1. **Problem:** Verify if the matrix \( R = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \) is a solution matrix or not. Since the problem is ambiguous without a system, we assume checking if \( R \) is invertible or not by calculating its determinant. \[ \det(R) = 2 \begin{vmatrix} 4 & 5 \\ -3 & -4 \end{vmatrix} - (-3) \begin{vmatrix} -1 & 5 \\ 1 & -4 \end{vmatrix} + (-5) \begin{vmatrix} -1 & 4 \\ 1 & -3 \end{vmatrix} \] Calculate minors: \[ = 2(4 \times -4 - 5 \times -3) + 3(-1 \times -4 - 5 \times 1) - 5(-1 \times -3 - 4 \times 1) \] \[ = 2(-16 + 15) + 3(4 - 5) - 5(3 - 4) = 2(-1) + 3(-1) - 5(-1) = -2 - 3 + 5 = 0 \] Since \( \det(R) = 0 \), matrix \( R \) is singular and not invertible. 2. **Problem:** Given a point with global coordinates \( (5, a) \) and Cartesian coordinates \( (3, b) \), find the distance from the point to the X-axis. The distance from a point \( (x, y) \) to the X-axis is the absolute value of the \( y \)-coordinate. Since the Cartesian coordinate is \( (3, b) \), the distance to the X-axis is \( |b| \). 3. **Problem:** If \( \tan \theta = \frac{y}{x} \), find \( x \cos 2\theta + y \sin 2\theta \). Use double angle formulas: \[ \cos 2\theta = \frac{x^2 - y^2}{x^2 + y^2}, \quad \sin 2\theta = \frac{2xy}{x^2 + y^2} \] Calculate: \[ x \cos 2\theta + y \sin 2\theta = x \frac{x^2 - y^2}{x^2 + y^2} + y \frac{2xy}{x^2 + y^2} = \frac{x(x^2 - y^2) + 2xy^2}{x^2 + y^2} = \frac{x^3 - x y^2 + 2 x y^2}{x^2 + y^2} = \frac{x^3 + x y^2}{x^2 + y^2} = \frac{x(x^2 + y^2)}{x^2 + y^2} = x \] 4. **Problem:** Given \( A = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 4 \\ 1 & -3 \end{bmatrix} \), find \( (A^{-1} B^{-1})^{-1} \). Recall the property: \[ (A^{-1} B^{-1})^{-1} = B A \] Calculate \( B A \): \[ B A = \begin{bmatrix} 2 & 4 \\ 1 & -3 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2 \times 1 + 4 \times (-2) & 2 \times 3 + 4 \times 5 \\ 1 \times 1 + (-3) \times (-2) & 1 \times 3 + (-3) \times 5 \end{bmatrix} = \begin{bmatrix} 2 - 8 & 6 + 20 \\ 1 + 6 & 3 - 15 \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ 7 & -12 \end{bmatrix} \] 5. **Problem:** Given points \( A(4,7) \), \( B(-2,4) \), and a point \( C \) on segment \( AB \) such that \( BC = 3 \) and \( AC = 3\sqrt{1} = 3 \), find the coordinates of \( C \). Length \( AB = 3\sqrt{5} \). Since \( BC + AC = AB \), and \( BC = AC = 3 \), the sum is \( 6 \), but \( AB = 3\sqrt{5} \approx 6.708 \), so the given lengths are consistent. Let \( C \) divide \( AB \) in ratio \( AC : CB = 3 : 3 = 1 : 1 \), so \( C \) is the midpoint. Midpoint formula: \[ C = \left( \frac{4 + (-2)}{2}, \frac{7 + 4}{2} \right) = (1, 5.5) \] 6. **Problem:** If \( A + B = \frac{\pi}{4} \), find the value of \( (1 + \tan A)(1 + \tan B) \). Use the tangent addition formula: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan \frac{\pi}{4} = 1 \] So, \[ 1 = \frac{\tan A + \tan B}{1 - \tan A \tan B} \implies 1 - \tan A \tan B = \tan A + \tan B \implies 1 = \tan A + \tan B + \tan A \tan B \] Now, \[ (1 + \tan A)(1 + \tan B) = 1 + \tan A + \tan B + \tan A \tan B = 1 + (\tan A + \tan B) + \tan A \tan B \] From above, \[ \tan A + \tan B + \tan A \tan B = 1 \implies (1 + \tan A)(1 + \tan B) = 1 + 1 = 2 \] **Final answers:** 1. \( \det(R) = 0 \), matrix \( R \) is singular. 2. Distance to X-axis = \( |b| \). 3. \( x \cos 2\theta + y \sin 2\theta = x \). 4. \( (A^{-1} B^{-1})^{-1} = B A = \begin{bmatrix} -6 & 26 \\ 7 & -12 \end{bmatrix} \). 5. Coordinates of \( C \) are \( (1, 5.5) \). 6. \( (1 + \tan A)(1 + \tan B) = 2 \).