Math 240 Homework Assignment Four Do the following problems showing all your work. Be sure to submit your solutions in PDF format to the Assignment area of LEO. (Point values are given in the square brackets.) 1. [6] Show (that is, verify all the axioms) that the set of second degree polynomials defined as, ! = = ! + + , , } is a vector space. What is the dimension of this vector space? 2. [4] Consider the set S of 23 matrices defined by, = 1 0 0 0 1 0 0 0 0 0 0 0 , , , 0 0 0 0 0 0 0 1 0 0 0 1 Find the span of S. 3. [3] Show that ! is a subspace of ! (See Problem #1 for the definition of the polynomial vector spaces.) 4. [3] Do the polynomials ! + 2 + 1, ! + 2, ! + 2, and ! + ! 5 + 2 span ! ? (Give support for your conclusion.) 5. [4] Let S be the set of vectors in ! ( = , , , , ) where, = = = = = , , , , , , , , , , , , , , , Find a subset of S that is a basis for the span(S). \f\f\fAnswer to the question no-1 Answer to the question no-2 Answer to the question no-3 Answer to the question no-4 Answer to the question no-5 1st we will set up all in matrix form, try to put in write v1 , v2 as column in stead of row and reduce it we will get . X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 2 0 1 3 3 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Find the pivot in the 1st column in the 1st row X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 2 0 1 3 3 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Multiply the 1st row by 2 X1 X2 X3 X4 b 1 2 -6 4 -6 18 2 2 0 1 3 3 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Subtract the 1st row from the 2nd X1 X2 X3 X4 b 1 2 -6 4 -6 18 2 0 6 -3 9 -15 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Multiply the 1st row by 1 X1 X2 X3 X4 b 1 -2 6 -4 6 -18 2 0 6 -3 9 -15 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Subtract the 1st row from the 3rd row and restore it X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 6 -3 9 -15 3 0 -10 5 -15 25 4 1 3 -1 6 -6 Subtract the 1st row from the 4th X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 6 -3 9 -15 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Make the pivot in the 2nd column by dividing the 2nd row by 6 X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 1 -1/2 3/2 -5/2 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Multiply the 2nd row by 3 X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 -3 3/2 -9/2 15/2 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Subtract the 2nd row from the 1st row and restore it X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 1 -1/2 3/2 -5/2 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Multiply the 2nd row by 10 X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 -10 5 -15 25 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Subtract the 2nd row from the 3rd row and restore it X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 1 -1/2 3/2 -5/2 3 0 0 0 0 0 4 0 6 -3 9 -15 Multiply the 2nd row by 6 1 X1 1 X2 0 X3 1/2 X4 3/2 b 3/2 2 0 6 -3 9 -15 3 0 0 0 0 0 4 0 6 -3 9 -15 Subtract the 2nd row from the 4th row and restore it, since we see that last two rows are zers, it means set of basis is v1 and v2 X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 1 -1/2 3/2 -5/2 3 0 0 0 0 0 4 0 0 0 0 0 \f\f\fAnswer to the question no-1 Answer to the question no-2 Answer to the question no-3 Answer to the question no-4 Answer to the question no-5 1st we will set up all in matrix form, try to put in write v1 , v2 as column in stead of row and reduce it we will get . X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 2 0 1 3 3 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Find the pivot in the 1st column in the 1st row X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 2 0 1 3 3 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Multiply the 1st row by 2 X1 X2 X3 X4 b 1 2 -6 4 -6 18 2 2 0 1 3 3 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Subtract the 1st row from the 2nd X1 X2 X3 X4 b 1 2 -6 4 -6 18 2 0 6 -3 9 -15 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Multiply the 1st row by 1 X1 X2 X3 X4 b 1 -2 6 -4 6 -18 2 0 6 -3 9 -15 3 -2 -4 1 -9 7 4 1 3 -1 6 -6 Subtract the 1st row from the 3rd row and restore it X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 6 -3 9 -15 3 0 -10 5 -15 25 4 1 3 -1 6 -6 Subtract the 1st row from the 4th X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 6 -3 9 -15 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Make the pivot in the 2nd column by dividing the 2nd row by 6 X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 1 -1/2 3/2 -5/2 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Multiply the 2nd row by 3 X1 X2 X3 X4 b 1 1 -3 2 -3 9 2 0 -3 3/2 -9/2 15/2 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Subtract the 2nd row from the 1st row and restore it X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 1 -1/2 3/2 -5/2 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Multiply the 2nd row by 10 X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 -10 5 -15 25 3 0 -10 5 -15 25 4 0 6 -3 9 -15 Subtract the 2nd row from the 3rd row and restore it X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 1 -1/2 3/2 -5/2 3 0 0 0 0 0 4 0 6 -3 9 -15 Multiply the 2nd row by 6 1 X1 1 X2 0 X3 1/2 X4 3/2 b 3/2 2 0 6 -3 9 -15 3 0 0 0 0 0 4 0 6 -3 9 -15 Subtract the 2nd row from the 4th row and restore it, since we see that last two rows are zers, it means set of basis is v1 and v2 X1 X2 X3 X4 b 1 1 0 1/2 3/2 3/2 2 0 1 -1/2 3/2 -5/2 3 0 0 0 0 0 4 0 0 0 0 0