Explanation Given information: The system of equations is { 2 x + y + 5 z = 1 3 x + 2 y + 2 z = 1 4 x + 3 z = 1 6 . Concept used: Matrix equation for system of equations: For a system of equations { a x + b y = m c x + d y = n , the corresponding matrix equation is given by, [ a b c d ] [ x y ] = [ m n ] Here, [ a b c d ] is the coefficient matrix, [ x y ] is the variable matrix and [ m n ] is the known matrix. Solution of a matrix equation: If A is a square n × n matrix that has inverse A − 1 and if x is a variable matrix and B is a known matrix both with n rows then the solution of the matrix equation A X = B is, X = A − 1 B ...... (1) Inverse of a n × n matrix: For a n × n matrix, construct a n × 2 n augmented matrix such that it has entries of A on the left and Identity matrix I n on the right. A = [ a 11 a 12 … a 1 n 1 0 … 0 a 21 d … a 2 n 0 1 … 0 ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 … a n n 0 0 … 1 ] Now, use row elementary operation on the above new matrix to change it the left side to identity matrix and hence the right side is the inverse matrix A − 1 . Calculation: The given system of equations can be written in the form of matrix equation as, [ 2 1 5 1 2 2 1 0 3 ] [ x y z ] = [ 1 3 1 4 1 6 ] Let the coefficient matrix [ 2 1 5 1 2 2 1 0 3 ] be A, variable matrix [ x y z ] as X and [ 1 3 1 4 1 6 ] as the known matrix B. Now, to obtain the solution of the matrix equation, first calculate the inverse of the coefficient matrix as shown below. The augmented matrix [ A I ] for the matrix A is constructed as shown below. [ A I ] = [ 2 1 5 1 0 0 1 2 2 0 1 0 1 0 3 0 0 1 ] Apply row-transformation R 2 − 1 2 R 1 → R 2 to the above augmented matrix then the matrix becomes, [ 2 1 5 1 0 0 0 3 2 − 1 2 − 1 2 1 0 1 0 3 0 0 1 ] Now, apply row-transformation R 3 − 1 2 R 1 → R 3 then the matrix becomes, [ 2 1 5 1 0 0 0 3 2 − 1 2 − 1 2 1 0 1 − 1 2 1 2 − 1 2 0 1 ] Now, apply row-transformation R 3 + 1 3 R 2 → R 3 then the matrix becomes, [ 2 1 5 1 0 0 0 3 2 − 1 2 − 1 2 1 0 0 0 1 3 − 2 3 1 3 1 ] Now, apply row-transformation 3R 3 → R 3 then the matrix becomes, [ 2 1 5 1 0 0 0 <

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ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 6, Problem 63E

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The matrix equation for the system of equations {2x+y+5z=13x+2y+2z=14x+3z=16 and then find the solution of the matrix equation.

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Chapter 6, Problem 62E

Chapter 6, Problem 64E

Chapter 6 Solutions

College Algebra

Ch. 6.1 - If a system of linear equations has infinitely...Ch. 6.1 - Write the augmented matrix of the following system...Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Dimension of a Matrix State the dimension of the...Ch. 6.1 - Prob. 10E

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The matrix equation for the system of equations { 2 x + y + 5 z = 1 3 x + 2 y + 2 z = 1 4 x + 3 z = 1 6 and then find the solution of the matrix equation.

Solution Summary: The author describes the matrix equation for the system of equations c2x+y+5z=13

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To find:

The matrix equation for the system of equations {2x+y+5z=13x+2y+2z=14x+3z=16 and then find the solution of the matrix equation.

Want to see the full answer?

Chapter 6 Solutions

The matrix equation for the system of equations { 2 x + y + 5 z = 1 3 x + 2 y + 2 z = 1 4 x + 3 z = 1 6 and then find the solution of the matrix equation.

Solution Summary: The author describes the matrix equation for the system of equations c2x+y+5z=13

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To find: The matrix equation for the system of equations {2x+y+5z=13x+2y+2z=14x+3z=16 and then find the solution of the matrix equation.

Want to see the full answer?

Chapter 6 Solutions

To find:

The matrix equation for the system of equations {2x+y+5z=13x+2y+2z=14x+3z=16 and then find the solution of the matrix equation.