# solq2 - x 2 x 3 x 4 x 5 = changes to x 1 x 2 x 3 x 4 x 5 =...

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Solutions to quiz # 2 (September 16) 1. Writing the augmented matrix of the system and then bringing it to the reduced row-echelon form, we obtain 1 0 2 2 | 7 2 1 5 5 | 19 1 0 2 3 | 10 -→ 1 0 2 2 | 7 0 1 1 1 | 5 0 0 0 1 | 3 -→ 1 0 2 0 | 1 0 1 1 0 | 2 0 0 0 1 | 3 . Hence x 3 is a free variable, x 1 + 2 x 3 = 1, x 2 + x 3 = 2, and x 4 = 3, from which we obtain Answer: x 1 = 1 - 2 t , x 2 = 2 - t , x 3 = t , and x 4 = 3, where t can be any number. 2. Let us reduce the matrix of the original system to the reduced row-echelon form. Since there are inFnitely many solutions, there has to be a free variable. When I change the right hand side of the system, the reduced row echelon form of the coe±cient matrix of the system doesn’t change, so the new system has at least one free variable. Therefore, the new system cannot have a unique solution. It can have no solutions, for example, when x 1 + x 2 + x 3 + x 4 + x 5 = 0 x 1 + x 2 + x 3 + x 4 + x 5 = 0 x 1 + x 2 + x 3 + x 4 + x 5 = 0 x 1 + x 2 + x 3 + x 4 + x 5 = 0 x 1 +
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Unformatted text preview: x 2 + x 3 + x 4 + x 5 = changes to x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = 1 or it can have inFnitely many solutions, for example, when x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = x 1 + x 2 + x 3 + x 4 + x 5 = changes to x 1 + x 2 + x 3 + x 4 + x 5 = 1 x 1 + x 2 + x 3 + x 4 + x 5 = 1 x 1 + x 2 + x 3 + x 4 + x 5 = 1 x 1 + x 2 + x 3 + x 4 + x 5 = 1 x 1 + x 2 + x 3 + x 4 + x 5 = 1 . Answer: The new system can have no solutions or inFnitely many solutions. 1...
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