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MATH 304–503 Spring 2010Sample problems for Test 1: SolutionsAny 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) = ax2+ bx + c. Then p(1) = a + b + c, p(2) = 4a + 2b + c, and p(3) = 9a + 3b + c. Thecoefficients a, b, and c have to b e chosen so thata + b + c = 1,4a + 2b + c = 3,9a + 3b + c = 7.We solve this system of linear equations using elementary operations:a + b + c = 14a + 2b + c = 39a + 3b + c = 7⇐⇒a + b + c = 13a + b = 29a + 3b + c = 7⇐⇒a + b + c = 13a + b = 28a + 2b = 6⇐⇒a + b + c = 13a + b = 24a + b = 3⇐⇒a + b + c = 13a + b = 2a = 1⇐⇒a + b + c = 1b = −1a = 1⇐⇒c = 1b = −1a = 1Thus the desired polynomial is p(x) = x2− x + 1.Problem 2 (25 pts.) Let A =1 −2 4 12 3 2 02 0 −1 12 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:1 −2 4 12 3 2 02 0 −1 12 0 0 1=−1 −2 4 12 3 2 00 0 −1 10 0 0 1.Now the determinant can be easily expanded by the fourth row:−1 −2 4 12 3 2 00 0 −1 10 0 0 1=−1 −2 42 3 20 0 −1.The 3 × 3 determinant is easily expanded by the third row:−1 −2 42 3 20 0 −1= (−1)−1 −22 3.1Thusdet A = −−1 −22 3= −1.Another way to evaluate det A is to r educe the matrix A to the identity matrix using elementaryrow operations (see below). This requires much more work but we are going to do it anyway, to findthe inverse of A.(ii) Find the inverse matrix A−1.First we merge the matrix A with the identity matrix into one 4 × 8 matrix(A | I) =1 −2 4 11 0 0 02 3 2 00 1 0 02 0 −1 10 0 1 02 0 0 10 0 0 1.Then we apply elementary row operations to this matrix until the left part becomes the identitymatrix.Subtract 2 times the first row from the second row:1 −2 4 11 0 0 02 3 2 00 1 0 02 0 −1 10 0 1 02 0 0 10 0 0 1→1 −2 4 11 0 0 00 7 −6 −2−2 1 0 02 0 −1 10 0 1 02 0 0 10 0 0 1.Subtract 2 times the first row from the third row:1 −2 4 11 0 0 00 7 −6 −2−2 1 0 02 0 −1 10 0 1 02 0 0 10 0 0 1→1 −2 4 11 0 0 00 7 −6 −2−2 1 0 00 4 −9 −1−2 0 1 02 0 0 10 0 0 1.Subtract 2 times the first row from the fourth row:1 −2 4 11 0 0 00 7 −6 −2−2 1 0 00 4 −9 −1−2 0 1 02 0 0 10 0 0 1→1 −2 4 11 0 0 00 7 −6 −2−2 1 0 00 4 −9 −1−2 0 1 00 4 −8 −1−2 0 0 1.Subtract 2 times the fourth row from the second row:1 −2 4 11 0 0 00 7 −6 −2−2 1 0 00 4 −9 −1−2 0 1 00 4 −8 −1−2 0 0 1→1 −2 4 11 0 0 00 −1 10 02 1 0 −20 4 −9 −1−2 0 1 00 4 −8 −1−2 0 0 1.Subtract the fourth row from the third row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 4 −9 −1−2 0 1 00 4 −8 −1−2 0 0 1→1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 4 −8 −1−2 0 0 1.2Add 4 times the second row to the fourth row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 4 −8 −1−2 0 0 1→1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 0 32 −16 4 0 −7.Add 32 times the third row to the fourth row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 0 32 −16 4 0 −7→1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 0 0 −16 4 32 −39.Add 10 times the third row to the second row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 0 0 −16 4 32 −39→1 −2 4 11 0 0 00 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39.Add the fourth row to the first row:1 −2 4 11 0 0 00 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39→1 −2 4 07 4 32 −390 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39.Add 4 times the third row to the first row:1 −2 4 07 4 32 −390 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39→1 −2 0 07 4 36 −430 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39.Subtract 2 times the second row from the first row:1 −2 0 07 4 36 −430 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39→1 0 0 03 2 16 −190 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39.Multiply the second, the third, and the fourth rows by −1:1 0 0 03 2 16 −190 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39→1 0 0 03 2 16 −190 1 0 0−2 −1 −10 120 0 1 00 0 −1 10 0 0 1−6 −4 −32 39.Finally the left part of our 4 × 8 matrix is transformed into the identity matrix. Therefore thecurrent right part is the inverse matrix of A. ThusA−1=1 −2 4 12 3 2 02 0 −1 12 0 0 1−1=3 2 16 −19−2 −1 −10 120 0 −1 1−6 −4 −32 39.3As a byproduct, we can evaluate the determinant of A. We have transformed A into the identitymatrix using elementary row operations. These included no row exchanges and three row multiplica-tions, each time by −1. It follows that det I = (−1)3det A. Hence det A = …


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TAMU MATH 304 - Test1samplesolved

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