lab3_sol_f_08

lab3_sol_f_08 - Math 115 - Lab 3 - Fall 2008 Topics: •...

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Unformatted text preview: Math 115 - Lab 3 - Fall 2008 Topics: • Gaussian elimination • Homogeneous equations • Matrix addition, scalar multiplication and transposition • Matrix multiplication • Matrix inversion You are to provide full solutions to the following problems. You are allowed, and encouraged to collaborate with your classmates, use your notes and textbook and ask the TA for guidance. Direct copying of solutions is not encouraged, nor is it allowed or ethical. Last name: Student number: First name: Math 115 - Lab 3 - Fall 2008 Student number: 1. Find all solutions to the following homogeneous systems. If there is more than one solution, then please express the solution set in vector equation form (i.e. a linear combination of vectors). a) 5x +z=0 x + 2y =0 y +z=0 Solution: 501 120 1 2 0 ∼ 0 1 1 , swapping rows 011 501 1 20 ∼ 0 1 1 R 3 = R 3 − 5R 1 0 −10 1 12 0 ∼ 0 1 1 R3 = R3 + 10R2 0 0 11 120 1 R3 ∼ 0 1 1 R3 = 11 001 and from this it is clear that this system has only the trivial solution. 1 b) x + 2y + 3z = 0 x + 2z = 0 2x + 6y + 7z = 0 Solution: 123 102 1 0 2 ∼ 1 2 3 267 267 102 0 2 1 ∼ 063 102 0 1 1 ∼ 2 063 102 0 1 1 ∼ 2 000 , swapping rows R 2 = R 2 − R 1 ; R 3 = R 3 − 2R 1 R2 = 1 R2 2 R 3 = R 3 − 6R 2 So letting z = t, we obtain the following. 1 R2 reads y + t = 0 2 1 =⇒ y = − t 2 R1 reads x + 2t = 0 =⇒ x = −2t Therefore the solution set is −2 −2t 0 −4 − 1 t = 0 + t − 1 = t −1 , for t ∈ R 2 2 t 0 1 2 2 2. Compute the following sums, or indicate ‘cannot add’. a) 107 3 −2 1 2 −4 4 0 1 3 −3 2 + Solution: 107 −3 1 2 2 + −4 4 0 1 3 −3 2 = −3 4 7 −1 4 −1 b) sin 1 e−3 + Solution: sin 1 + e−3 = sin 1 + e−3 c) 107 3 −2 1 2 1 −4 2 + 4 3 0 −3 Solution: Cannot add, since the matrices have different sizes. 3 3. Compute the following if possible, or indicate ‘cannot compute’. a) π 2 π 3 1 π 2 3 108 Solution: π 2 π 3 = 2 3 108 1 (π)π 2 1π (π) 3 = 1 π 1 2 1 3 1 (π)2 3 1 ( π )108 2 3 108 π b) T −5 −1 7 0 8 1 Solution: T −5 −1 7 0 8 1 4 = −5 7 8 −1 0 1 c) T 11 −1 1 2 1 1 −1 11 0 1 Solution: T 11 −1 1 2 1 1 −1 = 11 0 1 = 121 111 10 01 −1 1 1 −1 0 1 d) T 11 2 1 11 −1 10 1 −1 1 Solution: Cannot compute, since the matrices are the wrong size to multiply together. 5 12 1 −1 0 , C = 3 4 . 0 11 56 211 ,B= 130 e) (A + B )C , where A = Solution 1: (A + B )C = 211 130 = 301 141 = 8 12 18 24 + 1 3 5 1 −1 0 0 11 2 4 6 12 3 4 56 Solution 2: (A + B )C = AC + BC = 10 14 10 14 = 12 3 4 + 56 211 130 = 8 12 18 24 + 6 −2 −2 8 10 1 −1 0 0 11 12 3 4 56 f) −1 1 −1 0 2 Solution: 1 −1 0 2 −1 1 1 · 2 − 0 · (−1) 1 21 2 01 = = 1 0 = 21 01 1 2 1 2 Check: 1 −1 0 2 1 0 1 2 1 2 = 10 01 as required. g) 1 −1 −1 1 −1 Solution: The determinant of this matrix is 1 · 1 − (−1) · (−1) = 0 and therefore this matrix has no inverse. 7 4. Compute the inverses for the following matrices, or prove that no inverse exists. a) 3 12 1 −1 3 1 24 Solution: 3 1 1 1 ∼1 3 1 ∼0 0 1 ∼0 0 1 ∼0 0 1 0 ∼ 0 1 0 ∼ 0 1 0 ∼ 0 12 −1 3 24 100 010 001 −1 3 24 12 010 001 100 , swapping rows −1 3 3 1 4 −7 0 10 0 −1 1 1 −3 0 −1 3 1 1 3 4 −7 0 1 0 −1 3 1 −3 −1 3 1 1 3 0 − 25 3 −1 1 0 3 1 3 1 −1 0 10 01 00 10 01 0 R 2 = R 2 − R 1 ; R 3 = R 3 − 3R 1 R2 = 1 R2 3 0 0 1 0 1 R 3 = R 3 − 4R 2 0 −1 3 3 5 4 1 −3 −3 0 10 1 R 3 = (− 3 )R 3 0 −1 3 3 25 3 5 4 − 25 25 25 9 10 12 25 25 − 25 1 10 7 R 1 = R 1 − 3R 3 ; R 2 = R 2 − 1 R 3 25 − 25 25 3 3 5 4 − 25 25 25 10 5 0 − 25 25 10 7 1 R1 = R1 + R2 25 − 25 25 3 − 25 5 25 1 3 4 25 Check: 3 12 10 0 −5 25 0 0 100 1 1 1 −1 3 1 −10 7 = 0 25 0 = 0 1 0 25 25 1 24 −3 5 4 0 0 25 001 8 b) −1 1 2 4 0 7 1 3 13 Solution: ∼ ∼ ∼ ∼ −1 1 2 40 7 1 3 13 1 −1 −2 4 0 7 1 3 13 1 −1 −2 0 4 15 0 4 15 1 −1 −2 0 1 15 4 0 4 15 1 −1 −2 0 1 15 4 0 0 0 100 010 001 −1 0 0 010 001 −1 0 0 410 101 −1 1 1 00 1 40 01 −1 00 1 1 40 −3 −1 1 R1 = (−1)R1 R 2 = R 2 − 4R 1 ; R 3 = R 3 − R 1 R2 = 1 R2 4 R 3 = R 3 − 4R 2 So we can now see that no inverse exists for the given matrix. 9 Bonus Question: Compute the values of a, b, c, given the following matrices. 1 0 −1 0 b2 A= 3 2 0 , A−1 = 0 −1 c −1 −1 a −1 12 Solution: Since AA−1 1 0 0 = In for any invertible n × n matrix A, we have 00 1 0 −1 0 b2 1 0 = 3 2 0 0 −1 c 01 −1 −1 a −1 12 1 b−1 0 = 0 3b − 2 2c + 6 −a a − b + 1 2a − c − 2 • From row 3, column 1, we obtain that 0 = −a ⇐⇒ a = 0. • From row 1, column 2, we obtain that 0 = b − 1 ⇐⇒ b = 1. • From row 2, column 3, we obtain that 0 = 2c + 6 ⇐⇒ c = −3. These substitutions satisfy the remaining identities, so they are correct and we are done. 10 ...
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This note was uploaded on 09/17/2011 for the course MATH 115 taught by Professor Dunbar during the Fall '07 term at Waterloo.

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