midterm2 2007 - METU Department of Mathematics Group BASIC...

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Group List No. Last Name Name Student No. Department Section Signature : : : : : : : : : : : : : Code Acad. Year Semester Coordinator Date Time Duration M E T U Department of Mathematics BASIC LINEAR ALGEBRA Midterm II Math 260 2007-2008 Fall S.F, S.O, A.S. November 28 2007 17:40 90 minutes 4 QUESTIONS ON 4 PAGES TOTAL 100 POINTS 1 2 3 4 Show your work ! Partial credits will not be given for correct answers if they are not justified. Question 1 (12+6+6=24 points) The product of two (3 × 3)-matrices A and B is known to be AB = 1 2 3 0 4 5 0 0 6 , and the adjoint matrix Adj(A) has determinant 4. a) Find the determinants det( A - 1 ) = det A = det B = Solution: (det A )(det B ) = det 1 2 3 0 4 5 0 0 6 = 24 and Adj(A) = (det A)A - 1 Since A is 3 × 3-matrix, det(Adj(A)) = (det A) 3 det(A - 1 ) = (det A) 3 det A = (det A) 2 = 4 There are two possibilities: either det A = 2, det B = 12, det A - 1 = 1 2 , or det A = - 2, det B = - 12, det A - 1 = - 1 2 . b) Can we conclude that the row h 1 2 3 i is a linear combination of the rows of matrix B ? (Present complete and detailed arguments !) Solution: Yes. The rows of matrix B form a basis of R 3 , because det B 6 = 0 (a theorem). So, any row and in particular h 1 2 3 i is a linear combination of the rows of matrix B .
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