math 304 practice final exam solution

math 304 practice final exam solution - MATH 304–505...

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Unformatted text preview: MATH 304–505 Spring 2011 Sample problems for the final exam: Solutions Any problem may be altered or replaced by a different one! Problem 1 (15 pts.) Find a quadratic polynomial p ( x ) such that p ( − 1) = p (3) = 6 and p ′ (2) = p (1). Let p ( x ) = ax 2 + bx + c . Then p ( − 1) = a − b + c , p (1) = a + b + c , and p (3) = 9 a + 3 b + c . Also, p ′ ( x ) = 2 ax + b and p ′ (2) = 4 a + b . The coefficients a , b , and c have to be chosen so that a − b + c = 6 , 9 a + 3 b + c = 6 , 4 a + b = a + b + c. This is a system of linear equations in variables a,b,c . To solve the system, let us convert the third equation to the standard form and add it to the first and the second equations: a − b + c = 6 9 a + 3 b + c = 6 3 a − c = 0 ⇐⇒ 4 a − b = 6 9 a + 3 b + c = 6 3 a − c = 0 ⇐⇒ 4 a − b = 6 12 a + 3 b = 6 3 a − c = 0 Now divide the second equation by 3, add it to the first equation, and find the solution by back substitution: 4 a − b = 6 4 a + b = 2 3 a − c = 0 ⇐⇒ 8 a = 8 4 a + b = 2 3 a − c = 0 ⇐⇒ a = 1 4 a + b = 2 3 a − c = 0 ⇐⇒ a = 1 b = − 2 3 a − c = 0 ⇐⇒ a = 1 b = − 2 c = 3 Thus the desired polynomial is p ( x ) = x 2 − 2 x + 3. Problem 2 (20 pts.) Let v 1 = (1 , 1 , 1), v 2 = (1 , 1 , 0), and v 3 = (1 , , 1). Let L : R 3 → R 3 be a linear operator on R 3 such that L ( v 1 ) = v 2 , L ( v 2 ) = v 3 , L ( v 3 ) = v 1 . (i) Show that the vectors v 1 , v 2 , v 3 form a basis for R 3 . Let U be a 3 × 3 matrix such that its columns are vectors v 1 , v 2 , v 3 : U = 1 1 1 1 1 0 1 0 1 . To find the determinant of U , we subtract the second row from the first one and then expand by the first row: det U = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 0 1 1 1 0 1 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 1 1 1 0 vextendsingle vextendsingle vextendsingle vextendsingle = − 1 . Since det U negationslash = 0, the vectors v 1 , v 2 , v 3 are linearly independent. It follows that they form a basis for R 3 . 1 (ii) Find the matrix of the operator L relative to the basis v 1 , v 2 , v 3 . Let A denote the matrix of L relative to the basis v 1 , v 2 , v 3 . By definition, the columns of A are coordinates of vectors L ( v 1 ) ,L ( v 2 ) ,L ( v 3 ) with respect to the basis v 1 , v 2 , v 3 . Since L ( v 1 ) = v 2 = v 1 + 1 v 2 + 0 v 3 , L ( v 2 ) = v 3 = 0 v 1 + 0 v 2 + 1 v 3 , L ( v 3 ) = v 1 = 1 v 1 + 0 v 2 + 0 v 3 , we obtain A = 0 0 1 1 0 0 0 1 0 ....
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This note was uploaded on 02/13/2012 for the course MATH 304 taught by Professor Hobbs during the Spring '08 term at Texas A&M.

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math 304 practice final exam solution - MATH 304–505...

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