
Problem 1 (1): The cross product is [2,1,2]^{T,} and the unit vectors are [2/3,1/3,2/3]^{T,} and [2/3,1/3,2/3]^{T,}.
One can partially verify his results by calculating the inner product with the given vectors, although it does not totally guarantee the correctness of the cross product.

Problem 1 (2): The minimal polynomial is t^{3} as E, A, A^{2} are linearly independent and A^{3}=O. The powers of A are given by
A^{0} =

 ,
A^{1} =

 ,
A^{2} =

 ,
A^{n} =

 ,
(n ≥ 3).


Problem 2 (1): The angle is 3π/4, i.e., 135°.
Some students errorniously considered the absolute value of what is to be equal to cosine of the angle.

Problem 2 (2): The orthogonal vectors obtained by Gram–Schmidt process are
[1,2,1]^{T},
[1,1,1]^{T},
[1,0,1]^{T}.
One can partially verify his results by calculating the inner products of the resulting vectors with each other, although it does not totally guarantee the correctness of the results of Gram–Schmidt process.

Problem 3 (1): The inverse is given by
One can verify his result by calculating the product with the given matrix.

Problem 3 (2): rank A=3 and dim Ker A^{T}=1.

Problem 4 (1): The row reduced form is given by

Problem 4 (2): A basis of Im A is given by [1,2,3,2]^{T}, [0,1,1,2]^{T}.
As the first and the second columns of the row reduced form give a basis of the image of the row reduced form, the first and the second columns of A gives a basis of the image of A.
Some students errorniously answered the first and the second columns of the row reduced form instead of those of A.

Problem 4 (3): A basis of Ker A is given by [1,1,1,0]^{T}, [1,1,0,1]^{T}.
Some students used properties of the dimension instead of the definition of linear dependence or spanning. Such a solution does not fulfill the requiremen of the problem.

Problem 5: No one received a positive point.