Elementary linear algebra: solutions to problems by Matthews K.R.

By Matthews K.R.

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8 1. (i) (−3 + i)(14 − 2i) = (−3)(14 − 2i) + i(14 − 2i) = {(−3)14 − (−3)(2i)} + i(14) − i(2i) = (−42 + 6i) + (14i + 2) = −40 + 20i. (ii) (iii) 2 + 3i (2 + 3i)(1 + 4i) = 1 − 4i (1 − 4i)(1 + 4i) ((2 + 3i) + (2 + 3i)(4i) = 12 + 4 2 −10 11 −10 + 11i = + i. = 17 17 17 (1 + 2i)2 1 + 4i + (2i)2 = 1−i 1−i −3 + 4i 1 + 4i − 4 = = 1−i 1−i (−3 + 4i)(1 + i) −7 + i 7 1 = = = − + i. 2 2 2 2 2. (i) iz + (2 − 10i)z = 3z + 2i ⇔ z(i + 2 − 10i − 3) = 2i −2i 1 + 9i −18 − 2i −9 − i −2i(1 − 9i) = = . 1 + 81 82 41 =⇔ z(−1 − 9i) = 2i ⇔ z = = (ii) The coefficient determinant is 1+i 2−i 1 + 2i 3 + i = (1 + i)(3 + i) − (2 − i)(1 + 2i) = −2 + i = 0.

Xm−1 = am − am−1 , xm = am . a b c . If [a, b, c] is a multiple of [1, 1, 1], (that is, 1 1 1 a = b = c), then rank A = 1. For if 10. Let A = [a, b, c] = t[1, 1, 1], then R(A) = [a, b, c], [1, 1, 1] = t[1, 1, 1], [1, 1, 1] = [1, 1, 1] , 38 so [1, 1, 1] is a basis for R(A). However if [a, b, c] is not a multiple of [1, 1, 1], (that is at least two of a, b, c are distinct), then the left–to–right test shows that [a, b, c] and [1, 1, 1] are linearly independent and hence form a basis for R(A).

E. At = A and that AB is defined. Then (B t AB)t = B t At (B t )t = B t AB, so B t AB is also symmetric. 25. Let A be m × n and B be n × m, where m > n. Then the homogeneous system BX = 0 has a non–trivial solution X0 , as the number of unknowns is greater than the number of equations. Then (AB)X0 = A(BX0 ) = A0 = 0 30 and the m × m matrix AB is therefore singular, as X0 = 0. 26. (i) Let B be a singular n × n matrix. Then BX = 0 for some non–zero column vector X. Then (AB)X = A(BX) = A0 = 0 and hence AB is also singular.

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