Explanation Given Information: The system of equation is, x − y + z − u + v = 0 y − z + u − v = − 2 x − 2 v = − 2 2 x − y + z − u − 3 v = − 2 And, 4 x − y + z − u − 7 v = − 6 Formula Used: Elementary row operations Type 1: Replacing the row R i by a R i , where a is a nonzero number. Type 2: Replacing the row R i by a R i ± b R j , where a is a nonzero number. Gauss Jordan reduction method Step 1: First clear the fractions or decimals if any, using operations of type 1. Step 2: Select the first nonzero element of the second row as pivot. Step 3: Use the pivot to clear its column using operations of type 2. Step 4: Select the first nonzero element in the third row a pivot and clear its column. Step 5: Select the first nonzero element in the fourth row a pivot and clear its column. Step 6: Select the first nonzero element in the fifth row a pivot and clear its column. Calculation: Consider the system of equation, x − y + z − u + v = 0 y − z + u − v = − 2 x − 2 v = − 2 2 x − y + z − u − 3 v = − 2 And, 4 x − y + z − u − 7 v = − 6 The augmented matrix for the given system of equations is, [ 1 − 1 1 − 1 1 0 0 1 2 4 1 0 − 1 − 1 − 1 0 1 1 1 0 − 1 − 1 − 1 − 2 − 3 − 7 − 2 − 2 − 2 − 6 ] Apply Gauss Jordan reduction method to get the solution of the given system of equation. Begin by, the selection of the pivot as the first nonzero element of the first row and clear its column. Perform the operation R 3 → R 3 − R 1 and R 4 → R 4 − 2 R 1 [ 1 − 1 1 − 1 1 0 0 1 2 4 1 0 − 1 − 1 − 1 0 1 1 1 0 − 1 − 1 − 1 − 2 − 3 − 7 − 2 − 2 − 2 − 6 ] → [ 1 − 1 1 − 1 1 0 0 0 0 4 1 1 1 − 1 − 1 − 1 − 1 1 1 1 1 − 1 − 1 − 3 − 5 − 7 − 2 − 2 − 2 − 6 ] Next clear the leftover element of that column. Perform the operation R 5 → R 5 − 4 R 1 [ 1 − 1 1 − 1 1 0 0 0 0 4 1 1 1 − 1 − 1 − 1 − 1 1 1 1 1 − 1 − 1 − 3 − 5 − 7 − 2 − 2 − 2 − 6 ] → [ 1 − 1 1 − 1 1 0 0 0 0 0 1 1 1 3 − 1 − 1 − 1 − 3 1 1 1 3 − 1 − 3 − 5 − 11 − 2 − 2 − 2 − 6 ] Next pivot the first nonzero element of the second row and clear its column. Perform the operation R 1 → R 1 + R 2 and R 3 → R 3 − R 2 , [ 1 − 1 1 − 1 1 0 0 0 0 0 1 1 1 3 − 1 − 1 − 1 − 3 1 1 1 3 − 1 − 3 − 5 − 11 − 2 − 2 − 2 − 6 ] → [ 1 0 0 0 0 − 2 0 0 0 0 1 0 1 3 − 1 0 − 1 − 3 1 0 1 3 − 1 − 2 − 5 − 11 − 2 0 − 2 − 6 ] Next clear the leftover element of that column. Perform the operation R 4 → R 4 − R 2 and R 5 → R 5 − 3 R 2 [ 1 0 0 0 − 2 − 9 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 2 1 − 1 1 12 12 − 1 5 ] → [ 1 0 0 0 − 2 − 9 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 2 2 − 1 1 12 13 − 1 5 ] Next simplify the third and fourth row

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Chapter 3.2, Problem 41E

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To calculate: The solution of the given system of equations x−y+z−u+v=0,y−z+u−v=−2,x−2v=−2,2x−y+z−u−3v=−2,4x−y+z−u−7v=−6 by the use of Gauss Jordan row reduction.

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Chapter 3.2, Problem 40E

Chapter 3.2, Problem 42E

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The solution of the given system of equations x − y + z − u + v = 0 , y − z + u − v = − 2 , x − 2 v = − 2 , 2 x − y + z − u − 3 v = − 2 , 4 x − y + z − u − 7 v = − 6 by the use of Gauss Jordan row reduction.

Solution Summary: The author calculates the solution of the given system of equations by the use of Gauss Jordan row reduction.

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To calculate: The solution of the given system of equations x−y+z−u+v=0,y−z+u−v=−2,x−2v=−2,2x−y+z−u−3v=−2,4x−y+z−u−7v=−6 by the use of Gauss Jordan row reduction.

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