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MAT1332W10A5S

# MAT1332W10A5S - 1 MAT1332 Assignment#5 solutions Question 1...

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1 MAT1332 Assignment #5 solutions Question 1 Determine the solution of the following systems : a ) x + 2 y + z = 2 x + 3 y + 2 z = 5 2 x + 9 y + 7 z = 1 Solution : The augmented matrix associated to this system is 1 2 1 2 1 3 2 5 2 9 7 1 . Replace the second row R2 with R2 - R1, and replace the third row R3 with R3 - 2R1 (2R1 means 2 times row 1) : 1 2 1 2 0 1 1 3 0 5 5 - 3 . Replace the third row R3 with R3 - 5R2 : 1 2 1 2 0 1 1 3 0 0 0 - 15 . The bottom row, when expressed as equations, says 0 = - 15. Since that never holds, the original system of equations has no solution. b ) 2 x + 2 y + z = 6 x + 3 y + 2 z = 18 2 x + 2 y + 7 z = 12 Solution : The augmented matrix associated to this system is 2 2 1 6 1 3 2 18 2 2 7 12 . For convenience, we switch the first two rows R1 R2. This way the leading entry in the first row is a 1 instead of a 2. (It is also correct to avoid this step.) 1 3 2 18 2 2 1 6 2 2 7 12 .

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2 Replace R2 with R2 - 2R1 and R3 with R3 - 2R1, to clear our the rest of the third column : 1 3 2 18 0 - 4 - 3 - 30 0 - 4 3 - 24 . The leading entry in the second row is in the second column. We next clear out the entries below it, by replacing R3 with R3 - R2 : 1 3 2 18 0 - 4 - 3 - 30 0 0 6 6 . Our matrix is now in row-echelon form. We now continue through to reduced row-echelon form. We first make all leading entries equal to one, by scaling each row. We leave R1 alone, multiply R2 by - 1 / 4, and multiply R3 by 1 / 6, to get : 1 3 2 18 0 1 3 / 4 30 / 4 0 0 1 1 .
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