Linear Algebra Solutions 47

# Linear Algebra Solutions 47 - Conversely, suppose that det...

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Unformatted text preview: Conversely, suppose that det A = 0. Then the homogeneous system AX = 0 has a non–trivial solution X = [x1 , . . . , xn ]t . So x1 A∗1 + · · · + xn A∗n = 0. Suppose for example that x1 = 0. Then − A ∗1 = x2 x1 + ··· + − xn x1 A ∗n and the ﬁrst column of A is a linear combination of the remaining columns. 14. Consider the system −2x + 3y − z = 1 x + 2y − z = 4 −2x − y + z = −3 0 7 −3 −2 3 −1 7 −3 1 2 −1 = 1 2 −1 = − = −2 = 0. Let ∆ = 3 −1 0 3 −1 −2 −1 1 Hence the system has a unique solution which can be calculated using Cramer’s rule: ∆2 ∆3 ∆1 , y= , z= , x= ∆ ∆ ∆ where ∆1 = ∆2 = ∆3 = Hence x = −4 −2 = 2, y = −6 −2 1 3 −1 4 2 −1 −3 −1 1 −2 1 −1 1 4 −1 −2 −3 1 −2 3 1 1 2 4 −2 −1 −3 = 3, z = −8 −2 = −4, = −6, = −8. = 4. 15. In Remark 4.0.4, take A = In . Then we deduce (a) det Eij = −1; (b) det Ei (t) = t; 50 ...
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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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