Explanation Approach: A system of linear equations can be written in the form of A X = B where A is called the coefficient matrix, B is called the known matrix, and X is called the variable matrix. If a square matrix A of dimension 3 × 3 has an 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 given by: X = A − 1 B . Take a 4 × 8 matrix with left half as matrix B and right half as I 4 and transform the left half into identity matrix by performing elementary row operations on the entire matrix, the right half becomes inverse of the matrix B . Calculation: The system of linear equations { x + 2 y + 3 w = 0 y + z + w = 1 y + w = 2 x + 2 y + 2 w = 3 can be written as a single matrix equation A X = B as follows. [ 1 2 0 3 0 1 1 1 0 1 0 1 1 2 0 2 ] [ x y z w ] = [ 0 1 2 3 ] Here A = [ 1 2 0 3 0 1 1 1 0 1 0 1 1 2 0 2 ] , B = [ 0 1 2 3 ] , and X = [ x y z w ] . The solution of this matrix equation is given by X = A − 1 B . First find A − 1 , Take a matrix of dimension 4 × 8 with left half as matrix A and right half as I 4 and perform row operations to transform the left half into I 4 . [ 1 2 0 3 1 0 0 0 0 1 1 1 0 1 0 0 0 1 0 1 0 0 1 0 1 2 0 2 0 0 0 1 ] R 4 → R 4 − R 1 → [ 1 2 0 3 1 0 0 0 0 1 1 1 0 1 0 0 0 1 0 1 0 0 1 0 0 0 0 − 1 − 1 0 0 1 ] R 3 → R 3 − R 2 → [ 1 2 0 3 1 0 0 0 0 1 1 1 0 1 0 0 0 0 − 1 0 0 − 1 1 0 0 0 0 − 1 − 1 0 0 1 ] R 1 → R 1 − 2 R 2 → [ 1 0 − 2 1 1 − 2 0 0 0 1 1 1 0 1 0 0 0 0 − 1 0 0 − 1 1 0 0 0 0 − 1 − 1 0 0 1 ] R 3 → − R 3 → [ 1 0 − 2 1 1 − 2 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 − 1 0 0 0 0 − 1 − 1 0 0 1 ] R 2 → R 2 − R 3 → [

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ISBN: 8220100548512

Author: Watson

Publisher: CENGAGE L

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Chapter 11.3, Problem 46E

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The system of equations {x+2y+3w=0y+z+w=1y+w=2x+2y+2w=3 by converting it into a matrix equation and find the values of x, y, z, and w.

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Chapter 11.3, Problem 45E

Chapter 11.3, Problem 47E

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Chapter 11 Solutions

EBK ALGEBRA AND TRIGONOMETRY

Ch. 11.1 - If a system of linear equations has infinitely...Ch. 11.1 - Write the augmented matrix of the following system...Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - 5-10 Dimension of a Matrix State the dimension of...Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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The system of equations { x + 2 y + 3 w = 0 y + z + w = 1 y + w = 2 x + 2 y + 2 w = 3 by converting it into a matrix equation and find the values of x , y , z , and w .

Solution Summary: The author explains how the system of linear equations cx+2y+3w is written as a single matrix equation AX=B.

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To solve:

The system of equations {x+2y+3w=0y+z+w=1y+w=2x+2y+2w=3 by converting it into a matrix equation and find the values of x, y, z, and w.

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Chapter 11 Solutions

The system of equations { x + 2 y + 3 w = 0 y + z + w = 1 y + w = 2 x + 2 y + 2 w = 3 by converting it into a matrix equation and find the values of x , y , z , and w .

Solution Summary: The author explains how the system of linear equations cx+2y+3w is written as a single matrix equation AX=B.

Videos

To solve: The system of equations {x+2y+3w=0y+z+w=1y+w=2x+2y+2w=3 by converting it into a matrix equation and find the values of x, y, z, and w.

Want to see the full answer?

Chapter 11 Solutions

To solve:

The system of equations {x+2y+3w=0y+z+w=1y+w=2x+2y+2w=3 by converting it into a matrix equation and find the values of x, y, z, and w.