 Inverse Matrices and Systems

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Inverse Matrices and Systems

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

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

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.

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).