**Unformatted text preview: **. But ±-5 2 3-1 ²± 1 2 3 5 ² M = IM = M so M = ± 0 0 0 0 ² . 18 ) T ³± a b c d ²´ = ± 1 0 0 0 ²± a b c d ²± 0 0 0 1 ² = ± a b 0 0 ²± 0 0 0 1 ² = ± b 0 0 ² so as long as the upper right entry is not 0, the image under T will not be zero. 31 ) Consider the vectors ~u and ~v deﬁning two sides of a rectangle. (The vertices are ~ 0, ~u , ~v and ~u + ~v .) Then the images of these four points under T , using the linearity are ~ 0, T ( ~u ), T ( ~v ) and T ( ~u )+ T ( ~v ), which form a parallelogram (squashed is { T ( ~u ) , T ( ~v ) } are dependent) by the rules of vector addition....

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