How to find the accurate determinant of this matrix?

Question

I'm solving the determinant of this matrix $\begin{bmatrix}1 & 2 & 3\\ 2& 2 &5\\ 3& 5& 1\end{bmatrix}$. I got two different answers by Gaussian Elimination. What's wrong with the second path?enter image description here

Answer 1

From wikipedia, the third rule says "Adding to one row a scalar multiple of another does not change the determinant." So adding half of row two to the third row will not change the determinant. But first multiplying row 3 by a factor of $2$ will double the determinant (rule 2). Subtracting now row 2 will not change the determinant anymore, but it will be doubled at the previous step. So the only operation you are allowed to do on some row $n$ is $$r_n\to r_n+\sum_{m\ne n}c_m r_m$$

Answer 2

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$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 2 & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 2 & 5 \\ 3 & 5 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 2 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - \frac{ 15 }{ 2 } \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - \frac{ 15 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 2 & 5 \\ 3 & 5 & 1 \\ \end{array} \right) $$

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