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Math 415 Old Exam # 1 and Solutions
- (12 points)
(a) Find an LU factorization of the matrix A =
Answer: (5 points) A =
= U
L =
= L
(b) If a matrix B has the LU factorization B =
solve Bx = (^ − 3 10 − 7 )T^ without reducing the augmented matrix. Answer: (6 points) b =
Write Bx = b as LU x = b and this as U x = c where Lc = b. Solve first for c:
c 1 c 2 c 3
c 1 = − 3 , c 2 = 10 + 4c 1 = − 2 , c 3 = − 7 − 3 c 2 = −1. Now solve for x
x 1 x 2 x 3
x 3 = 1, x 2 = −2 + 2x 3 = 0, x 1 = 12 (− 3 − 2 x 2 + 5x 3 ) = 12 (− 3 − 0 + 5) = 1. So x =
(c) What is the value of det B? Answer: (1 point) -2, namely the product of the diagonal members of U.
- (12 points)
(a) Give the definition of the inverse of a matrix A.
Answer: (3 points) The inverse of A is a matrix X for which XA = AX = I (b) Write down a 3 × 3 permutation matrix P. Also write down its inverse. Answer: (2 points) Any matrix which permutes two rows of the identity will do, so try P =
(^). Applying P again changes the rows back, so P −^1 = P
(c) Write down a 3 × 3 matrix E that performs an elementary row operation through matrix multiplication. Also write down its inverse. Answer: (3 points) There are many choices. Here is one E =
(^). Its inverse is E−^1 =
(d) Compute explicitly the inverse of the product P E. Answer: (3 points) (P E)−^1 = E−^1 P −^1 =
(e) What does the value of the determinant of a 3 × 5 matrix tell you about the existence of an inverse of that matrix? Answer: (1 points) Nothing since the determinant and the inverse are only defined for square matrices
- (8 points)
(a) Give a definition of the rank of a matrix A. Answer: (2 points) The rank of A is the number of pivots in the echelon form U of A. (b) Find the rank of the matrix
A =
Answer: (5 points)
so the rank of A is 2 (c) What is the rank of AT^? Answer: (1 points) rank AT^ = rank A =2 (^2)
a
(^) + b
a + b a − 2 b −a + b
Now verify that this vector satisfies the equation for the plane for all values of a and b: (a + b) + 2(a − 2 b) + 3(−a + b) = a + b + 2a − 4 b − 3 a + 3b = 0. Yes! (c) What is the dimension of the subspace W =
x y z )T^ for which x + 2y + 3z = 0
given the facts outlined in part b)? Be specific. Answer: (4 points) If the two vectors are linearly independent, then they form a basis for the plane and so the dimension of the plane is 2. To test linear independence either note that neither vactor is a multiple of the other, or reduce the matrix with these as columns and show there are two pivots:
(^) Done!
- (10 points) The three vectors
v 1 =
(^) , v 2 =
(^) and v 3 =
form a basis for R^3 (you do NOT need to verify this!) (a) Express the vector u = (^5 − 8 − 1 )T^ as a linear combination of these three vectors. Answer: (8 points) We need to find numbers x, y, and z such that u =
(^) = x
(^) + y
(^) + z
x y z
By Gaussian elimination we have
−^143
so z = − 1 , y = −^13 (− 13 − z) = 4, x = 5 − y − z = 2 (b) What are the “coordinates” of u relative to this basis? Answer: (2 points) ( 2 4 − 1 )T
