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math 304 Test1sample solution

math 304 Test1sample solution - MATH 304505 Spring 2011...

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MATH 304–505 Spring 2011 Sample problems for Test 1: 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) = 1, p (2) = 3, and p (3) = 7. Let p ( x ) = ax 2 + bx + c . Then p (1) = a + b + c , p (2) = 4 a + 2 b + c , and p (3) = 9 a + 3 b + c . The coefficients a , b , and c have to be chosen so that a + b + c = 1 , 4 a + 2 b + c = 3 , 9 a + 3 b + c = 7 . We solve this system of linear equations using elementary operations: a + b + c = 1 4 a + 2 b + c = 3 9 a + 3 b + c = 7 ⇐⇒ a + b + c = 1 3 a + b = 2 9 a + 3 b + c = 7 ⇐⇒ a + b + c = 1 3 a + b = 2 8 a + 2 b = 6 ⇐⇒ a + b + c = 1 3 a + b = 2 4 a + b = 3 ⇐⇒ a + b + c = 1 3 a + b = 2 a = 1 ⇐⇒ a + b + c = 1 b = 1 a = 1 ⇐⇒ c = 1 b = 1 a = 1 Thus the desired polynomial is p ( x ) = x 2 x + 1. Problem 2 (25 pts.) Let A = 1 2 4 1 2 3 2 0 2 0 1 1 2 0 0 1 . (i) Evaluate the determinant of the matrix A . First let us subtract 2 times the fourth column of A from the first column: vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 2 4 1 2 3 2 0 2 0 1 1 2 0 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 2 4 1 2 3 2 0 0 0 1 1 0 0 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle . Now the determinant can be easily expanded by the fourth row: vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 2 4 1 2 3 2 0 0 0 1 1 0 0 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 2 4 2 3 2 0 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle . The 3 × 3 determinant is easily expanded by the third row: vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 2 4 2 3 2 0 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = ( 1) vextendsingle vextendsingle vextendsingle vextendsingle 1 2 2 3 vextendsingle vextendsingle vextendsingle vextendsingle . 1
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Thus det A = vextendsingle vextendsingle vextendsingle vextendsingle 1 2 2 3 vextendsingle vextendsingle vextendsingle vextendsingle = 1 . Another way to evaluate det A is to reduce the matrix A to the identity matrix using elementary row operations (see below). This requires much more work but we are going to do it anyway, to find the inverse of A .
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