Test 2. Fall 1996.



1 (20 points). Prove that if A and B are square matrices of the same size, B is not 0, and AB=0, then det(A)=0.

Solution

By contradiction, suppose that det(A) is not 0. Then A is invertible. Therefore AB=0 implies B=(A-1*A)*B=0, a contradiction (we assumed that B is not 0). This completes the proof.

2 (15 points). Let L be the set of all pairs (x,y) where x, y are real numbers. Define two operations + and scalar multiplication, on L:

(x,y)+(x',y')=(x+y', x'+y)
k(x,y)=(kx,ky)

Check if L is a vector space.

Solution

It is not a vector space. Let (x,y)=(1,2), (x',y')=(3,5). Then (x,y)+(x',y')=(6,5) and (3,5)+(1,2)=(5,6). So the operation + is not commutative (the addition in vector spaces must be commutative).

3 (15 points). Find the determinant of the following n by n matrix: 

                        2 3 0 0 ..... 0
                        0 2 3 0 ..... 0
                        ...............
                        3 0 0 0 ......2

Solution.

Expand along the first column. We get 2 non-zero co-factors. So the determinant is equal to 2 times (-1)2 times the determinant of the matrix obtained by removing the first row and the first column plus 3 times (-1)(1+n) times the determinant of the matrix obtained by removing the first column and the last row. The first of these two matrices is upper triangular, has size n-1 and 2 on the diagonal. So it's determinant is 2(n-1). The second matrix is lower triangular, size n-1, and with 3 on the diagonal. So its determinant is 3(n-1). Therefore the determinant of the original matrix is 2*2(n-1)+(-1)(1+n)*3*3(n-1)=2n+(-1)(1+n)3n. So it is either 2n+3n (if n is even) or 2n-3n (if n is odd).

4 (20 points). Find x from the following system of equations using Cramer's method: 

                        x+y+z=5
                       2x+3y-z=1
                        x-y+z=3
(you should use Cramer's method, any other method is not allowed.)

Solution

By Cramer's rule x=det(A1)/det(A) where A is the matrix of coefficients and A1 is the matrix obtained from A by replacing the first column with the column of right sides (5,1,3).

The answer: x=2/3.

5 (20 points). Find the matrix of the linear operator in R2 which first rotates through 45 degrees counterclockwise and then reflects about the x-axis.

Solution


The matrix of rotation through 45 degrees counterclockwise is

[cos 45 -sin 45] sqrt(2) [1 -1] A=[ ] = ------- [ ] [sin 45 cos 45] 2 [1 1]

The matrix of reflection about the x-axis is

[1 0] B=[ ] [0 -1]

So the matrix of our transformation is

sqrt(2) [ 1 -1] BA= ------ [ ] 2 [-1 -1]

6 (10 points). Find the standard matrix of the following linear transformation:

(x,y,z) ---> (5x+y-z, x-z, y+z)


Solutions

The matrix consists of coefficients in front of x, y, z in the formula of the linear transformation: 

                    [5 1 -1]
                    [1 0 -1]
                    [0 1  1]