Questions: Solve the determinant: 1 6 2 3 8 4 1 0 2

Solution Steps

We start with the matrix for which we need to calculate the determinant:

\[ \left|\begin{matrix}1 & 6 & 2\\3 & 8 & 4\\1 & 0 & 2\end{matrix}\right| \]

To simplify the calculation, we perform a row swap to position the largest leading coefficient at the top:

\[ \Rightarrow \left|\begin{matrix}3 & 8 & 4\\1 & 6 & 2\\1 & 0 & 2\end{matrix}\right| \]

Next, we apply row operations to convert the matrix into an upper triangular form:

1. Subtract \(\frac{1}{3}\) of the first row from the second row: \[ \Rightarrow \left|\begin{matrix}3 & 8 & 4\\0 & \frac{10}{3} & \frac{2}{3}\\1 & 0 & 2\end{matrix}\right| \]

2. Subtract \(\frac{1}{3}\) of the first row from the third row: \[ \Rightarrow \left|\begin{matrix}3 & 8 & 4\\0 & \frac{10}{3} & \frac{2}{3}\\0 & -\frac{8}{3} & \frac{2}{3}\end{matrix}\right| \]

3. Finally, eliminate the entry below the second pivot: \[ \Rightarrow \left|\begin{matrix}3 & 8 & 4\\0 & \frac{10}{3} & \frac{2}{3}\\0 & 0 & \frac{6}{5}\end{matrix}\right| \]

The determinant of an upper triangular matrix is the product of its diagonal elements, adjusted for any row swaps. Here, we have one row swap, so we calculate:

\[ Det = (-1)^1 \times a_{1,1} \times a_{2,2} \times a_{3,3} = (-1)^1 \times (3) \times \left(\frac{10}{3}\right) \times \left(\frac{6}{5}\right) \]

Calculating this gives:

\[ Det = -1 \times 3 \times \frac{10}{3} \times \frac{6}{5} = -12.00 \]

Final Answer