Algebra 2Algebra 2

Inverse Matrices and SystemsInverse Matrices and Systems

Algebra 2Algebra 2

Write the system

as a matrix equation.

Lesson 4-7

Inverse Matrices and SystemsInverse Matrices and Systems

–3x – 4y + 5z = 11–2x + 7y = –6–5x + y – z = 20

Then identify the coefficient matrix, the variable matrix, and the constant matrix.

Matrix equation: =–3 –4 5–2 7 0–5 1 –1

xyz

11–6 20

Coefficient matrix

–3 –4 5–2 7 0–5 1 –1

xyz

Variable matrix

11–6 20

Constant matrix

Additional Examples

Algebra 2Algebra 2

Solve the system.

Lesson 4-7

Inverse Matrices and SystemsInverse Matrices and Systems

2x + 3y = –1 x – y = 12

2 31 –1

xy

–112

= Write the system as a matrix equation.

= A–1B = = Solve for the variable matrix.

xy

15

35

15

25

–

–112

7–5

Additional Examples

Algebra 2Algebra 2

(continued)

Lesson 4-7

Inverse Matrices and SystemsInverse Matrices and Systems

The solution of the system is (7, –5).

Check: 2x + 3y = –1 x – y = 12 Use theoriginalequations.

2(7) + 3(–5) –1 (7) – (–5) 12 Substitute.

14 – 15 = –1 7 + 5 = 12 Simplify.

Additional Examples

Algebra 2Algebra 2

Solve the system .

Step 1: Write the system asa matrix equation.

Lesson 4-7

Inverse Matrices and SystemsInverse Matrices and Systems

7x + 3y + 2z = 13–2x + y – 8z = 26 x – 4y +10z = –13

Step 2: Store the coefficientmatrix as matrix Aand the constantmatrix as matrix B. 7 3 2

–2 1 –8 1 –4 10

13 26–13

xyz

=

The solution is (9, –12, –7).

Additional Examples