Explanation Given: The following system of equation: x + y + 2 w = 3 2 x − y + z − w = 3 3 x + 3 y + 2 z − 2 w = 5 x + 2 y + z = 3 Formula used A system of equations with the same number of equations as variables is written in matrix form as AX = B, where A is the square matrix of coefficients, X is the column matrix of variables, and B is the column matrix of constants. If A has an inverse, then the unique solution of the system is X = A -1 B. Calculation: Consider the provided system of equations: x + y + 2 w = 3 2 x − y + z − w = 3 3 x + 3 y + 2 z − 2 w = 5 x + 2 y + z = 3 Now create a 4 × 4 matrix with the coefficients of the variable provided in the system of equation. Such that: A = [ 1 1 0 2 2 − 1 1 − 1 3 3 2 − 2 1 2 1 0 ] Similarly, form a 4 × 1 matrix of the constants written on the right hand side of the provided system of equations. Such that: B = [ 3 3 5 3 ] Now rewrite the equation as of the form of: A X = B Where, X = [ x y z w ] X = A − 1 B To solve this matrix equation, need to calculate the inverse of the matrix A i.e. Calculate A − 1 Now put the values of A and B in the above equation. Such that: A A − 1 = Ι Now, [ 1 1 0 2 2 − 1 1 − 1 3 3 2 − 2 1 2 1 0 ] A − 1 = [ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] Now replace: R 2 → R 2 + R 1 [ 1 1 0 2 2 + 1 − 1 + 1 1 + 0 − 1 + 2 3 3 2 − 2 1 2 1 0 ] A − 1 = [ 1 0 0 0 0 + 1 1 + 0 0 + 0 0 + 0 0 0 1 0 0 0 0 1 ] [ 1 1 0 2 3 0 1 1 3 3 2 − 2 1 2 1 0 ] A − 1 = [ 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 ] Now replace: R 2 → R 2 − R 3 [ 1 1 0 2 3 − 3 0 − 3 1 − 2 1 + 2 3 3 2 − 2 1 2 1 0 ] A − 1 = [ 1 0 0 0 1 − 0 1 − 0 0 − 1 0 − 0 0 0 1 0 0 0 0 1 ] [ 1 1 0 2 0 − 3 − 1 3 3 3 2 − 2 1 2 1 0 ] A − 1 = [ 1 0 0 0 1 1 − 1 0 0 0 1 0 0 0 0 1 ] Now replace: R 4 → R 4 − R 1 [ 1 1 0 2 0 − 3 − 1 3 3 3 2 − 2 1 − 1 2 − 1 1 − 0 0 − 2 ] A − 1 = [ 1 0 0 0 1 1 − 1 0 0 0 1 0 0 − 1 0 − 0 0 − 0 1 − 0 ] [ 1 1 0 2 0 − 3 − 1 3 3 3 2 − 2 0 1 1 − 2 ] A − 1 = [ 1 0 0 0 1 1 − 1 0 0 0 1 0 − 1 0 0 1 ] Now replace: R 2 → R 2 + R 4 [ 1 1 0 2 0 + 0 − 3 + 1 − 1 + 1 3 − 2 3 3 2 − 2 0 1 1 − 2 ] A − 1 = [ 1 0 0 0 1 − 1 1 + 0 − 1 + 0 0 + 1 0 0 1 0 − 1 0 0 1 ] [ 1 1 0 2 0 − 2 0 1 3 3 2 − 2 0 1 1 − 2 ] A − 1 = [ 1 0 0 0 0 1 − 1 1 0 0 1 0 − 1 0 0 1 ] Now replace: R 3 → R 3 − R 4 [ 1 1 0 2 0 − 2 0 1 3 − 0 3 − 1 2 − 1 − 2 + 2 0 1 1 − 2 ] A − 1 = [ 1 0 0 0 0 1 − 1 1 0 + 1 0 − 0 1 − 0 0 − 1 − 1 0 0 1 ] [ 1 1 0 2 0 − 2 0 1 3 2 1 0 0 1 1 − 2 ] A − 1 = [ 1 0 0 0 0 1 − 1 1 1 0 1 − 1 − 1 0 0 1 ] Replace: R 1 → R 1 − 2 R 2 [ 1 − 0 1 + 4 0 − 0 2 − 2 0 − 2 0 1 3 2 1 0 0 1 1 − 2 ] A − 1 = [ 1 − 0 0 − 2 0 + 2 0 − 2 0 1 − 1 1 1 0 1 − 1 − 1 0 0 1 ] [ 1 5 0 0 0 − 2 0 1 3 2 1 0 0 1 1 − 2 ] A − 1 = [ 1 − 2 2 − 2 0 1 − 1 1 1 0 1 − 1 − 1 0 0 1 ] Replace: R 3 → R 3 − 3 R 1 [ 1 5 0 0 0 − 2 0 1 3 − 3 2 − 15 1 − 0 0 − 0 0 1 1 − 2 ] A − 1 = [ 1 − 2 2 − 2 0 1 − 1 1 1 − 3 0 + 6 1 − 6 − 1 + 6 − 1 0 0 1 ] [ 1 5 0 0 0 − 2 0 1 0 − 13 1 0 0 1 1 − 2 ] A − 1 = [ 1 − 2 2 − 2 0 1 − 1 1 − 2 6 − 5 5 − 1 0 0 1 ] Replace: R 4 → R 4 − R 3 [ 1 5 0 0 0 − 2 0 1 0 − 13 1 0 0 − 0 1 + 13 1 − 1 − 2 − 0 ] A − 1 = [ 1 − 2 2 − 2 0 1 − 1 1 − 2 6 − 5 5 − 1 + 2 0 − 6 0 + 5 1 − 5 ] [ 1 5 0 0 0 − 2 0 1 0 − 13 1 0 0 14 0 − 2 ] A − 1 = [ 1 − 2 2 − 2 0 1 −

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

11th Edition

ISBN: 9781323751671

Author: Lial

Publisher: PEARSON C

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Chapter 6.6, Problem 13E

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To calculate: The solution of each system using the inverse of the coefficient matrix.

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Chapter 6.6, Problem 12E

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Chapter 6 Solutions

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

Ch. 6.1 - Checkpoint1 Use the substitution method tosolve...Ch. 6.1 - Checkpoint 2 Use the elimination method to solve...Ch. 6.1 - Checkpoint 3 Solve the system of equations . Draw...Ch. 6.1 - Checkpoint 4 Solve the following system: Ch. 6.1 - Checkpoint 5 Solve the system Draw the graph of...Ch. 6.1 - Prob. 6CP Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Use substitution to solve each system. (See...Ch. 6.1 - Use substitution to solve each system. (See...

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