Explanation Given Information: The provided system of equations is: a − b + 3 c = 2 2 a + b + 2 c = 2 − 2 a − 2 b + c = 3 Formula used: Let A be an n × n matrix for which A − 1 exists and let I n be the n × n identity matrix. To find A − 1 , Step 1: Write a matrix of the form [ A | I n ] . Step 2: Perform row operations to write the matrix in the form [ I n | B ] . Step 3: The matrix B is A − 1 . If A X = B represents a system of n linear equations in n variables with a unique solution. Then, X = A − 1 B Where, A is the coefficient matrix, X is the column matrix of variables and B is the column matrix of constants. Calculation: Consider the provided system of equations: a − b + 3 c = 2 2 a + b + 2 c = 2 − 2 a − 2 b + c = 3 Convert this system of equation into matrix equation of the form A X = B as: [ 1 − 1 3 2 1 2 − 2 − 2 1 ] [ a b c ] = [ 2 2 3 ] Now, consider the coefficient matrix: A = [ 1 − 1 3 2 1 2 − 2 − 2 1 ] Now, write the matrix in the form [ A | I 3 ] as: [ A | I 3 ] : [ 1 − 1 3 1 0 0 2 1 2 0 1 0 − 2 − 2 1 0 0 1 ] Now, perform the row operations to transform left side of the matrix to the 3 × 3 identity matrix or to get [ I 3 | B ] as: Apply the operation R 2 − 2 R 1 → R 2 and R 3 + 2 R 1 → R 3 in above matrix and simplify as, [ 1 − 1 3 1 0 0 2 1 2 0 1 0 − 2 − 2 1 0 0 1 ] → R 2 − 2 R 1 → R 2 R 3 + 2 R 1 → R 3 [ 1 − 1 3 1 0 0 0 3 − 4 − 2 1 0 0 − 4 7 2 0 1 ] Apply the operation R 2 3 → R 2 in above matrix and simplify further as, [ 1 − 1 3 1 0 0 0 3 − 4 − 2 1 0 0 − 4 7 2 0 1 ] → R 2 3 → R 2 [ 1 − 1 3 1 0 0 0 1 − 4 3 − 2 3 1 3 0 0 − 4 7 2 0 1 ] Apply the operation R 3 + 4 R 2 → R 3 in above matrix and simplify further as, [ 1 − 1 3 1 0 0 0 1 − 4 3 − 2 3 1 3 0 0 − 4 7 2 0 1 ] → R 3 + 4 R 2 → R 3 [ 1 − 1 3 1 0 0 0 1 − 4 3 − 2 3 1 3 0 0 0 5 3 − 2 3 4 3 1 ] Apply the operation 3 5 R 3 → R 3 in above matrix and simplify further as, [ 1 − 1 3 1 0 0 0 1 − 4 3 − 2 3 1 3 0 0 0

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ISBN: 9781266242946

Author: Miller

Publisher: MCG

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Chapter 6.4, Problem 48PE

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To calculate: The solution of the provided system of equations,

a−b+3c=22a+b+2c=2−2a−2b+c=3

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Chapter 6.4, Problem 47PE

Chapter 6.4, Problem 49PE

Chapter 6 Solutions

COLLEGE ALGEBRA (LL)18 WK ALEKS 360

Ch. 6.1 - Skill Practice 1. a. Write the augmented...Ch. 6.1 - Skill Practice Use the matrix from Example 2 to...Ch. 6.1 - Skill Practice 3. Solve the system by using...Ch. 6.1 - Skill Practice Solve the system by using...Ch. 6.1 - Prob. 5SP Ch. 6.1 - 1. A rectangle array of elements is called a...Ch. 6.1 - Prob. 2PE Ch. 6.1 - Explain the meaning of the notation R2R3.Ch. 6.1 - Explain the meaning of the notation R2R3.Ch. 6.1 - Explain the meaning of the notation 3R1R1.

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The solution of the provided system of equations, a − b + 3 c = 2 2 a + b + 2 c = 2 − 2 a − 2 b + c = 3

Solution Summary: The author explains how to calculate the solution of the provided system of equations.

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To calculate: The solution of the provided system of equations,

a−b+3c=22a+b+2c=2−2a−2b+c=3

Want to see the full answer?

Chapter 6 Solutions

The solution of the provided system of equations, a − b + 3 c = 2 2 a + b + 2 c = 2 − 2 a − 2 b + c = 3

Solution Summary: The author explains how to calculate the solution of the provided system of equations.

Videos

To calculate: The solution of the provided system of equations, a−b+3c=22a+b+2c=2−2a−2b+c=3

Want to see the full answer?

Chapter 6 Solutions

To calculate: The solution of the provided system of equations,

a−b+3c=22a+b+2c=2−2a−2b+c=3