\begin{tabular}{|c|c|c|}
\hline
[tex]$\frac{1}{3}$[/tex] & [tex]$\overline{4}$[/tex] & [tex]$\frac{1}{1}$[/tex] \\
\hline
- & [tex]$\frac{5}{6}$[/tex] & - \\
\hline
- & - & [tex]$\frac{4}{3}$[/tex] \\
\hline
\end{tabular}



Answer :

Sure, let's go through the values given in the matrix step-by-step:

1. Consider the cell at position (0, 0), which is in the first row and first column. The value there is [tex]\(\frac{1}{3}\)[/tex]. In decimal form, [tex]\(\frac{1}{3}\)[/tex] is approximately [tex]\(0.3333333333333333\)[/tex].

2. Move to the cell at position (0, 1), which is in the first row and second column. The value provided there is [tex]\(4\)[/tex]. This remains [tex]\(4\)[/tex] as it is already in its simplest form.

3. Next, check the cell at position (0, 2), which is in the first row and third column. The value given here is [tex]\(\frac{1}{1}\)[/tex], which simplifies to [tex]\(1\)[/tex].

4. Now, let's move to the cell at position (1, 1), which is in the second row and second column (as some values were unspecified and treated as '-'). The value here is [tex]\(\frac{5}{6}\)[/tex]. In decimal form, [tex]\(\frac{5}{6}\)[/tex] is approximately [tex]\(0.8333333333333334\)[/tex].

5. Finally, examine the cell at position (2, 2), which is in the third row and third column. The value provided here is [tex]\(\frac{4}{3}\)[/tex]. In decimal form, [tex]\(\frac{4}{3}\)[/tex] is approximately [tex]\(1.3333333333333333\)[/tex].

So, the detailed solution to the matrix yields the values:
[tex]\[ (0.3333333333333333, 4, 1, 0.8333333333333334, 1.3333333333333333) \][/tex]