To determine To calculate: Use the inverse of the coefficient matrix to solve each system of equation. Explanation Given: The system of equation is − x + y − 3 z = 3 2 x + 4 y − 4 z = 6 − x + y + 4 z = − 1 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 − 3 z = 3 2 x + 4 y − 4 z = 6 − x + y + 4 z = − 1 Now create a 3 × 3 matrix with the coefficients of the variable provided in the system of equation. Therefore, A = [ − 1 1 − 3 2 4 − 4 − 1 1 4 ] Similarly, form a matrix 3 × 1 of the constants written on the right-hand side of the provided system of equations. Therefore, B = [ 3 6 − 1 ] Now rewrite the equation as, A X = B Where X = [ x y z ] So, X = A − 1 B . Now to solve this matrix equation, calculate the inverse of the matrix A . As, A A − 1 = Ι , So, [ − 1 1 − 3 2 4 − 4 − 1 1 4 ] A − 1 = [ 1 0 0 0 1 0 0 0 1 ] Use the transformation R 3 → R 3 − R 1 , [ − 1 1 − 3 2 4 − 4 − 1 − ( − 1 ) 1 − 1 4 − ( − 3 ) ] A − 1 = [ 1 0 0 0 1 0 0 − 1 0 − 0 1 − 0 ] [ − 1 1 − 3 2 4 − 4 − 1 + 1 0 4 + 3 ] A − 1 = [ 1 0 0 0 1 0 − 1 0 1 ] [ − 1 1 − 3 2 4 − 4 0 0 7 ] A − 1 = [ 1 0 0 0 1 0 − 1 0 1 ] Use the transformation R 2 → R 2 + R 1 , [ − 1 1 − 3 2 − 1 4 + 1 − 4 − 3 0 0 7 ] A − 1 = [ 1 0 0 0 + 1 1 + 0 0 + 0 − 1 0 1 ] [ − 1 1 − 3 1 5 − 7 0 0 7 ] A − 1 = [ 1 0 0 1 1 0 − 1 0 1 ] Use the transformation R 2 → R 2 + R 3 , [ − 1 1 − 3 1 + 0 5 + 0 − 7 + 7 0 0 7 ] A − 1 = [ 1 0 0 1 − 1 1 + 0 0 + 1 − 1 0 1 ] [ − 1 1 − 3 1 5 0 0 0 7 ] A − 1 = [ 1 0 0 0 1 1 − 1 0 1 ] Use the transformation R 1 → 7 R 1 + 3 R 3 [ − 7 + 0 7 + 0 − 21 + 21 1 5 0 0 0 7 ] A − 1 = [ 7 − 3 0 + 0 0 + 3 0 1 1 − 1 0 1 ] [ − 7 7 0 1 5 0 0 0 7 ] A − 1 = [ 4 0 3 0 1 1 − 1 0

Mathematics with Applications In the Management, Natural and Social Sciences (11th Edition)

11th Edition

ISBN: 9780321931078

Author: Margaret L. Lial, Thomas W. Hungerford, John P. Holcomb, Bernadette Mullins

Publisher: PEARSON

See similar textbooks

Concept explainers

Matrices

‘Matrices’ is the plural form of the word matrix, and it is basically a spreadsheet in the form of a box. In mathematics, various functions can be carried out with matrices. Generally, a matrix comes in the shape of a square or rectangle. The elements ar…

Videos

Question

Chapter 6.6, Problem 8E

To determine

To calculate: Use the inverse of the coefficient matrix to solve each system of equation.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6.6, Problem 7E

Chapter 6.6, Problem 9E

Students have asked these similar questions

Assume {u1, U2, u3, u4} does not span R³. Select the best statement. A. {u1, U2, u3} spans R³ if u̸4 is a linear combination of other vectors in the set. B. We do not have sufficient information to determine whether {u₁, u2, u3} spans R³. C. {U1, U2, u3} spans R³ if u̸4 is a scalar multiple of another vector in the set. D. {u1, U2, u3} cannot span R³. E. {U1, U2, u3} spans R³ if u̸4 is the zero vector. F. none of the above

Select the best statement. A. If a set of vectors includes the zero vector 0, then the set of vectors can span R^ as long as the other vectors are distinct. n B. If a set of vectors includes the zero vector 0, then the set of vectors spans R precisely when the set with 0 excluded spans Rª. ○ C. If a set of vectors includes the zero vector 0, then the set of vectors can span Rn as long as it contains n vectors. ○ D. If a set of vectors includes the zero vector 0, then there is no reasonable way to determine if the set of vectors spans Rn. E. If a set of vectors includes the zero vector 0, then the set of vectors cannot span Rn. F. none of the above

Chapter 6 Solutions

Mathematics with Applications In the Management, Natural and Social Sciences (11th Edition)

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.