M597K: Solution to Homework 3
Date: Sept. 20, Friday
Solutions 1-2: Omitted.
3. (Summation convention) Expand the terms Ai B k Ci and aij bj . Is there a summation in ai + bi ?
Solution: (20 points) Ai B k Ci = B k (A1 C1 + A2 C2 + A3 C3 ).
aij bj = ai1 b1 + ai2 b2 + ai3 b3 .
There is no summation in ai + bi since the repeated index i is not in a product.
4. The new coordinate system K 0 is obtained by rotating the ii coordinate system
an angle θ about the x3 axis counterclockwise. Find the coefficients (i.e., αi0 j and
αj 0 i ) in the equations:
x0i = αi0 j xj ,
xi = αj 0 i x0j .
Solution: (20 points) Once you draw the three-dimensional graphs of the coordinate unit vectors, you can find
i01 = cos θi1 + sin θi2 ,
i02 = − sin θi1 + cos θi2 ,
i03 = i3 .
Then by definition of the αi0 j and αj 0 i , one can find
x01 = cos θx1 + sin θx2 ;
x02 = − sin θx1 + cos θx2 ;
x03 = x3 .
And by rotating backward, one finds
x1 = cos θx01 − sin θx02 ;
x2 = sin θx01 + cos θx02 ;
x3 = x03 .
5. The unit base vectors i0i of a new coordinate system K 0 are given by
i2
i1
i1
2i3
i2
i3
i2
i3
i01 = √ + √ , i02 = √ − √ + √ , i03 = √ + √ − √ .
3
6
2
3
6
2
3
6
Find the coefficients in the equation: x0i = αi0 j xj .
Solution: (20 points) One can use the formula αi0 j = i0i · ij or use the similarity
between the expressions of i0i and x0i to draw conclusion

√1
3
− √13
√1
3
0



√1
2
√1
2
(αi0 j ) = 
√2
6
√1
6
− √16



.

6. Given the transformation of coordinates
x0i = αi0 j xj

where



(αi0 j ) = 
√
2
√2
− 33
√
6
6
0
√
3
√3
6
3
√
2
√2
3
√3
− 66



.

If a vector v has the components (2, −3, 6) with respect to the xi -coordinate system,
find its components in the x0i -system.
Solution: (20 points) One can use the formula x0i = αi0 j xj or the matrix notation
√
√ √
(αi0 j )vT to find x0 = (4 2, 33 , − 5 3 6 ).
7. Given the second-order tensor





0 −1 3 

1
0 2 .
(aij ) = 

−3 −2 0
Find the components a021 and a033 of this tensor in the coordinate system x0i defined
by x0i = αi0 j xj where


0 1 0 



(αi0 j ) =  −1 0 0  .


0 0 1
If you can use computer, then find all the components a0ij . (The formula in matrix
notation is (A0 ) = (α)(A)(α)T .)
Solution: (20 points) One can use a0ij = αi0 k αj 0 m akm or the matrix way to find




(a0ij ) = 

0 −1
1
−2
So a021 = 1 and a033 = 0.
2
2 

0 −3  .

3
0