1. **State the problem:** We are given an augmented matrix representing a system of linear equations (SLE) with parameter $k$. We want to find the value of $k$ such that the system has a 2-parameter family of solutions. 2. **Recall the concept:** A 2-parameter family of solutions means the system has two free variables, so the rank of the coefficient matrix is less than the number of variables by 2. 3. **Matrix setup:** The matrix has 4 variables (columns before the augmented part) and 3 equations (rows): $$\begin{bmatrix} 1 & 0 & 25 & k-3 \\ 0 & 1 & 0 & k-18 \\ 0 & -1 & 0 & 14 \end{bmatrix} \quad \text{with augmented column} \quad \begin{bmatrix} (k-3)(k+3) \\ -21 \\ 21 \end{bmatrix}$$ 4. **Analyze rows 2 and 3:** Add Row 2 and Row 3: $$ (0,1,0,k-18) + (0,-1,0,14) = (0,0,0,k-4) $$ Augmented part: $$ -21 + 21 = 0 $$ 5. **For consistency and free variables:** The new row after addition is: $$ (0,0,0,k-4 | 0) $$ - If $k \neq 4$, this row implies $0x + 0y + 0z + (k-4)w = 0$ which forces $w=0$ (no free variable here). - If $k = 4$, the row becomes all zeros, reducing the rank by 1 and increasing free variables. 6. **Check rank and free variables:** - Number of variables = 4 - Number of equations = 3 - For 2 free variables, rank must be $4 - 2 = 2$ 7. **Rank when $k=4$:** - Row 1 and Row 2 are independent. - Row 3 becomes dependent (sum of Row 2 and Row 3 is zero row). - So rank = 2, giving 2 free variables. **Final answer:** $$ k = 4 $$