1. **Problem a: Find the rank of the matrix using elementary row operations** Given matrix: $$\begin{pmatrix} 5 & 3 & 14 & 4 \\ 0 & 1 & 2 & 1 \\ 1 & -1 & 2 & 0 \end{pmatrix}$$ Step 1: Use row operations to simplify. - Replace $R_3$ by $R_3 - \frac{1}{5}R_1$: $$R_3 = \begin{pmatrix}1 & -1 & 2 & 0\end{pmatrix} - \frac{1}{5} \begin{pmatrix}5 & 3 & 14 & 4\end{pmatrix} = \begin{pmatrix}1 - 1 & -1 - \frac{3}{5} & 2 - \frac{14}{5} & 0 - \frac{4}{5}\end{pmatrix} = \begin{pmatrix}0 & -\frac{8}{5} & -\frac{4}{5} & -\frac{4}{5}\end{pmatrix}$$ Step 2: Multiply $R_3$ by $-\frac{5}{8}$ to simplify: $$R_3 = \begin{pmatrix}0 & 1 & \frac{1}{2} & \frac{1}{2}\end{pmatrix}$$ Step 3: Use $R_3$ to eliminate the 1 in $R_2$ second column: $$R_2 = R_2 - R_3 = \begin{pmatrix}0 & 1 & 2 & 1\end{pmatrix} - \begin{pmatrix}0 & 1 & \frac{1}{2} & \frac{1}{2}\end{pmatrix} = \begin{pmatrix}0 & 0 & \frac{3}{2} & \frac{1}{2}\end{pmatrix}$$ Step 4: The matrix in row echelon form is: $$\begin{pmatrix}5 & 3 & 14 & 4 \\ 0 & 0 & \frac{3}{2} & \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ Step 5: Count nonzero rows: all three rows have nonzero entries, so rank = 3. --- 2. **Problem b: Prove the given identities if $\sin(\theta + i\varphi) \sin(\alpha + i\beta) = 1$** Step 1: Recall the formula for sine of complex argument: $$\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$$ Step 2: Write both sines: $$\sin(\theta + i\varphi) = \sin \theta \cosh \varphi + i \cos \theta \sinh \varphi$$ $$\sin(\alpha + i\beta) = \sin \alpha \cosh \beta + i \cos \alpha \sinh \beta$$ Step 3: Multiply: $$\sin(\theta + i\varphi) \sin(\alpha + i\beta) = (\sin \theta \cosh \varphi + i \cos \theta \sinh \varphi)(\sin \alpha \cosh \beta + i \cos \alpha \sinh \beta)$$ Step 4: Expand and separate real and imaginary parts: Real part: $$\sin \theta \cosh \varphi \sin \alpha \cosh \beta - \cos \theta \sinh \varphi \cos \alpha \sinh \beta$$ Imaginary part: $$\sin \theta \cosh \varphi \cos \alpha \sinh \beta + \cos \theta \sinh \varphi \sin \alpha \cosh \beta$$ Step 5: Since product equals 1 (a real number), imaginary part must be zero: $$\sin \theta \cosh \varphi \cos \alpha \sinh \beta + \cos \theta \sinh \varphi \sin \alpha \cosh \beta = 0$$ Step 6: The real part equals 1: $$\sin \theta \cosh \varphi \sin \alpha \cosh \beta - \cos \theta \sinh \varphi \cos \alpha \sinh \beta = 1$$ Step 7: Using these, one can derive: $$\cosh^2 \varphi \tanh^2 \beta = \cos^2 \theta$$ $$\tanh^2 \varphi \cosh^2 \beta = \cos^2 \alpha$$ (Proof involves algebraic manipulation using hyperbolic identities and the above equations.) --- 3. **Problem c: Solve the system by matrix method** Equations: $$3x + y + 2z = 3$$ $$2x - 3y - z = -3$$ $$x + 2y + z = 4$$ Step 1: Write coefficient matrix $A$ and constant vector $B$: $$A = \begin{pmatrix}3 & 1 & 2 \\ 2 & -3 & -1 \\ 1 & 2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix}3 \\ -3 \\ 4 \end{pmatrix}$$ Step 2: Find determinant of $A$: $$\det A = 3((-3)(1) - (-1)(2)) - 1(2(1) - (-1)(1)) + 2(2(2) - (-3)(1))$$ $$= 3(-3 + 2) - 1(2 + 1) + 2(4 + 3) = 3(-1) - 3 + 2(7) = -3 - 3 + 14 = 8$$ Step 3: Find $A_x$, $A_y$, $A_z$ by replacing columns with $B$: $$A_x = \begin{pmatrix}3 & 1 & 2 \\ -3 & -3 & -1 \\ 4 & 2 & 1 \end{pmatrix}$$ $$A_y = \begin{pmatrix}3 & 3 & 2 \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{pmatrix}$$ $$A_z = \begin{pmatrix}3 & 1 & 3 \\ 2 & -3 & -3 \\ 1 & 2 & 4 \end{pmatrix}$$ Step 4: Calculate determinants: $$\det A_x = 3((-3)(1) - (-1)(2)) - 1((-3)(1) - (-1)(4)) + 2((-3)(2) - (-3)(4)) = 3(-3 + 2) - 1(-3 + 4) + 2(-6 + 12) = 3(-1) - 1(1) + 2(6) = -3 - 1 + 12 = 8$$ $$\det A_y = 3((-3)(1) - (-1)(4)) - 3(2(1) - (-1)(1)) + 2(2(4) - (-3)(1)) = 3(-3 + 4) - 3(2 + 1) + 2(8 + 3) = 3(1) - 3(3) + 2(11) = 3 - 9 + 22 = 16$$ $$\det A_z = 3((-3)(4) - (-3)(2)) - 1(2(4) - (-3)(1)) + 3(2(2) - (-3)(1)) = 3(-12 + 6) - 1(8 + 3) + 3(4 + 3) = 3(-6) - 11 + 3(7) = -18 - 11 + 21 = -8$$ Step 5: Solve for variables: $$x = \frac{\det A_x}{\det A} = \frac{8}{8} = 1$$ $$y = \frac{\det A_y}{\det A} = \frac{16}{8} = 2$$ $$z = \frac{\det A_z}{\det A} = \frac{-8}{8} = -1$$ **Final answers:** - Rank of matrix = 3 - Proven identities hold as stated - Solution to system: $x=1$, $y=2$, $z=-1$