1. **Problem Statement:** We have a rigid joint held in equilibrium by three actuators A, B, and C with unknown forces $x$, $y$, and $z$ respectively. The resultant force projections on the coordinate axes are given as 5, 7, and -15 units. The contribution of each actuator per unit force along the three coordinates are: $$\text{A} = \begin{pmatrix}1 \\ 2 \\ 4\end{pmatrix}, \quad \text{B} = \begin{pmatrix}-1 \\ -4 \\ -9\end{pmatrix}, \quad \text{C} = \begin{pmatrix}3 \\ 7 \\ 2\end{pmatrix}$$ 2. **Formulating the system:** The total force projection on each coordinate is the sum of contributions from each actuator multiplied by their forces: $$\begin{cases} 1x - 1y + 3z = 5 \\ 2x - 4y + 7z = 7 \\ 4x - 9y + 2z = -15 \end{cases}$$ 3. **Matrix form:** $$\begin{pmatrix}1 & -1 & 3 \\ 2 & -4 & 7 \\ 4 & -9 & 2\end{pmatrix} \begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}5 \\ 7 \\ -15\end{pmatrix}$$ 4. **Solving the system:** We use methods like substitution, elimination, or matrix inversion. 5. **Step 1: Calculate determinant of coefficient matrix $A$:** $$A = \begin{vmatrix}1 & -1 & 3 \\ 2 & -4 & 7 \\ 4 & -9 & 2\end{vmatrix}$$ Calculate: $$= 1 \times (-4 \times 2 - 7 \times -9) - (-1) \times (2 \times 2 - 7 \times 4) + 3 \times (2 \times -9 - (-4) \times 4)$$ $$= 1 \times (-8 + 63) + 1 \times (4 - 28) + 3 \times (-18 + 16)$$ $$= 1 \times 55 + 1 \times (-24) + 3 \times (-2) = 55 - 24 - 6 = 25$$ Since determinant $\neq 0$, the system has a unique solution. 6. **Step 2: Find $x$ using Cramer's rule:** Replace first column with constants: $$A_x = \begin{vmatrix}5 & -1 & 3 \\ 7 & -4 & 7 \\ -15 & -9 & 2\end{vmatrix}$$ Calculate: $$= 5 \times (-4 \times 2 - 7 \times -9) - (-1) \times (7 \times 2 - 7 \times -15) + 3 \times (7 \times -9 - (-4) \times -15)$$ $$= 5 \times (-8 + 63) + 1 \times (14 + 105) + 3 \times (-63 - 60)$$ $$= 5 \times 55 + 1 \times 119 + 3 \times (-123) = 275 + 119 - 369 = 25$$ So, $$x = \frac{25}{25} = 1$$ 7. **Step 3: Find $y$ using Cramer's rule:** Replace second column with constants: $$A_y = \begin{vmatrix}1 & 5 & 3 \\ 2 & 7 & 7 \\ 4 & -15 & 2\end{vmatrix}$$ Calculate: $$= 1 \times (7 \times 2 - 7 \times -15) - 5 \times (2 \times 2 - 7 \times 4) + 3 \times (2 \times -15 - 7 \times 4)$$ $$= 1 \times (14 + 105) - 5 \times (4 - 28) + 3 \times (-30 - 28)$$ $$= 119 - 5 \times (-24) + 3 \times (-58) = 119 + 120 - 174 = 65$$ So, $$y = \frac{65}{25} = 2.6$$ 8. **Step 4: Find $z$ using Cramer's rule:** Replace third column with constants: $$A_z = \begin{vmatrix}1 & -1 & 5 \\ 2 & -4 & 7 \\ 4 & -9 & -15\end{vmatrix}$$ Calculate: $$= 1 \times (-4 \times -15 - 7 \times -9) - (-1) \times (2 \times -15 - 7 \times 4) + 5 \times (2 \times -9 - (-4) \times 4)$$ $$= 1 \times (60 + 63) + 1 \times (-30 - 28) + 5 \times (-18 + 16)$$ $$= 123 - 58 + 5 \times (-2) = 123 - 58 - 10 = 55$$ So, $$z = \frac{55}{25} = 2.2$$ 9. **Final answer:** $$\boxed{x = 1, \quad y = 2.6, \quad z = 2.2}$$ These are the actuator forces required to maintain equilibrium with the given projections.