Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer continue on the back of the page. No notes, books, or other aids may be used on the exam.
Student Id:
Answer Key
1. (10 points)
2. (10 points)
3. (10 points)
4. (10 points)
5. (10 points)
6. (10 points)
7. (10 points)
8. (10 points)
9. (10 points)
10. (10 points)
Total (100 points)
1. (10 points) Determine whether each of the following statements is true or false. No justification is required. (a) A linear system with fewer unknowns than equations must have infinitely many solutions or none.
Solution: False; the systemx = 2, y= 3, x+y = 5 has a unique solution.
(b) IfAandBare matrices of the same size, the formula rank(A+B) = rank(A) + rank(B) must hold. Solution: False; letA=B=I2 for example.
(c) The functionT x y = (y+ 1)2 −(y−1)2 (x−3)2−(x+ 3)2
represents a linear transformation.
Solution: True; simplify to see thatT
x y = 4y −12x = 0 4 −12 0 x y . (d) The matrix k −2 5 k−6
is invertible for all real numbersk. Solution: True; det(A) =k2
−6k+ 10, which has no real solution.
(e) The vectors
1 0 0 , 2 1 0 , 3 2 1
form a basis forR3.
2. (10 points) Find all solutions to the set of equations x1 + x2 = b1 3x1 + 3x2 = b2 .
Solution: By inspection, we see that the system is consistent only ifb2= 3b1.
3. (10 points) Find all vectors inR4 that are perpendicular to the three vectors
1 1 1 1 , 1 2 3 4 , 1 9 9 7 .
Solution: We need to find a vector~x such that~v1·~x=v~2·~x=~v3·~x= 0. We can represent this as the system A~x=~0, where the rows ofAare the three vectors given. We can easily compute:
rref 1 1 1 1 0 1 2 3 4 0 1 9 9 7 0 = 1 0 0 0.25 0 0 1 0 −1.50 0 0 0 1 2.25 0
Lettingx4=t, we see that any vector of the form~x =t
−0.25 1.50 −2.25 1.00 is a solution.
4. (10 points) LetB−1= 1 2 3 5 and (AB)−1= 1 3 2 5 . FindA. Solution: A= (AB)B−1= ((AB)−1)−1B−1= 1 3 2 5 −1 1 2 3 5 = −5 3 2 −1 1 2 3 5 = 4 5 −1 −1
5. (10 points) Describe each of the following linear transformations as a well-known geometric transformation combined with a scaling. Give the scaling factor in each case.
(a)
1 1 1 1
Solution: This matrix has the form of a projection: k
u2 1 u1u2 u1u2 u22 = 1 1 1 1 , or equivalently u2 1 u1u2 u1u2 u22 = 1/k 1/k 1/k 1/k , where u2 1+u 2 2= 1 k + 1 k = 1. Thus,k = 2 andu~= u1 u2 = " 1 √ 2 1 √ 2 # . (b) 3 0 −1 3
Solution: This matrix has the form of a vertical shear: k
1 0 c 1 = 3 0 −1 3 , or equivalently 1 0 c 1 = 3/k 0 −1/k 3/k . Thus,k = 3 and c=−13. (c) 3 4 4 −3
Solution: This matrix has the form of a reflection: k
a b b −a = 3 4 4 −3 , or equivalently a b b −a = 3/k 4/k 4/k −3/k , wherea2 +b2 = 3k 2 + 4k 2 = 1. Thus,k = 5.
6. (10 points) The color of light can be represented as a vector R G B ,
whereR= amount of red,G = amount of green, andB= amount of blue. The human eye and the brain transform the incoming signal into the signal
I L S , where intensity I = R+G+B 3 long-wave signal L = R−G short-wave signal S = B−R+G 2 . (a) Find the matrixP representing the transformation from
R G B to I L S . Solution: P = 1 3 1 3 1 3 1 −1 0 −21 − 1 2 1
(b) Consider a pair of yellow sunglasses for water sports that cuts out all blue light and passes all red and green light. Find the 3×3 matrixA that represents the transformation incoming light undergoes as it passes through the sunglasses.
Solution: A= 1 0 0 0 1 0 0 0 0
(c) Find the matrix for the composite transformation that light undergoes as it first passes through the sunglasses and then the eye.
Solution: P A= 1 3 1 3 0 1 −1 0 −1 −1 0
7. (10 points) Find vectors that span the kernel ofA= 1 2 3 4 0 1 2 3 0 0 0 1 .
Solution: Reducing the matrix to reduced row-echelon form:
1 2 3 4 0 1 2 3 0 0 0 1 → 1 0 −1 −2 0 1 2 3 0 0 0 1 → 1 0 −1 0 0 1 2 0 0 0 0 1
Lettingx3=t, we see that any vector of the formt
1 −2 1 0
is a solution. Thus, ker(A) = span
1 −2 1 0 .
8. (10 points) Express the kernel ofA=
1 2 0 0 3 0 0 0 1 0 2 0 0 0 0 1 1 0 0 0 0 0 0 0
as the image of another matrixB.
Solution: This matrix is already in reduced row-echelon form and has three free variables. Lettingx2=r,
x5=s andx6=t, we see that solutions are of the form:
−2r−3s r −2s −s s t =r −2 1 0 0 0 0 +s −3 0 −2 −1 1 0 +t 0 0 0 0 0 1 = span −2 1 0 0 0 0 , −3 0 −2 −1 1 0 , 0 0 0 0 0 1
Since the image of a matrix is simply the span of its columns, we have:
B= −2 −3 0 1 0 0 0 −2 0 0 −1 0 0 1 0 0 0 1
9. (10 points) Find a basis of the image ofA= 1 1 1 1 2 5 1 3 7 .
Solution: The first two columns are linearly independent, since the second is not a scalar multiple of the first. The third column is not a linear combination of the first two, which can be seen from:
rref 1 1 1 1 2 5 1 3 7 = 1 0 0 0 1 0 0 0 1
Therefore, a basis for the image is
1 1 1 , 1 2 3 , 1 5 7 .
10. (10 points) Verify that the image of ann×mmatrixAis a subspace ofRn .
Solution: We need to prove that im(A) meets the three necessary conditions: 1. A~0 =~0 for anyA. Therefore,~0 is in im(A).
2. Let~x, ~ybe in im(A). Then, by definition of the image, there exist vectors~v andw~ such thatA~v=~x
andA ~w =~y. We see thatx~+y~=A~v+A ~w =A(~v+w~). Therefore,~x+~y is in im(A).
3. Let~x be in im(A). Then, by definition of the image, there exists a vector~v such thatA~v =~x. We see that k~x =kA~v =A(k~v). Therefore,k~x is in im(A).
