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Und f(b) = z mit y 6= z Dann gehört zwar das Element y zu f(A)∩f(B) = {y}, es gehört aber nicht zu f(A ∩ B) = f({b}) = {z} ( ) Nach der Definition von f −1ist f (f(A)) = {x ∈ X f(x) ∈ f(A)} Gehört nun ein Element x zu der Menge A, so gehört auch f(x) zu dem Bild B = f(A) von A Dann muss aber das Element x zu dem Urbild f−1(B) = f−1(f(A)) gehören Die umgekehrte Inklusio. A V b N X f B X X j J ̒ʔ̂ ߂͂ I A V b N X f B X X j J ̂ Љ Ă ܂ B TOP y W ̃y W A V b N X @ASICS Gel180?. J l ׂĎ u ɂ jaa a } t f c v b n x f z ł a w a c ؂𔭍s v ܂ b ɂ ajaa a } r f b l ^ ̎ { afr c z x f 莎 ̎󌱎 i ^ ܂ b.

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X ·A = B genug¨ t es, nur die Form A·X = B (mit X rechts) zu betrachten Transponierte Matrixgleichung XT · A = B ⇔ AT · X = BT Wenn X L¨osung von AT ·X = BT ist, so ist Y = XT L¨osung von Y ·A = B Bsp7 Um y1,1 y1,2 y1,3 y2,1 y2,2 y2,32×3 · 12 33 3×2 = 112×2 zu l¨osen, wird also die transponierte Gleichung 13 2 23 02×3 · x1,1 x1,2 x2,1 x2,2 x3,1. T b N X Z s ́A X y C h q F \ Z t F C V t N \ W j t b g t N \ W ̃e N j b N g ݍ 킹 R X ł B Ƒ 瓭 ܂ B f 3,240 i ō j 8,640 i ō j L b Y i w ȉ j \5,400 i.