If `A=[(1,k,3),(3,k,-2),(2,3,-4)]` is singular then K=?

Options

A. `(16)/3`
B. `(34)/3`
C. `(33)/3`
D. none of these

Explanation

To find the value of \( k \) such that the matrix \( A = \begin{pmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{pmatrix} \) is singular, we need to calculate the determinant of the matrix and set it equal to zero.

### Step-by-Step Solution:

1. **Write the Determinant:**
The determinant of matrix \( A \) can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ A = \begin{pmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{pmatrix} \]
The determinant can be expanded along the first row:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} k & -2 \\ 3 & -4 \end{vmatrix} - k \cdot \begin{vmatrix} 3 & -2 \\ 2 & -4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 3 & k \\ 2 & 3 \end{vmatrix} \]

2. **Calculate the 2x2 Determinants:**
- For the first determinant:
\[ \begin{vmatrix} k & -2 \\ 3 & -4 \end{vmatrix} = k \cdot (-4) - (-2) \cdot 3 = -4k + 6 \]
- For the second determinant:
\[ \begin{vmatrix} 3 & -2 \\ 2 & -4 \end{vmatrix} = 3 \cdot (-4) - (-2) \cdot 2 = -12 + 4 = -8 \]
- For the third determinant:
\[ \begin{vmatrix} 3 & k \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 - k \cdot 2 = 9 - 2k \]

3. **Substitute Back into the Determinant:**
Now substituting these values back into the determinant:
\[ \text{det}(A) = 1 \cdot (-4k + 6) - k \cdot (-8) + 3 \cdot (9 - 2k) \]
Simplifying this:
\[ \text{det}(A) = -4k + 6 + 8k + 27 - 6k \]
Combine like terms:
\[ \text{det}(A) = (-4k + 8k - 6k) + (6 + 27) = -2k + 33 \]

4. **Set the Determinant to Zero:**
For the matrix to be singular, we set the determinant equal to zero:
\[ -2k + 33 = 0 \]

5. **Solve for \( k \):**
Rearranging gives:
\[ -2k = -33 \implies k = \frac{33}{2} \]

### Final Answer:
Thus, the value of \( k \) such that the matrix \( A \) is singular is:
\[ k = \frac{33}{2} \]