Chapter 5, Problem 18RE

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Chapter
Section

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

# In Exercises 17–20, write the given system of linear equations as a matrix equation, and solve by inverting the coefficient matrix. x + y + z = 3 y + 2 z = 4 y − z = 1

To determine

To calculate: The matrix equation of the system of linear equations and solve by inverting the coefficient matrix

x+y+z=3y+2z=4yz=1

Explanation

Given Information:

The system of linear equations are

x+y+z=3y+2z=4yz=1.

Formula used:

In the matrix equation AX=B, the matrix A has entries equal to the coefficients of the left side of the system of equations and the matrix X is the column matrix of the unknowns in the system and the matrix B is the column matrix and has entries equal to the coefficients of the right side of the system of equations.

AA1=I and A1A=I, where I is the n×n identity matrix.

The two matrices A and B are equal if they have same dimensions, also their corresponding entries are equal.

For matrices A have dimension m×n and B have dimension n×k, the product AB is the matrix of dimension m×k and ijth entry of AB is the sum of product of corresponding entries of the row i of A and the column j of B.

Calculation:

Consider the system of linear equations,

x+y+z=3y+2z=4yz=1

Recall that in the matrix equation AX=B, the matrix A has entries equal to the coefficients of the left side of the system of equations and the matrix X is the column matrix of the unknowns in the system and the matrix B is the column matrix and has entries equal to the coefficients of the right side of the system of equations.

Thus, A=[111012011], X=[xyz] and B=[341].

Hence, the matrix equation is [111012011][xyz]=[341].

Solve the equation for X.

Consider the equation AX=B.

Multiply both sides on the left of the equation by A1.

A1AX=A1B

Recall that AA1=I and A1A=I, where I is the n×n identity matrix.

Substitute I for A1A in A1AX=A1B and simplify.

A1AX=A1BIX=A1BX=A1B

Thus, the equation is [xyz]=[111012011]1[341].

Consider the matrix [111012011].

Recall that a matrix with m rows and n columns is of dimension m×n, where m and n are positive integers.

Since, number of rows in [111012011] is 3 and number of columns is 3,

Substitute 3 for m and 3 for n in m×n.

Thus, dimension of [111012011] is 3×3.

Augment the matrix with the 3×3 identity matrix [100010001].

Thus, the augmented matrix is [111100012010011001].

Row reduce the matrix [111100012010011001].

Perform the row operations R1R2 and R3R2.

[111100012010011001][101110012010003011]

Perform the row operations 3R1R3 and 3R2+2R3.

[101110012010003011][300321030012003011]

Perform the row operations (13)R1, (13)R2 and (13)R3.

[300321030012003011][100123130100132300101313]

Thus, [111012011]1=[123130132301313]

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