Let `A=[{:(x^(2),6,8),(3,y^(2),9),(4,5,z^(2)):}],B=[{:(2x,3,5),(2,2y,6),(1,4,2z-3):}]`. If trace A = trace B, then `x+y+z` is equal to ________

Answer

To solve the problem, we need to find the values of \( x \), \( y \), and \( z \) such that the trace of matrix \( A \) is equal to the trace of matrix \( B \).

### Step-by-step Solution:

1. **Define the matrices**:
- Matrix \( A \) is given as:
\[ A = \begin{pmatrix} x^2 & 6 & 8 \\ 3 & y^2 & 9 \\ 4 & 5 & z^2 \end{pmatrix} \]
- Matrix \( B \) is given as:
\[ B = \begin{pmatrix} 2x & 3 & 5 \\ 2 & 2y & 6 \\ 1 & 4 & 2z - 3 \end{pmatrix} \]

2. **Calculate the trace of matrix \( A \)**:
- The trace of a matrix is the sum of its diagonal elements. Thus, for matrix \( A \):
\[ \text{trace}(A) = x^2 + y^2 + z^2 \]

3. **Calculate the trace of matrix \( B \)**:
- Similarly, for matrix \( B \):
\[ \text{trace}(B) = 2x + 2y + (2z - 3) = 2x + 2y + 2z - 3 \]

4. **Set the traces equal to each other**:
- Since we are given that \( \text{trace}(A) = \text{trace}(B) \):
\[ x^2 + y^2 + z^2 = 2x + 2y + 2z - 3 \]

5. **Rearrange the equation**:
- Rearranging gives:
\[ x^2 - 2x + y^2 - 2y + z^2 - 2z + 3 = 0 \]

6. **Complete the square for each variable**:
- Completing the square for \( x \), \( y \), and \( z \):
\[ (x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 0 \]

7. **Analyze the equation**:
- The sum of squares equals zero only when each square is zero:
\[ (x - 1)^2 = 0 \implies x = 1 \]
\[ (y - 1)^2 = 0 \implies y = 1 \]
\[ (z - 1)^2 = 0 \implies z = 1 \]

8. **Calculate \( x + y + z \)**:
- Therefore:
\[ x + y + z = 1 + 1 + 1 = 3 \]

### Final Answer:
Thus, \( x + y + z \) is equal to **3**.