1. **Stating the problem:** We are given three matrices multiplied together and asked to find the final result vector $\mathbf{x} = (x_1, x_2, x_3, x_4)$ by performing Gaussian elimination (OBE) to find the leading 1. 2. **Understanding the matrices:** The matrices are: $$ A = \begin{bmatrix} \frac{1}{\sqrt{14}} & \frac{1}{2} & -\frac{5}{\sqrt{42}} & -\frac{1}{\sqrt{12}} \\ \frac{2}{\sqrt{14}} & \frac{1}{2} & \frac{4}{\sqrt{42}} & -\frac{1}{\sqrt{12}} \\ -\frac{3}{\sqrt{14}} & \frac{1}{2} & \frac{1}{\sqrt{42}} & -\frac{1}{\sqrt{12}} \\ 0 & \frac{1}{2} & 0 & \frac{3}{\sqrt{12}} \end{bmatrix} $$ $$ B = \begin{bmatrix} \sqrt{28} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{12}} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ $$ C = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 1 & 1 & 1 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ 1 & 1 & -2 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} \end{bmatrix} $$ 3. **Matrix multiplication order:** We first multiply $A \cdot B$, then multiply the result by $C$. 4. **Calculate $A \cdot B$:** Since $B$ is mostly zeros except for the first two diagonal elements, the multiplication simplifies: - The first column of $A \cdot B$ is the first column of $A$ multiplied by $\sqrt{28}$. - The second column of $A \cdot B$ is the second column of $A$ multiplied by $\frac{1}{\sqrt{12}}$. - The third and fourth columns of $A \cdot B$ are zero vectors. Calculate each element: - $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$ - $\frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}$ So, $$ A \cdot B = \begin{bmatrix} \frac{1}{\sqrt{14}} \times 2\sqrt{7} & \frac{1}{2} \times \frac{1}{2\sqrt{3}} & 0 & 0 \\ \frac{2}{\sqrt{14}} \times 2\sqrt{7} & \frac{1}{2} \times \frac{1}{2\sqrt{3}} & 0 & 0 \\ -\frac{3}{\sqrt{14}} \times 2\sqrt{7} & \frac{1}{2} \times \frac{1}{2\sqrt{3}} & 0 & 0 \\ 0 & \frac{1}{2} \times \frac{1}{2\sqrt{3}} & 0 & 0 \end{bmatrix} $$ Simplify each element: - $\frac{1}{\sqrt{14}} \times 2\sqrt{7} = 2 \times \frac{\sqrt{7}}{\sqrt{14}} = 2 \times \frac{\sqrt{7}}{\sqrt{14}} = 2 \times \frac{\sqrt{7}}{\sqrt{14}}$ Note $\sqrt{14} = \sqrt{7 \times 2} = \sqrt{7} \sqrt{2}$, so $$ \frac{\sqrt{7}}{\sqrt{14}} = \frac{\sqrt{7}}{\sqrt{7} \sqrt{2}} = \frac{1}{\sqrt{2}} $$ Thus, $$ \frac{1}{\sqrt{14}} \times 2\sqrt{7} = 2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \frac{2 \sqrt{2}}{2} = \sqrt{2} $$ Similarly, - $\frac{2}{\sqrt{14}} \times 2\sqrt{7} = 2 \times \frac{2}{\sqrt{14}} \sqrt{7} = 2 \times 2 \times \frac{\sqrt{7}}{\sqrt{14}} = 4 \times \frac{1}{\sqrt{2}} = 2 \sqrt{2}$ - $-\frac{3}{\sqrt{14}} \times 2\sqrt{7} = 2 \times -3 \times \frac{1}{\sqrt{2}} = -6 \times \frac{1}{\sqrt{2}} = -3 \sqrt{2}$ - $0 \times 2\sqrt{7} = 0$ For the second column: - $\frac{1}{2} \times \frac{1}{2\sqrt{3}} = \frac{1}{4\sqrt{3}}$ So the matrix $A \cdot B$ is: $$ \begin{bmatrix} \sqrt{2} & \frac{1}{4\sqrt{3}} & 0 & 0 \\ 2\sqrt{2} & \frac{1}{4\sqrt{3}} & 0 & 0 \\ -3\sqrt{2} & \frac{1}{4\sqrt{3}} & 0 & 0 \\ 0 & \frac{1}{4\sqrt{3}} & 0 & 0 \end{bmatrix} $$ 5. **Multiply $(A \cdot B) \cdot C$:** Dimensions: $(4 \times 4) \cdot (5 \times 3)$ is not defined, so likely a typo or the last matrix $C$ is $4 \times 3$ or $5 \times 3$. Assuming $C$ is $4 \times 3$ (taking first 4 rows), then: $$ C = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 1 & 1 & 1 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ 1 & 1 & -2 \end{bmatrix} $$ Multiply $4 \times 4$ matrix by $4 \times 3$ matrix to get $4 \times 3$ matrix. Calculate each element of the product $D = (A \cdot B) \cdot C$: For row 1: $$ D_{11} = \sqrt{2} \times \frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 + 0 + 0 = 1 + \frac{1}{4\sqrt{3}} = \frac{4\sqrt{3}}{4\sqrt{3}} + \frac{1}{4\sqrt{3}} = \frac{4\sqrt{3} + 1}{4\sqrt{3}} $$ $$ D_{12} = \sqrt{2} \times -\frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 + 0 + 0 = -1 + \frac{1}{4\sqrt{3}} = \frac{-4\sqrt{3} + 1}{4\sqrt{3}} $$ $$ D_{13} = \sqrt{2} \times 0 + \frac{1}{4\sqrt{3}} \times 1 + 0 + 0 = \frac{1}{4\sqrt{3}} $$ Similarly for row 2: $$ D_{21} = 2\sqrt{2} \times \frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 = 2 + \frac{1}{4\sqrt{3}} = \frac{8\sqrt{3} + 1}{4\sqrt{3}} $$ $$ D_{22} = 2\sqrt{2} \times -\frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 = -2 + \frac{1}{4\sqrt{3}} = \frac{-8\sqrt{3} + 1}{4\sqrt{3}} $$ $$ D_{23} = 2\sqrt{2} \times 0 + \frac{1}{4\sqrt{3}} \times 1 = \frac{1}{4\sqrt{3}} $$ Row 3: $$ D_{31} = -3\sqrt{2} \times \frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 = -3 + \frac{1}{4\sqrt{3}} = \frac{-12\sqrt{3} + 1}{4\sqrt{3}} $$ $$ D_{32} = -3\sqrt{2} \times -\frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 = 3 + \frac{1}{4\sqrt{3}} = \frac{12\sqrt{3} + 1}{4\sqrt{3}} $$ $$ D_{33} = -3\sqrt{2} \times 0 + \frac{1}{4\sqrt{3}} \times 1 = \frac{1}{4\sqrt{3}} $$ Row 4: $$ D_{41} = 0 \times \frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 = \frac{1}{4\sqrt{3}} $$ $$ D_{42} = 0 \times -\frac{1}{\sqrt{2}} + \frac{1}{4\sqrt{3}} \times 1 = \frac{1}{4\sqrt{3}} $$ $$ D_{43} = 0 \times 0 + \frac{1}{4\sqrt{3}} \times -2 = -\frac{1}{2\sqrt{3}} $$ 6. **Final vector $\mathbf{x}$:** Since the problem asks for $x_1, x_2, x_3, x_4$ presumably from the resulting matrix or vector, and the last matrix multiplication yields a $4 \times 3$ matrix, we interpret $x_i$ as the first column of $D$: $$ \mathbf{x} = \begin{bmatrix} \frac{4\sqrt{3} + 1}{4\sqrt{3}} \\ \frac{8\sqrt{3} + 1}{4\sqrt{3}} \\ \frac{-12\sqrt{3} + 1}{4\sqrt{3}} \\ \frac{1}{4\sqrt{3}} \end{bmatrix} $$ 7. **Summary:** We multiplied the matrices step-by-step, simplified radicals, and found the resulting vector $\mathbf{x}$ as above. This completes the solution.