Let t be the trace of matrix A`{:[((|x+y|)/(|x| +|y|),alpha_1,beta_1),(alpha_2,(|y+z|)/(|y| +|z|),beta_2), (alpha_3,beta_3,((|z+x|)/(|z| +|x|]) ] ` then

Options

A. A) `0letlt3`
B. B) `1letle2`
C. C) `1letle3`
D. D) `-1letle1`

Answer

To find the trace \( T \) of the given matrix \( A \), we will follow these steps:

### Step 1: Write down the matrix
The matrix \( A \) is given as:
\[ A = \begin{bmatrix} \frac{|x+y|}{|x| + |y|} & \alpha_1 & \beta_1 \\ \alpha_2 & \frac{|y+z|}{|y| + |z|} & \beta_2 \\ \alpha_3 & \beta_3 & \frac{|z+x|}{|z| + |x|} \end{bmatrix} \]

### Step 2: Identify the elements of the trace
The trace \( T \) of a matrix is the sum of the diagonal elements. Therefore, we need to sum the diagonal elements of matrix \( A \):
\[ T = \frac{|x+y|}{|x| + |y|} + \frac{|y+z|}{|y| + |z|} + \frac{|z+x|}{|z| + |x|} \]

### Step 3: Analyze the terms
Using the triangle inequality, we know that:
\[ |x+y| \leq |x| + |y|, \quad |y+z| \leq |y| + |z|, \quad |z+x| \leq |z| + |x| \]
This implies that:
\[ \frac{|x+y|}{|x| + |y|} \leq 1, \quad \frac{|y+z|}{|y| + |z|} \leq 1, \quad \frac{|z+x|}{|z| + |x|} \leq 1 \]

### Step 4: Establish bounds for the trace
From the above inequalities, we can conclude that:
\[ T \leq 1 + 1 + 1 = 3 \]

### Step 5: Consider the lower bound
For the lower bound, we can also note that:
\[ \frac{|x+y|}{|x| + |y|} \geq 0, \quad \frac{|y+z|}{|y| + |z|} \geq 0, \quad \frac{|z+x|}{|z| + |x|} \geq 0 \]
Thus, we have:
\[ T \geq 0 \]

### Step 6: Combine the results
From the analysis, we find that:
\[ 0 \leq T \leq 3 \]

### Step 7: Final conclusion
The trace \( T \) can take values in the range:
\[ 1 \leq T \leq 3 \]
This means that the trace \( T \) is bounded between 1 and 3.

### Final Answer
Thus, we conclude that:
\[ 1 \leq T \leq 3 \]

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