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ÆÄÀÏ : ¿¬¼¼´ë ¼±Çü´ë¼öÇÐ Á·º¸ 2Çбâ-¼±´ë½ÃÇè-1Â÷Áß°£-¸ð¹ü~.pdf   [Size : 102 Kbyte ]
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Problem 1. Indicate whether the statement is true(T) or false(F). Justify your answer. [each 3pt] (1) If A and B are invertible matrices, then A + B is also invertible. (F) Solve Take B = A, and A is an any invertible matrix. Then, A and B are invertible, but A + B = O is not invertible.
Problem 2. Indicate whether the statement is true(T) or

(1) If x0 is a vector in Rn , and if v1 and v2 are nonzero vectors in Rn , then the set of all vectors x = x0 + t1 v1 + t2 v2 (t1 , t2 ¡ô R) is a plane. (F)

(2) For nonzero vector a and b in Rn , if a¡Ñ = b¡Ñ , then a = b. (F) Solve In R2 , take a = (1, 0) and b = (1, 0). Then, a¡Ñ = b¡Ñ is y-axis, but a = b. In general, if a¡Ñ = b¡Ñ , then a and b are parallel vectors. (2) A homogeneous linear system with more equations than unknowns has innitely many solutions. (F)

(3) For nonzero vector b ¡ô Rn , if Ax = b has innitely many solutions, then so does Ax = 0. (T) Solve If Ax = 0 has only the trivial solution, then Ax = b has a unique solution or is inconsistent. It is impossible, so Ax = 0 has also innitely many solutions.

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(4) For any m ¡¿ n matrix A, AT A is an m ¡¿ m symmetric (4) If Ax = b and Ax = c are consistent, then so is matrix. (F) Ax = b + c. (T) Solve Since A is an m¡¿n, AT A is an n¡¿n symmetric matrix.

(5) If B has a column of zeros, then so does AB if this product is dened. (T) Solve Put B = b1 Ab1 ... Abi ... b2 . . . bn as a column partition. If bi = 0 for some 1 ¡Â i ¡Â n. Since AB = Abn and Abi = A0 = 0, AB has a column of zeros. (5) If span(S1 ) = span(S2 ), then S1 = S2 . (Si is the set of vectors in Rn ) (F)



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