The problem consists of two parts. Part 1: Convert the given matrix $\begin{bmatrix} 5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & 7 \end{bmatrix}$ into an upper triangular matrix. Part 2: Using properties of determinants, prove that $\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^2$.

1. Problem Description

The problem consists of two parts.

Part 1: Convert the given matrix $\begin{bmatrix} 5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & 7 \end{bmatrix}$ into an upper triangular matrix.

Part 2: Using properties of determinants, prove that $\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^2$.

2. Solution Steps

Part 1: Converting to Upper Triangular Matrix

Let $A = \begin{bmatrix} 5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & 7 \end{bmatrix}$. We want to transform this matrix into an upper triangular matrix.

Step 1: Make the element in the (2,1) position zero.

Perform the operation $R_2 \rightarrow R_2 - \frac{6}{5}R_1$.

$R_2 = [6, -8, -1] - \frac{6}{5}[5, -7, 1] = [6-6, -8 + \frac{42}{5}, -1 - \frac{6}{5}] = [0, \frac{2}{5}, -\frac{11}{5}]$.

The new matrix becomes $\begin{bmatrix} 5 & -7 & 1 \\ 0 & \frac{2}{5} & -\frac{11}{5} \\ 3 & 2 & 7 \end{bmatrix}$.

Step 2: Make the element in the (3,1) position zero.

Perform the operation $R_3 \rightarrow R_3 - \frac{3}{5}R_1$.

$R_3 = [3, 2, 7] - \frac{3}{5}[5, -7, 1] = [3-3, 2 + \frac{21}{5}, 7 - \frac{3}{5}] = [0, \frac{31}{5}, \frac{32}{5}]$.

The new matrix becomes $\begin{bmatrix} 5 & -7 & 1 \\ 0 & \frac{2}{5} & -\frac{11}{5} \\ 0 & \frac{31}{5} & \frac{32}{5} \end{bmatrix}$.

Step 3: Make the element in the (3,2) position zero.

Perform the operation $R_3 \rightarrow R_3 - \frac{31}{2}R_2$.

$R_3 = [0, \frac{31}{5}, \frac{32}{5}] - \frac{31}{2}[0, \frac{2}{5}, -\frac{11}{5}] = [0, \frac{31}{5} - \frac{31}{5}, \frac{32}{5} + \frac{341}{10}] = [0, 0, \frac{64+341}{10}] = [0, 0, \frac{405}{10}] = [0, 0, \frac{81}{2}]$.

The new matrix becomes $\begin{bmatrix} 5 & -7 & 1 \\ 0 & \frac{2}{5} & -\frac{11}{5} \\ 0 & 0 & \frac{81}{2} \end{bmatrix}$.

Part 2: Proving the determinant identity

Let $D = \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}$.

Apply $C_1 \rightarrow C_1 + C_2 + C_3$.

$D = \begin{vmatrix} x+4+2x+2x & 2x & 2x \\ 2x+x+4+2x & x+4 & 2x \\ 2x+2x+x+4 & 2x & x+4 \end{vmatrix} = \begin{vmatrix} 5x+4 & 2x & 2x \\ 5x+4 & x+4 & 2x \\ 5x+4 & 2x & x+4 \end{vmatrix}$.

Factor out $(5x+4)$ from the first column:

$D = (5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{vmatrix}$.

Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:

$D = (5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 0 & x+4-2x & 0 \\ 0 & 0 & x+4-2x \end{vmatrix} = (5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 0 & 4-x & 0 \\ 0 & 0 & 4-x \end{vmatrix}$.

Now, we can compute the determinant by expanding along the first column:

$D = (5x+4) \cdot 1 \cdot \begin{vmatrix} 4-x & 0 \\ 0 & 4-x \end{vmatrix} = (5x+4) \cdot (4-x)(4-x) = (5x+4)(4-x)^2$.

3. Final Answer

Part 1: The upper triangular matrix is $\begin{bmatrix} 5 & -7 & 1 \\ 0 & \frac{2}{5} & -\frac{11}{5} \\ 0 & 0 & \frac{81}{2} \end{bmatrix}$.

Part 2: The proof is complete, $\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^2$.