# In Problems 41-43, solve each system of equations using inverse matrices. { x + 3 y + z = 0 x + 4 y + 3 z = 2 2 x − y − 11 z = − 12

### Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
Publisher: Cengage Learning
ISBN: 9781337625340

Chapter
Section

### Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
Publisher: Cengage Learning
ISBN: 9781337625340
Chapter 3, Problem 43RE
Textbook Problem
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## In Problems 41-43, solve each system of equations using inverse matrices. {     x +       3 y +     z = 0       x +     4 y + 3 z = 2       2 x −   y −     11 z = − 12

To determine

The solution of system of linear equations using inverse matrix.

### Explanation of Solution

Given information:

{x+3y+z=0x+4y+3z=22xy11z=12

Formula used:

Solve the system of equations using inverse matrix.

AX=B

A1(AX)=A1(B).

X=A1(B).

Calculation:

From the system of equations, the associated matrix equation is AX=B.

It can be written as [1311432111][xyz]=[0212].

Where A=[1311432111],B=[0212].

To solve the given system of linear equations, first find the inverse of the coefficient matrix.

From the system of linear equations, the coefficient matrix is,

A=[1311432111].

Find the inverse of the coefficient matrix A by proceeding as follows:

[1311432111|100010001].

Then, perform elementary row operations such that left side matrix becomes an identity matrix and the inverse of the matrix will be in augmented part.

The procedure is as follows,

The entry in row 1 column 1 is 1.

To get 0 in row 2 and column 1, add 1 times row 1 to row 2, and put the result in row 2.

1R1+R2R2[131(1)×1+1(1)×3+4(1)×1+32111|100(1)×1+0(1)×0+1(1)×0+0001]

=[1310122111|100110001].

To get 0 in row 3 and column 1, add 2 times row 1 to row 3, and put the result in row 3.

2R1+R3R3[131012(2)×1+2(2)×3+(1)(2)×1+(11)|100110(2)×1+0(2)×0+0(2)×0+1]

=[1310120713|100110201].

To get 0 in row 3 and column 2, add 7 times row 2 to row 3, and put the result in row 3.

7R2+R3R3[1310127×0+07×1+(7)7×2+(13)|1001107×(1)+(2)7×1+07×0+1]

=[131012001|100110971]

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