To determine the trace [tex]\( t \)[/tex] of the given matrix [tex]\( A \)[/tex], we need to look at its diagonal elements. The trace of a matrix is the sum of its diagonal elements.
Given the matrix
[tex]\[
A=\begin{pmatrix}
\frac{|x+y|}{|x|+|y|} & \alpha_1 & \beta_1 \\
\alpha_2 & \frac{|y+z|}{|y|+|z|} & \beta_2 \\
\alpha_3 & \beta_3 & \frac{|z+x|}{|z|+|x|}
\end{pmatrix}
\][/tex]
we need to compute:
[tex]\[
t = \frac{|x+y|}{|x|+|y|} + \frac{|y+z|}{|y|+|z|} + \frac{|z+x|}{|z|+|x|}
\][/tex]
We are told that the solution to this problem results in [tex]\((0, 0)\)[/tex]. Treating this given result as true, we infer that the sum of the diagonal elements, which is the trace [tex]\( t \)[/tex], must be zero.
So:
[tex]\[
t = \frac{|x+y|}{|x|+|y|} + \frac{|y+z|}{|y|+|z|} + \frac{|z+x|}{|z|+|x|} = 0
\][/tex]
Now we need to find the correct range for [tex]\( t \)[/tex] from the given options. Since we have [tex]\( t = 0 \)[/tex], we evaluate each option:
(a) [tex]\( 0 \leq t \leq 3 \)[/tex] - This is true because [tex]\( 0 \)[/tex] lies within the interval [tex]\([0, 3]\)[/tex].
(b) [tex]\( 1 \leq t \leq 2 \)[/tex] - This is false because [tex]\( 0 \)[/tex] does not lie within the interval [tex]\([1, 2]\)[/tex].
(c) [tex]\( 1 \leq 1 \leq 3 \)[/tex] - This is incorrectly formatted and does not make sense as an interval.
(d) [tex]\( -1 \leq 1 \leq 1 \)[/tex] - Assuming there’s a typographical error, it should probably be [tex]\( -1 \leq t \leq 1 \)[/tex]. This interval would also include [tex]\( 0 \)[/tex].
Given these observations, the correct answer is:
(a) [tex]\( 0 \leq t \leq 3 \)[/tex]
