If `2a x-2y+3z=0,x+a y+2z=0,a n d2+a z=0` have a nontrivial solution, find the value of `adot`

Explanation

To find the value of \( a \) for which the given system of equations has a nontrivial solution, we need to set up the equations and find the determinant of the coefficient matrix. The equations given are:

1. \( 2a x - 2y + 3z = 0 \)
2. \( x + a y + 2z = 0 \)
3. \( 2 + a z = 0 \)

### Step 1: Write the coefficient matrix
The coefficient matrix \( A \) can be formed from the coefficients of \( x, y, z \) in the equations:

\[ A = \begin{bmatrix} 2a & -2 & 3 \\ 1 & a & 2 \\ 0 & 0 & a \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix
To find the value of \( a \) for which the system has a nontrivial solution, we need to set the determinant of \( A \) equal to zero:

\[ \text{det}(A) = 2a \cdot \begin{vmatrix} a & 2 \\ 0 & a \end{vmatrix} - (-2) \cdot \begin{vmatrix} 1 & 2 \\ 0 & a \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & a \\ 0 & 0 \end{vmatrix} \]

Calculating the smaller determinants:

1. \( \begin{vmatrix} a & 2 \\ 0 & a \end{vmatrix} = a \cdot a - 0 \cdot 2 = a^2 \)
2. \( \begin{vmatrix} 1 & 2 \\ 0 & a \end{vmatrix} = 1 \cdot a - 0 \cdot 2 = a \)
3. \( \begin{vmatrix} 1 & a \\ 0 & 0 \end{vmatrix} = 1 \cdot 0 - 0 \cdot a = 0 \)

Now substituting these back into the determinant:

\[ \text{det}(A) = 2a \cdot a^2 + 2a + 0 = 2a^3 + 2a \]

### Step 3: Set the determinant equal to zero
For a nontrivial solution, we set the determinant to zero:

\[ 2a^3 + 2a = 0 \]

Factoring out \( 2a \):

\[ 2a(a^2 + 1) = 0 \]

### Step 4: Solve for \( a \)
This gives us two cases:

1. \( 2a = 0 \) which implies \( a = 0 \)
2. \( a^2 + 1 = 0 \) which has no real solutions since \( a^2 = -1 \) is not possible for real \( a \).

Thus, the only real solution is:

\[ a = 0 \]

### Conclusion
The value of \( a \) for which the system of equations has a nontrivial solution is \( a = 0 \).

---