1. Inverse Matrices and Systems

2. Inverse Matrices and Systems
Lesson 4-7
Additional Examples
–3x – 4y + 5z = 11
–2x + 7y = –6
–5x + y – z = 20
Write the system
as a matrix equation.
Then identify the coefficient matrix, the variable matrix, and the constant matrix.
Matrix equation: =
–3 – –
– –1 x y z 11 –6 20
Coefficient matrix
–3 – –
– –1 x y z
Variable matrix 11 –6 20
Constant matrix

3. Inverse Matrices and Systems
Lesson 4-7
Additional Examples
Solve the system.
2x + 3y = –1
x – y = 12
1 –1 x y –1 12
= Write the system as a matrix equation.
= A–1B = = Solve for the
variable matrix. x y 1 5 3 2 – –1 12 7 –5

4. Inverse Matrices and Systems
Lesson 4-7
Additional Examples
(continued)
The solution of the system is (7, –5).
Check: x + 3y = – x – y = 12 Use the
original
equations.
2(7) + 3(–5) – (7) – (–5) Substitute.
14 – 15 = – = 12 Simplify.

5. Inverse Matrices and Systems
Lesson 4-7
Additional Examples
7x + 3y + 2z = 13
–2x + y – 8z = 26
x – 4y +10z = –13
Solve the system
Step 1: Write the system as
a matrix equation.
Step 2: Store the coefficient
matrix as matrix A
and the constant
matrix as matrix B.
– –8
1 – 13 26
–13 x y z =
The solution is (9, –12, –7).