1)
For the example problem given, use the remaining two eigenvalues, l2 and l3,
to diagonaliize the matrix and obtain solutions in

**c**,_{x}**c**and_{y}**c**_{z}.
Solution:

Using the eigenvalue

**l2 = 5**we arrive at:
For
which:

c

_{y}= c_{z}/**Ö**2 and c = 0**e**c_{1}^ -_{z}/**Ö**2**e**c_{2}^ +_{z }**e**_{3}^
Whence:

(c

_{z }^{2 }/ 2 + c_{z }^{2}) = 1 Þ c_{z}=**Ö**(2/ 3)
Further:

**e**= 1/

_{2}^’**Ö**2 (

**Ö**(2/ 3))

**e**

_{2}^” +**Ö**(2/ 3)

**e**=

_{3}^”1/ /

**Ö**3

**e**

_{2}^” +**Ö**(2/ 3)

**e**

_{3}^”
For the eigenvalue

**l3 = 14**:
We
arrive at:

Yielding: c

_{y}= -**Ö**2 c_{z}and c_{z }= 1/**Ö**3
Þ c
=

**Ö**2 c

_{z }

**e**+

_{2}’’’**c**

_{z }

**e**= -

_{3}^”’**Ö**(2/ 3)

**e**

_{3}^” +**Ö**(1/ 3)

**e**

_{3}^”
2)
Thereby obtain the principal axes in terms of:

**e^’**,

_{x}**e ^’**and

_{y}**e^’**.

_{z}**Solution**:

Note
conversions between primed and unprimed systems:

e

_{ j}= å^{3}_{i= 1}a_{i j}e_{ j }’ and: e_{ i}’ = å^{3}_{j= 1}a_{i j}e_{ j }^{}
Then:

**e**=

_{x}^”**- 1/**

**Ö**2 e

_{1}- 1/

**Ö**2 e

_{2}

So:

**e**=

_{x}^’**e**=

_{x}^”**1/**

**Ö**2 e

_{1}- 1/

**Ö**2 e

_{2}+ 0

Similarly:

**e**=

_{y}^”**1/**

**Ö**2 e^

_{1}+ 1/

**Ö**2 e^

_{2}+ 0

And:

**e**=_{z}^”**e^**_{3 }Þ**e**=_{y}^’

1/

**Ö**3 (1/**Ö**2)**e^**+_{1}**1/****Ö**3 (1/**Ö**2)**e^**_{2 }+**Ö**(2/ 3)**e**_{3}^
=

**e^**/_{1}**Ö**6 +**e^**/_{2}**Ö**6 + 2**e^**/_{3}**Ö**6
And:

**e**=_{z}^’
- 1/

**Ö**3**(e^**) - 1/_{1}**Ö**3 (**e^**) + 1/_{2}**Ö**3 (**e^**)_{3}
The Principal axes are:

**e**and_{x}^’ , e_{y}^’**e**_{z}^’
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