Let's analyze the system and the concept of a **zero pivot error** in naïve Gauss elimination. ## Given System \[ \begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & 2 \\ 3 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \] ## Naïve Gauss Elimination (No Pivoting) The **pivot** at each elimination stage is the diagonal element of the current row. ### Step 1: First Pivot The first pivot is \( a_{11} = 0 \). - This means the method will **immediately encounter a zero pivot** in the first step, and the method will fail (division by zero or cannot eliminate). ## The Question The question says: > Which of the following linear systems will **NOT** produce a "zero pivot" error if the naïve Gauss elimination method is used... But **only one system is shown**. There must be a list of options in your question (a), (b), (c), (d), etc. **However, for the system you gave:** \[ \begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & 2 \\ 3 & 1 & 2 \end{bmatrix} \] **Naïve Gauss elimination will produce a zero pivot error, because the top-left element is zero.** --- ## General Rule - If the leading diagonal element to be used as a pivot at any step is zero, naïve Gauss elimination will fail (without row exchanges). ## If you have multiple options... - The correct answer is the system whose **first element** (top left) is **not zero**, and after the first elimination step, the next pivot (in the reduced matrix) is also not zero, and so on. - For the system you gave, **it will produce a zero pivot error** at the first step. --- **Conclusion:** **The given system will produce a zero pivot error in naïve Gauss elimination.** If you want to know which system will NOT produce a zero pivot error, look for a system whose leading diagonal elements (after each elimination step, without row exchanges) are all nonzero. --- **If you provide the rest of the options, I can help you check them specifically!**