Is the matrix `[{:(pi,22),((1)/(7),1):}]` singular?

Answer

To determine if the matrix \(\begin{bmatrix} \pi & 22 \\ \frac{1}{7} & 1 \end{bmatrix}\) is singular, we need to find its determinant. A matrix is singular if its determinant is equal to zero.

### Step-by-Step Solution:

1. **Substitute the value of \(\pi\)**:
The value of \(\pi\) is approximately \(3.14\), but for the purpose of this problem, we can also use the fraction \(\frac{22}{7}\) as an approximation. Thus, we can rewrite the matrix as:
\[ \begin{bmatrix} \frac{22}{7} & 22 \\ \frac{1}{7} & 1 \end{bmatrix} \]

2. **Calculate the determinant**:
The determinant of a \(2 \times 2\) matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is calculated using the formula:
\[ \text{det} = ad - bc \]
For our matrix:
- \(a = \frac{22}{7}\)
- \(b = 22\)
- \(c = \frac{1}{7}\)
- \(d = 1\)

Plugging in these values, we get:
\[ \text{det} = \left(\frac{22}{7} \cdot 1\right) - \left(22 \cdot \frac{1}{7}\right) \]

3. **Simplify the determinant**:
\[ \text{det} = \frac{22}{7} - \frac{22}{7} = 0 \]

4. **Conclusion**:
Since the determinant is \(0\), the matrix is singular.

### Final Answer:
Yes, the matrix \(\begin{bmatrix} \pi & 22 \\ \frac{1}{7} & 1 \end{bmatrix}\) is singular.