SOLUTION: You have the following matrix, and perform the following 3 row operations. What is the resultant matrix after performing these 3 row operations? –R1 + R2 &#61672; R2 2R1 +

Algebra ->  Algebra  -> Matrices-and-determiminant -> SOLUTION: You have the following matrix, and perform the following 3 row operations. What is the resultant matrix after performing these 3 row operations? –R1 + R2 &#61672; R2 2R1 +      Log On

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 Question 622427: You have the following matrix, and perform the following 3 row operations. What is the resultant matrix after performing these 3 row operations? –R1 + R2  R2 2R1 + R3  R3 -4R2 + R3  R3 1 0 0 10 1 1 3 5 -2 2 0 4 Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website! –R1 + R2 -> R2 2R1 + R3 -> R3 -4R2 + R3 -> R3 ```[ 1 0 0 10] [ 1 1 3 5] [-2 2 0 4] R1 = the first (top) row = [ 1 0 0 10] R2 = the second (middle) row = [ 1 1 3 5] R3 = the third (bottom) row = [-2 2 0 4] First row operation: –R1 + R2 -> R2 Let's do the left side -R1 + R2 first. Substitute [ 1 0 0 10] for R1, [ 1 1 3 5] for R2 –R1 + R2 = -[ 1 0 0 10] + [ 1 1 3 5] = (distribute the - sign) [-1 0 0 -10] + [ 1 1 3 5] = (just combine corresponding elements) [ 0 1 3 -5] The " -> R2 " tells us to replace R2 (the 2nd row) by [ 0 1 3 -5], so the new matrix is like the old one with the middle row replaced by what we got: [ 1 0 0 10] [ 0 1 3 -5] [-2 2 0 4] Now: R1 = the first (top) row = [ 1 0 0 10] R2 = the second (middle) row = [ 0 1 3 -5] R3 = the third (bottom) row = [-2 2 0 4] Next row operation: 2R1 + R3 -> R3 As before, let's do the left side 2R1 + R3 first. Substitute [ 1 0 0 10] for R1, [-2 2 0 4] for R3 2R1 + R3 = 2[ 1 0 0 10] + [-2 2 0 4] = (distribute the 2) = [ 2 0 0 20] + [-2 2 0 4] = (combine corresponding elements) = [ 0 2 0 24] The " -> R3 " tells us to replace R3 (the 3rd row) by [ 0 2 0 24], so the new matrix is like the previous one with the bottom row replaced by what we got: [ 1 0 0 10] [ 0 1 3 -5] [ 0 2 0 24] Edwin```