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CMI-CE5S

# CMI-CE5S - CC AD 2011-12 242 College Mathematics I —...

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Unformatted text preview: CC: AD [ 2011-12 ] 242: College Mathematics I — Class Exercises [ Sol ] — Chapter 5: Linear Algebra ª 5.1 Matrix ª 5.2 Determinant ª 5.3 Solving System of Linear Equations ª 5.4 Powers of a Square Matrix ª 5.5 Applications Ch5: Linear Algebra [ CE5S-1 ] q 5.1 Matrix Class Ex. 1 If A = 3 0 1 4 ™ and B = 2 6 5 0 ™ . (a) Calculate A + B , A − B and ( A + B )( A − B ) . [ 3 ] (b) Calculate A 2 , B 2 and A 2 − B 2 . [ 3 ] Is A 2 − B 2 = ( A + B )( A − B ) ? [ 1 ] Solution (a) A + B = 5 6 6 4 ™ , A − B = 1 − 6 − 4 4 ™ and ( A + B )( A − B ) = 5 6 6 4 ™ 1 − 6 − 4 4 ™ = − 19 − 6 − 10 − 20 ™ . (b) A 2 = 9 7 16 ™ , B 2 = 34 12 10 30 ™ and A 2 − B 2 = − 25 − 12 − 3 − 14 ™ . A 2 − B 2 ̸ = ( A + B )( A − B ) . Class Ex. 2 If A = 3 − 4 1 − 1 ™ , then prove that A n = 1 + 2 n − 4 n n 1 − 2 n ™ where n is any positive integer. [ 6 ] Solution Use induction, when n = 1, A = 3 − 4 1 − 1 ™ = 1 + 2 ( 1 ) − 4 ( 1 ) ( 1 ) 1 − 2 ( 1 ) ™ . ∴ The statement is true for n = 1. Assume the statement is true for some integer k ≥ 1, i.e., A k = 1 + 2 k − 4 k k 1 − 2 k ™ . When n = k + 1, A k + 1 = A k · A = 1 + 2 k − 4 k k 1 − 2 k ™ 3 − 4 1 − 1 ™ = 3 + 6 k − 4 k − 4 − 8 k + 4 k 3 k + 1 − 2 k − 4 k − 1 + 2 k ™ = 3 + 2 k − 4 − 4 k k + 1 − 2 k − 1 ™ = 1 + 2 ( k + 1 ) − 4 ( k + 1 ) k + 1 1 − 2 ( k + 1 ) ™ ∴ The statement is also true for n = k + 1. Hence, by the principle of M.I., the statement is true for all positive integers n . AD: College Mathematics I [ 2011-12 ] Louis Lau 1 Ch5: Linear Algebra [ CE5S-2 ] q 5.2 Determinant Class Ex. 1 The 3 × 3 matrices A and B are such that AB =    3 − 1 2 4 2 3 2 4    . (a) Calculate det ( AB ) . [ 2 ] (b) Given that A =    1 1 − 2 1 1 0 1    , find (i) det A T , det ( 2 A ) , det A 2 , det A − 1 , [ 4 ] (ii) det B , det B − 1 , [ 2 ] (iii) | B T A T | , | B − 1 A − 1 | . [ 2 ] (iv) A − 1 . [ 3 ] Solution (a) det ( AB ) = 24 − 6 − 8 + 16 = 26. (b) | A | = 2. (i) det A T = det A = 2, det ( 2 A ) = 2 3 det A = 16, det A 2 = ( det A ) 2 = 4, det A − 1 = 1 det A = 1 2 . (ii) det B = 26 det A = 13, det B − 1 = 1 det B = 1 13 . (iii) | B T A T | = | ( AB ) T | = | AB | = 26, | B − 1 A − 1 | = | ( AB ) − 1 | = 1 | AB | = 1 26 . (iv) A − 1 = 1 | A | ( Cof ) T = 1 2    1 2 − 1 − 2 2    T =    1 2 − 1 2 1 − 1 1    AD: College Mathematics I [ 2011-12 ] Louis Lau 2 Ch5: Linear Algebra [ CE5S-3 ] q 5.3 Solving System of Linear Equations Class Ex. 1 Solve the system of linear equations in unknowns x , y and z :    x + y + 2 z = 9 2 x + 4 y − 3 z = 1 3 x + 6 y − 5 z = ....
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CMI-CE5S - CC AD 2011-12 242 College Mathematics I —...

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