=-16$.

$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-16)}}{2(1)}$$ $$x = \frac{-1 \pm \sqrt{1 + 64}}{2}$$ $$x = \frac{-1 \pm \sqrt{65}}{2}$$

Kwa hivyo, kuna thamani mbili za $x$:

$$x_1 = \frac{-1 + \sqrt{65}}{2}$$ $$x_2 = \frac{-1 - \sqrt{65}}{2}$$

Thamani za $x$ ni $\boxed{\frac{-1 + \sqrt{65}}{2}}$ na $\boxed{\frac{-1 - \sqrt{65}}{2}}$.