Linear Algebra Solutions 42

# Linear Algebra Solutions 42 - Hence A is non–singular and...

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Unformatted text preview: Hence A is non–singular and C11 C21 C31 −11 −4 2 1 1 1 C12 C22 C32 = 29 7 −10 . A−1 = adj A = −13 −13 −13 C13 C23 C33 1 −2 1 6. (i) 2a 2b b−c 2b 2a a + c a+b a+b b 2a + 2b 2b + 2a b + a 2b 2a a+c a+b a+b b R1 → R1 + R2 = 2 2 1 2b 2a a + c = (a+b) a+b a+b b = 2(a + b)(a − b) 0 2 1 C1 → C 1 − C 2 (a+b) 2(b − a) 2a a + c = 0 a+b b 2 1 a+b b = −2(a + b)(a − b)2 . (ii) b+c b c c c+a a b a a+b C3 → C 3 − C 1 = C1 → C 1 − C 2 = c b 0 −a c + a 2a b−a a 2a c b 0 R3 → R3 − R2 2a −a c + a 1 = b −c 0 c b c −a c + a a b−a a a+b = 2a = −2a c b 0 −a c + a 1 b−a a 1 c b b −c = 2a(c2 + b2 ). 7. Suppose that the curve y = ax2 + bx + c passes through the points (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), where xi = xj if i = j . Then ax2 + bx1 + c = y1 1 ax2 + bx2 + c = y2 2 ax2 + bx3 + c = y3 . 3 The coeﬃcient determinant is essentially a Vandermonde determinant: x2 x1 1 1 x2 x2 1 2 x2 x3 1 3 = x2 x2 x2 1 2 3 x1 x2 x3 111 111 = − x1 x2 x3 x2 x2 x2 3 1 2 45 = −(x2 −x1 )(x3 −x1 )(x3 −x2 ). ...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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