MATH311ansHW6

MATH311ansHW6 - Math 311, Homework 6 partial answers and...

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Unformatted text preview: Math 311, Homework 6 partial answers and solutions 3.2.7. (optional) Subspace, since s x y z x + t a b c a = sx + ta sy + tb sz + tc sx + ta . 3.2.8. Answer: subspace. Detailed solution. The subset S is the collection of all matrices of the form x y z- x , for all possible choices of numbers x, y, z . To check it S is a subspace, we need to check if any linear combination of any two elements of S is again in S , in other words if any such linear combination still satisfies the property the first entry in the first row is equal to minus the second entry in the second row. Two typical elements of S have the form x 1 y 1 z 1- x 1 and x 2 y 2 z 2- x 2 , where x 1 , y 1 , z 1 , x 2 , y 2 , z 2 are arbitrary numbers. For arbitrary constants s, t , s x 1 y 1 z 1- x 1 + t x 2 y 2 z 2- x 2 = sx 1 sy 1 sz 1- sx 1 + tx 2 ty 2 tz 2- tx 2 = sx 1 + tx 2 sy 1 + ty 2 sz 1 + tz 2- sx 1- tx 2 and- sx 1- tx 2 =- ( sx 1 + tx 2 ) , so this matrix is again of the same form, and belongs to S . As a result, S is a subspace. We could also have used the same argument in two steps: x 1 y 1 z 1- x 1 + x 2 y 2 z 2- x 2 = x 1 + x 2 y 1 + y 2 z 1 + z 2- x 1- x 2 with- x 1- x 2 =- ( x 1 + x 2 ) , and s x 1 y 1 z 1- x 1 = sx 1 sy 1 sz 1- sx 1 with- sx 1 =- ( sx 1...
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MATH311ansHW6 - Math 311, Homework 6 partial answers and...

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