In transformation an object of area x cm2 is mapped on to an image whose area is 13x cm2. given that the matrix of the transformation is ((x-1, 7), (x, 3x)). find the possible values of x.

Matrix ya mabadiliko imetolewa kama $M = \begin{pmatrix} x & 7 \\ x-1 & 3x \end{pmatrix}$.
Determinant ya matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ ni $ad - bc$.
Kwa hiyo, determinant ya matrix hii ni:

$$\det(M) = (x)(3x) - (7)(x-1)$$ $$\det(M) = 3x^2 - (7x - 7)$$ $$\det(M) = 3x^2 - 7x + 7$$

Formula inayohusisha eneo la kitu na eneo la picha baada ya mabadiliko ni:
Eneo la picha = $|\det(M)| \times$ Eneo la kitu
Tunapewa kuwa eneo la kitu ni $x \text{ cm}^2$ na eneo la picha ni $13x \text{ cm}^2$.
Kwa hiyo, tunaweza kuandika mlinganyo:

$$13x = |3x^2 - 7x + 7| \times x$$

Kwa kuwa eneo $x$ lazima liwe chanya ($x > 0$), tunaweza kugawanya pande zote mbili kwa $x$:

$$13 = |3x^2 - 7x + 7|$$

Hii inamaanisha tuna kesi mbili zinazowezekana:
Kesi 1: $3x^2 - 7x + 7 = 13$
Kesi 2: $3x^2 - 7x + 7 = -13$

Kesi 1:

$$3x^2 - 7x + 7 = 13$$ $$3x^2 - 7x - 6 = 0$$

Tunatumia formula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-6)}}{2(3)}$$ $$x = \frac{7 \pm \sqrt{49 + 72}}{6}$$ $$x = \frac{7 \pm \sqrt{121}}{6}$$ $$x = \frac{7 \pm 11}{6}$$

Hii inatoa thamani mbili za $x$:

$$x_1 = \frac{7 + 11}{6} = \frac{18}{6} = 3$$ $$x_2 = \frac{7 - 11}{6} = \frac{-4}{6} = -\frac{2}{3}$$

Kesi 2:

$$3x^2 - 7x + 7 = -13$$ $$3x^2 - 7x + 20 = 0$$

Tunatumia formula ya quadratic:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(20)}}{2(3)}$$ $$x = \frac{7 \pm \sqrt{49 - 240}}{6}$$ $$x = \frac{7 \pm \sqrt{-191}}{6}$$

Kwa kuwa discriminant ($\sqrt{-191}$) ni namba hasi, hakuna suluhisho halisi (real solutions) kwa $x$ katika kesi hii.

Kwa kuwa $x$ inawakilisha eneo, lazima iwe namba chanya. Kati ya suluhisho zilizopatikana kutoka Kesi 1, $x=3$ ni chanya na $x=-\frac{2}{3}$ ni hasi.
Kwa hiyo, thamani pekee inayowezekana ya $x$ ni 3.

Thamani inayowezekana ya $x$ ni $\boxed{\text{3}}$.