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Math 415 Exam # 1 Solutions
- (15 points) Let LD 2 × 2 denote the set of lower triangular 2 × 2 matrices, i.e. the set of matrices of the form
A =
x 1 0 x 2 x 3
(a) Show that LD 2 × 2 is a subspace of M 2 × 2 Solution: (5 points)
aA + bB = a
x 1 0 x 2 x 3
y 1 0 y 2 y 3
ax 1 + by 1 0 ax 2 + by 2 ax 3 + by 3
is lower triangular for any lower triangular matrices A and B and any scalars a and b so LD 2 × 2 is a subspace of M 2 × 2 (b) Write down an equation that demonstrates that LD 2 × 2 is spanned by the vectors v 1 =
, v 2 =
, v 3 =
Solution: (4 points) ( x 1 0 x 2 x 3
= x 1
(c) Are v 1 , v 2 , and v 3 linearly independent? Justify your response. Solution: (4 points) By the last equation, the only way that we can have c 1 v 1 + c 2 v 2 + c 3 v 3 = 0 is if ( c 1 0 c 2 c 3
and so we see immediately that c 1 = c 2 = c 3 = 0. (d) What is the dimension of LD 2 × 2 and why? Solution: (2 points) By parts b and c the vis form a basis of LD 2 × 2 and so the dimension is 3.
- (15 points) Let SLD 3 × 3 denote the set of special lower triangular 3×3 matrices, i.e. the set of matrices of the form
A =
x 1 1 0 x 2 x 3 1
(a) Is SLD 3 × 3 a subspace of M 3 × 3? Explain. Solution: (2 points) No. It is not closed under scalar multiplication since a multiple of A will not have 1’s along the diagonal.
(b) What is the permutted LU factorization of the elements of SLD 3 × 3 , i.e. of the matrix A? Solution: (3 points) A is already in factored form if we choose P = I, L = A and U = I. (c) What is the determinant of the elements of SLD 3 × 3 , i.e. of the matrix A? Solution: (2 points) det A is the product of the diagonal elements of U = I and hence is 1. (d) What is the inverse of
E 12 (2) =
Solution: (3 points)
E 12 (2)−^1 = E 12 (−2) =
(e) Compute the inverse of
B =
Solution: (5 points) Since B = E 12 (2)E 23 (3), we conclude that B−^1 = E 23 (3)−^1 E 12 (2)−^1 = E 23 (−3)E 12 (−2), that is
B−^1 =
- (10 points) What are the coordinates of the vector p(x) = x^2 + 1 in P (2)^ relative to the vectors
p 1 (x) = x^2 + x + 1 p 2 (x) = 2x^2 + 3x − 2 p 3 (x) = x^2 + 2x − 3
Explain and justify your answer carefully. Solution: We must find constants c 1 , c 2 , c 3 such that
x^2 + 1 = c 1 (x^2 + x + 1) + c 2 (2x^2 + 3x − 2) + c 3 (x^2 + 2x − 3) = (c 1 + 2c 2 + c 3 )x^2 + (c 1 + 3c 2 + 2c 3 )x + (c 1 − 2 c 2 − 3 c 3 )
(a) Find a basis for rngA for the matrix
A =
Solution: (9 points) By Gaussian elimination
A =
Since the pivots appear in columns 1, 2, and 4, a basis for the range of A consists of columns 1, 2, and 4 of A, namely
(^) , and
(b) What is rankA? Solution: (1 point) 3, the number of pivots.
