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Linear algebra questions 2010/12 Paper code：10043C Class Times ：45 ClassCourse Title：Linear Algebra Class：Elective classes【Note：Write your answers and the question numbers on your answer sheet ! Otherwise, invalid ! 】 1. Fill in the blank（） –1，0) and B = (1，1，1，1，2) , so Rank(ATB) = ( 1 ). Solution: (2) Let A is a (3×3) matrix, |A| = ?3，|?3A| = ( 81 ) (3) If A is an (n×n) matrix，XX2 are the solutions to the linear equations AX = B，X1 ≠ X2)，|A| = ( 0 ). (4) Let A be a linearly dependent the vectors set, A = { β1, β2, β3 }, here β1 = (1, 2, 3), β2 = (4, t, 3), β3 = (0, 0, 1), so t = ( 8 ). Solution: (5) There are n dimension vectors : α1，α2，α3 ， if (αi，αj) = i + j ，so we have ( α1 + α2，α1 – α3) = ( 4 ). Solution: For (αi，αj) = i + j ，so we have (6) If X1=(1, 0, 2)T and X2=(3, 4, 5)T are two solutions to the three variables non-homogeneous linear equations AX=B, so the corresponding homogeneous linear equations AX = 0 has solution ( (2, 4, 3)T ). Solution: X1 –X2 =(1, 0, 2)T – (3, 4, 5)T = (2, 4, 3)T is a solution to the equation AX = 0. (7) Let β = ( 1，–1，0，2 )T and γ = ( a，1，–1，1 )T are orthogonal vectors, so a = ( –1 ). Solution: β and γ are orthogonal vectors, so β·γ =0, that is (8) A，B(4×4) matrices，for A = (α，α1，α2，α3) and B = (β，α1，α2，α3)，if |A| = 1 and |B| = 2, so we can find that |A+B| =( 24 ). Solution: A+B = (α，α1，α2，α3) + (β，α1，α2，α3) = (α+β，2α1，2α2，2α3) , so |A+B| = det (α+β，2α1，2α2，2α3) = det (α，2α1，2α2，2α3)+ det (β，2α1，2α2，2α3) =23 det (α，α1，α2，α3)+ 23det (β，α1，α2，α3) =8+16=24 (9) Let A is (3×3) matrices, for |A| = 2， | A*－A－1| –1/2 ). Solution: for |A| = 2, |A－1|A* = A－1 |A|| A*－A－1| | A－1 |A|－A－1| | A－1|A|－|= |(－A－1|–1/2 (10) Let ，|A| ≠ 0,A?1 = ( ). Solution: A－1 A*/|A| Q.2 / Scores 2. Multiple Choice (1*20 = 20 scores) (1) Let A，B，CD are (n×n) matrices，(n×n) identity matrix，he following statement is correct ( D ). A．， B．， C．， A ≠