The value of `[{:(102,18,36),(1,3,4),(17,3,6):}]` is _____________

Answer

To find the value of the determinant of the matrix

\[ A = \begin{pmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{pmatrix} \]

we will follow these steps:

### Step 1: Write the Determinant
We need to calculate the determinant of the matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \]

### Step 2: Factor Out Common Elements
Notice that the first row can be factored. We can express the elements of the first row in terms of 6:

- \( 102 = 6 \times 17 \)
- \( 18 = 6 \times 3 \)
- \( 36 = 6 \times 6 \)

Thus, we can factor out 6 from the first row:

\[ \text{det}(A) = 6 \cdot \begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \]

### Step 3: Identify Duplicate Rows
Now, observe that the first row and the third row of the new determinant are identical:

\[ \begin{pmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{pmatrix} \]

### Step 4: Apply the Property of Determinants
According to the properties of determinants, if two rows (or columns) of a determinant are identical, the value of the determinant is 0. Therefore:

\[ \begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} = 0 \]

### Step 5: Calculate the Final Value
Now substituting back into our expression for the determinant:

\[ \text{det}(A) = 6 \cdot 0 = 0 \]

### Conclusion
Thus, the value of the determinant of the matrix \( A \) is:

\[ \text{Value} = 0 \]

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