Explanation Given information: The linear system of equations is, { x + y = 1 y + z = 2 z + w = 3 w − x = 4 Concept used : The general linear system of equations can be expressed as, { a 11 x + a 12 y + a 13 z + a 14 w = r a 21 x + a 22 y + a 23 z + a 24 w = s a 31 x + a 32 y + a 33 z + a 34 w = t a 41 x + a 42 y + a 43 z + a 44 w = u The solution of the above system of equation can be calculated as, x = | D x | | D | , y = | D y | | D | , z = | D z | | D | and w = | D w | | D | Here, the determinant is | D | = | a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 | ≠ 0 The matrices D , D x , D y , D z and D w are represented as follows, D = ( a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 ) , D x = ( r a 12 a 13 a 14 s a 22 a 23 a 24 t a 32 a 33 a 34 u a 42 a 43 a 44 ) D y = ( a 11 r a 13 a 14 a 21 s a 23 a 24 a 31 t a 33 a 34 a 41 u a 43 a 44 ) , D z = ( a 11 a 12 r a 14 a 21 a 22 s a 24 a 31 a 32 t a 34 a 41 a 42 u a 44 ) and D w = ( a 11 a 12 a 13 r a 21 a 22 a 23 s a 31 a 32 a 33 t a 41 a 42 a 43 u ) Calculation: The equations are { x + y = 1 y + z = 2 z + w = 3 w − x = 4 . For the system of equations, the coefficient matrix can be obtained as below, D = ( 1 1 0 0 0 1 1 0 0 0 1 1 − 1 0 0 1 ) The determinant of the coefficient matrix D can be calculated as, D = ( 1 1 0 0 0 1 1 0 0 0 1 1 − 1 0 0 1 ) | D | = 1 | 1 1 0 0 1 1 0 0 1 | − 1 | 0 1 0 0 1 1 − 1 0 1 | + 0 | 0 1 0 0 0 1 − 1 0 1 | − 0 | 0 1 1 0 0 1 − 1 0 0 | The determinant of the matrix leaving the zero cofactors is to be calculated as, | D | = 1 [ 1 ( 1 ⋅ 1 − 0 ⋅ 1 ) − 1 ( 0 ⋅ 1 − 0 ⋅ 1 ) ] − 1 [ − 1 ( 0 ⋅ 1 − ( − 1 ) ⋅ 1 ) ] + 0 + 0 | D | = 1 ( 1 ) − 1 ( − 1 ) | D | = 2 The matrix D x can be obtained by replacing the first column of the matrix by the values of the equations or the constants 1 , 2 , 3 , 4 as, D x = ( 1 1 0 0 2 1 1 0 3 0 1 1 4 0 0 1 ) To calculate the determinant of D x , expand the determinant by use of the first row as the first row contains two of the elements equals to zero. The determinant of the coefficient matrix D x can be calculated as, | D x | = | 1 1 0 0 2 1 1 0 3 0 1 1 4 0 0 1 | | D x | = 1 ⋅ A 11 − 1 ⋅ A 12 + 0 ⋅ A 13 − 0 ⋅ A 14 The determinant of the cofactor matrix A 11 , A 12 is calculated as, | D x | = 1 ⋅ | 1 1 0 0 1 1 0 0 1 | − 1 ⋅ | 2 1 0 3 1 1 4 0 1 | | D x | = 1 ( 1 ( 1 ⋅ 1 − 0 ⋅ 1 ) − 1 ( 0 ⋅ 1 − 0 ⋅ 1 ) + 0 ) − 1 ( 2 ( 1 ⋅ 1 − 0 ⋅ 1 ) − 1 ( 3 ⋅ 1 − 1 ⋅ 4 ) + 0 ) | D x | = 1 ( 1 ) − 1 ( 2 + 1 ) | D x | = − 2 The matrix D y can be obtained by replacing the second column of the coefficient matrix by the values of the equations or the constants 1 , 2 , 3 , 4 as, D y = ( 1 1 0 0 0 2 1 0 0 3 1 1 − 1 4 0 1 ) To calculate the determinant of D y , expand the determinant by use of the first row as the first row contains all except two of the elements is zero

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 6.4, Problem 56E

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The system of equations {x+y=1y+z=2z+w=3w−x=4 by the Cramer’s rule.

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Chapter 6.4, Problem 55E

Chapter 6.4, Problem 57E

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The system of equations { x + y = 1 y + z = 2 z + w = 3 w − x = 4 by the Cramer’s rule.

Solution Summary: The author explains the Cramer's rule to solve the linear system of equations: x=-1,y=2,z=0andw=3.

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To solve:

The system of equations {x+y=1y+z=2z+w=3w−x=4 by the Cramer’s rule.

Want to see the full answer?

Chapter 6 Solutions

The system of equations { x + y = 1 y + z = 2 z + w = 3 w − x = 4 by the Cramer’s rule.

Solution Summary: The author explains the Cramer's rule to solve the linear system of equations: x=-1,y=2,z=0andw=3.

Videos

To solve: The system of equations {x+y=1y+z=2z+w=3w−x=4 by the Cramer’s rule.

Want to see the full answer?

Chapter 6 Solutions

To solve:

The system of equations {x+y=1y+z=2z+w=3w−x=4 by the Cramer’s rule.