1. **State the problem:** We need to find the values of $\theta_1$, $\theta_2$, and $\theta_3$ that satisfy the matrix equation $$\begin{bmatrix} 40 & -20 & -10 \\ -20 & 50 & -20 \\ -10 & -20 & 45 \end{bmatrix} \begin{bmatrix} \theta_1 \\ \theta_2 \\ \theta_3 \end{bmatrix} = \begin{bmatrix} 150 \\ -80 \\ 120 \end{bmatrix}$$ 2. **Formula and method:** This is a system of linear equations represented as $A\mathbf{\theta} = \mathbf{b}$, where $A$ is the coefficient matrix, $\mathbf{\theta}$ is the vector of unknowns, and $\mathbf{b}$ is the constants vector. To solve for $\mathbf{\theta}$, we use the formula: $$\mathbf{\theta} = A^{-1} \mathbf{b}$$ provided $A$ is invertible. 3. **Calculate the inverse of $A$:** The matrix $A$ is $$A = \begin{bmatrix} 40 & -20 & -10 \\ -20 & 50 & -20 \\ -10 & -20 & 45 \end{bmatrix}$$ We find $A^{-1}$ using standard methods (e.g., adjoint and determinant or row reduction). For brevity, the inverse is computed as: $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$ 4. **Calculate determinant $\det(A)$:** $$\det(A) = 40(50 \times 45 - (-20)(-20)) - (-20)(-20 \times 45 - (-20)(-10)) + (-10)(-20 \times -20 - 50 \times -10)$$ Calculate each term: - $50 \times 45 = 2250$ - $(-20)(-20) = 400$ - $-20 \times 45 = -900$ - $(-20)(-10) = 200$ - $-20 \times -20 = 400$ - $50 \times -10 = -500$ So, $$\det(A) = 40(2250 - 400) - (-20)(-900 - 200) + (-10)(400 - (-500))$$ $$= 40(1850) - (-20)(-1100) + (-10)(900)$$ $$= 74000 - 22000 - 9000 = 43000$$ 5. **Calculate adjoint matrix $\mathrm{adj}(A)$:** Calculate cofactors and transpose: $$\mathrm{adj}(A) = \begin{bmatrix} 1850 & 1100 & 900 \\ 1100 & 1750 & 700 \\ 900 & 700 & 1800 \end{bmatrix}$$ 6. **Calculate $A^{-1}$:** $$A^{-1} = \frac{1}{43000} \begin{bmatrix} 1850 & 1100 & 900 \\ 1100 & 1750 & 700 \\ 900 & 700 & 1800 \end{bmatrix}$$ 7. **Multiply $A^{-1}$ by $\mathbf{b}$:** $$\mathbf{\theta} = A^{-1} \mathbf{b} = \frac{1}{43000} \begin{bmatrix} 1850 & 1100 & 900 \\ 1100 & 1750 & 700 \\ 900 & 700 & 1800 \end{bmatrix} \begin{bmatrix} 150 \\ -80 \\ 120 \end{bmatrix}$$ Calculate each component: - $\theta_1 = \frac{1}{43000}(1850 \times 150 + 1100 \times (-80) + 900 \times 120) = \frac{1}{43000}(277500 - 88000 + 108000) = \frac{297500}{43000} = 6.9209$ - $\theta_2 = \frac{1}{43000}(1100 \times 150 + 1750 \times (-80) + 700 \times 120) = \frac{1}{43000}(165000 - 140000 + 84000) = \frac{109000}{43000} = 2.5349$ - $\theta_3 = \frac{1}{43000}(900 \times 150 + 700 \times (-80) + 1800 \times 120) = \frac{1}{43000}(135000 - 56000 + 216000) = \frac{295000}{43000} = 6.8605$ 8. **Final answer:** $$\boxed{\theta_1 \approx 6.92, \quad \theta_2 \approx 2.53, \quad \theta_3 \approx 6.86}$$ These are the values of $\theta_1$, $\theta_2$, and $\theta_3$ that satisfy the given matrix equation.