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Inverse Matrices and Systems

Inverse Matrices and Systems. 4-7 Inverse Matrices And Systems. –3 –4 5 –2 7 0 –5 1 –1. x y z. 11 –6 20. Matrix equation: =. Coefficient matrix. Variable matrix. Constant matrix. –3 –4 5 –2 7 0 –5 1 –1.

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Inverse Matrices and Systems

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  1. Inverse Matrices and Systems 4-7 Inverse Matrices And Systems

  2. –3 –4 5 –2 7 0 –5 1 –1 x y z 11 –6 20 Matrix equation: = Coefficient matrix Variable matrix Constant matrix –3 –4 5 –2 7 0 –5 1 –1 x y z 11 –6 20 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.

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

  4. Check: 2x + 3y = –1 x – y = 12 Use the original equations. 2(7) + 3(–5) –1 (7) – (–5) 12 Substitute. 14 – 15 = –1 7 + 5 = 12 Simplify. Inverse Matrices and Systems Lesson 4-7 Additional Examples (continued) The solution of the system is (7, –5).

  5. 7 3 2 –2 1 –8 1 –4 10 x y z 13 26 –13 = 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. The solution is (9, –12, –7).

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