Suppose `a_(1),a_(2),a_(3),a_(4)gt0`. If periods of `sin(a_(1)pix+b_(1)),cos(a_(2)pix+b_(2)), tan (a_(3)pix+b_(3))` and `cot(a_(4)pix+b_(4))` are respectively `1/2,1/3,1/2` and `1/3` rthen `|(a_(1),a_(2)),(a_(3),a_(4))|^(2)` =_________

Answer

To solve the problem step by step, we need to find the values of \( a_1, a_2, a_3, \) and \( a_4 \) based on the periods of the trigonometric functions given. Let's break it down:

### Step 1: Determine \( a_1 \)
The period of \( \sin(a_1 \pi x + b_1) \) is given as \( \frac{1}{2} \). The period of \( \sin(kx) \) is \( \frac{2\pi}{k} \).

Setting up the equation:
\[ \frac{2\pi}{a_1 \pi} = \frac{1}{2} \]
Simplifying this gives:
\[ \frac{2}{a_1} = \frac{1}{2} \]
Cross-multiplying:
\[ 2 \cdot 2 = a_1 \implies a_1 = 4 \]

### Step 2: Determine \( a_2 \)
The period of \( \cos(a_2 \pi x + b_2) \) is given as \( \frac{1}{3} \). The period of \( \cos(kx) \) is also \( \frac{2\pi}{k} \).

Setting up the equation:
\[ \frac{2\pi}{a_2 \pi} = \frac{1}{3} \]
Simplifying this gives:
\[ \frac{2}{a_2} = \frac{1}{3} \]
Cross-multiplying:
\[ 2 \cdot 3 = a_2 \implies a_2 = 6 \]

### Step 3: Determine \( a_3 \)
The period of \( \tan(a_3 \pi x + b_3) \) is given as \( \frac{1}{2} \). The period of \( \tan(kx) \) is \( \frac{\pi}{k} \).

Setting up the equation:
\[ \frac{\pi}{a_3 \pi} = \frac{1}{2} \]
Simplifying this gives:
\[ \frac{1}{a_3} = \frac{1}{2} \]
Cross-multiplying:
\[ 1 \cdot 2 = a_3 \implies a_3 = 2 \]

### Step 4: Determine \( a_4 \)
The period of \( \cot(a_4 \pi x + b_4) \) is given as \( \frac{1}{3} \). The period of \( \cot(kx) \) is \( \frac{\pi}{k} \).

Setting up the equation:
\[ \frac{\pi}{a_4 \pi} = \frac{1}{3} \]
Simplifying this gives:
\[ \frac{1}{a_4} = \frac{1}{3} \]
Cross-multiplying:
\[ 1 \cdot 3 = a_4 \implies a_4 = 3 \]

### Step 5: Calculate the Determinant
Now we have:
- \( a_1 = 4 \)
- \( a_2 = 6 \)
- \( a_3 = 2 \)
- \( a_4 = 3 \)

We need to find the determinant:
\[ \left| \begin{array}{cc} a_1 & a_2 \\ a_3 & a_4 \end{array} \right| = \left| \begin{array}{cc} 4 & 6 \\ 2 & 3 \end{array} \right| \]
Calculating the determinant:
\[ = (4 \cdot 3) - (6 \cdot 2) = 12 - 12 = 0 \]

### Step 6: Square the Determinant
Finally, we need to square the determinant:
\[ |(a_1, a_2), (a_3, a_4)|^2 = 0^2 = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]