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Ma1502Te3ESolns-Spring2009

# Ma1502Te3ESolns-Spring2009 - MATH1502 Calculus II E Section...

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MATH1502 - Calculus II E Section TEST 3 Solutions - April 7, 2009 Question Grade Ex 1 20 2 12 3 26 Total 58 There are 3 questions and 58 marks on this paper. 50 marks =100%. Question 1 Let b be a real number. (a) Use row reduction to °nd all solutions of the system of linear equations x 1 ° x 2 ° 2 x 3 ° 2 x 4 +4 x 5 = 1 x 1 ° x 3 + x 5 = 0 4 x 1 ° 3 x 2 ° 7 x 3 ° 7 x 4 +14 x 5 = 3 2 x 1 ° 2 x 2 ° 4 x 3 ° 4 x 4 +8 x 5 = b + 1 (13 marks) Also identify all the pivotal columns, and pivotal variables. For which b is there no solution? For which b are there in°nitely many solutions? (b) From your solution in (a), write down Ker ( A ) and Im ( A ) for A , where A = 2 6 6 4 1 ° 1 ° 2 ° 2 4 1 0 ° 1 0 1 4 ° 3 ° 7 ° 7 14 2 ° 2 ° 4 ° 4 8 3 7 7 5 : (5 marks) (c) What is the rank of A ? From this decide whether or not A is invertible. (2 marks) Solutions The augmented matrix is 2 6 6 4 1 ° 1 ° 2 ° 2 4 1 1 0 ° 1 0 1 0 4 ° 3 ° 7 ° 7 14 3 2 ° 2 ° 4 ° 4 8 b + 1 3 7 7 5 1

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Step 1: Row 2 - Row 1; Row 3 - 4 ± Row 1; Row 4 - 2 ± Row 1 2 6 6 4 1 ° 1 ° 2 ° 2 4 1 0 1 1 2 ° 3 ° 1 0 1 1 1 ° 2 ° 1 0 0 0 0 0 b ° 1 3 7 7 5 Step 2: Row 3 -Row 2 2 6 6 4 1 ° 1 ° 2 ° 2 4 1 0 1 1 2 ° 3 ° 1 0 0 0 ° 1 1 0 0 0 0 0 0 b ° 1 3 7 7 5 This is now in row reduced form. Step 4: Back Substitution We see that columns 1, 2, 4 are pivotal, so x 1 ; x 2 ; x 4 are pivotal - but the 3rd, 5th columns are not pivotal, so x 3 ; x 5 are non-pivotal variables. We shall set x 3 = t 1 and x 5 = t 2 . Then the rank is 3. Since this is < 4, or because the matrix is not square, the matrix is not invertible. Now substitute back: x 1 ° x 2 ° 2 x 3 ° 2 x 4 +4 x 5 = 1 x 2 + x 3 +2 x 4 ° 3 x 5 = ° 1 ° x 4 + x 5 = 0 0 = b ° 1 (8 marks) We see that if b 6 = 1 , there is no solution.
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Ma1502Te3ESolns-Spring2009 - MATH1502 Calculus II E Section...

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