Answer:
(x, y, z) = (3, 1, 2)
Step-by-step explanation:
Solving using a calculator, I would enter the coefficients of 1/x, 1/y, 1/z as they are given. The augmented matrix in that case looks like ...
[tex]\left[\begin{array}{ccc|c}\frac{1}{2}&\frac{1}{4}&-\frac{1}{3}&\frac{1}{4}\\1&-\frac{1}{3}&0&0\\1&-\frac{1}{5}&4&\frac{32}{15}\end{array}\right][/tex]
My calculator shows the solution to this set of equations to be ...
- 1/x = 1/3
- 1/y = 1
- 1/z = 1/2
So, (x, y, z) = (3, 1, 2).
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Doing this by hand, I might eliminate numerical fractions. Then the augmented matrix for equations in 1/x, 1/y, and 1/z would be ...
[tex]\left[\begin{array}{ccc|c}6&3&-4&3\\3&-1&0&0\\15&-3&60&32\end{array}\right][/tex]
Adding 3 times the second row to the first, and adding the first row to the third gives ...
[tex]\left[\begin{array}{ccc|c}15&0&-4&3\\3&-1&0&0\\21&0&56&35\end{array}\right][/tex]
Then adding 14 times the first row to the third, and dividing that result by 77 yields equations that are easily solved in a couple of additional steps.
[tex]\left[\begin{array}{ccc|c}6&3&-4&3\\3&-1&0&0\\3&0&0&1\end{array}\right][/tex]
The third row tells you 3/x = 1, or x=3.
Then the second row tells you 3/3 -1/y = 0, or y=1.
Finally, the first row tells you 15/3 -4/z = 3, or z=2.
