Exam2MA262fall2011

# Exam2MA262fall2011 - MA 262 Exam 2 Instructor Raphael Hora...

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MA 262 Exam 2 Fall 2011 Instructor: Raphael Hora Name: Max Possible Student ID#: 1. No books or notes are allowed. 2. You CAN NOT USE calculators or any electronic devices. 3. Show all work to receive full credit. 4. Boa Sorte! (Good Luck in portuguese) Problem Max. Possible Points 1 15 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 15 Total 100 1

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1. Compute - 1 - 2 - 2 - 3 1 0 - 2 0 2 4 4 9 3 - 2 1 - 2 . (15 points) Solution. If we replace row-2 by (row - 2) + (row - 1), row-3 by (row - 3) + 2(row - 1) and row-4 by (row - 4) + 3(row - 1) , we get - 1 - 2 - 2 - 3 1 0 - 2 0 2 4 4 9 3 - 2 1 - 2 = - 1 - 2 - 2 - 3 0 - 2 - 4 - 3 0 0 0 3 0 - 8 - 5 - 11 = - - 1 - 2 - 2 - 3 0 - 2 - 4 - 3 0 - 8 - 5 - 11 0 0 0 3 , where in the last step we switched the fouth and third rows. Now we replace row-3 by (row - 3) - 4(row - 2) to get - - 1 - 2 - 2 - 3 0 - 2 - 4 - 3 0 0 11 1 0 0 0 3 = - ( - 1)( - 2)(11)(3) = - 66 . 2. If A , B and C are 2 x 2 matrices such that det( A ) = - 2, det( B ) = 1 2 and (10 points) det( C ) = 3, then compute det( - A T B - 1 C ). Solution. We have det( - A T B - 1 C ) = ( - 1) 2 det( A )(det( B )) - 1 det( C ) = ( - 2)(2)(3) = - 12 . 3. Find a such that the vector ( - 1 , 0 , 2) can be written as a linear combination (10 points) of (2 , - 1 , 3) and ( a - 1 , 1 , 1). Solution. We want to find c 1 and c 2 such that ( - 1 , 0 , 2) = c 1 (2 , - 1 , 3) + c 2 ( a - 1 , 1 , 1) 2 c 1 + ( a - 1) c 2 = - 1 - c 1 + c 2 = 0 3 c 1 + c 2 = 2 Then the last two equations imply c 1 = c 2 = 1 2 . Hence the first equation gives 2 1 2 + 1 2 ( a - 1) = - 1 2 + a - 1 = - 2 a = - 3 .
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