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# Matrix algebra determining errors

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Algebra

Algebra

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• 1. MATRIXALGEBRA
• 2. A systematic approach of theelimination method for solving asystem of linear equations providesanother method of solution thatinvolves a simplified notation. 3 ways in finding determinants: Criss-cross multiplication Row Column
• 3. DETERMINING THE ERROR OF 3X3 MATRIX
• 4. The Given Matrix: 3 1 1 A= 2 -4 -3 7 -2 0
• 5. 3 1 1 3 1 2 -4 -3 2 -4 7 -2 0 7 -2 (0 -21 -4) - (-28 +18+0 ) = -15Criss-cross multiplication
• 6. Cofactor: 3= -4 -3 -2 0 = -6 1= 2 -3 7 0 = 21 1= 2 -4 7 -2 = 24
• 7. Cofactor: 2= 1 1 -2 0 =2 -4= 3 1 7 0 = -7 -3= 3 1 7 -2 = -13
• 8. Cofactor: 7= 1 1 -4 -3 =1 -2= 3 1 2 -3 = -11 0= 3 1 2 -4 = -14
• 9. Inverse Matrix:A-1 = -1/15 -6 2 1 + - + 21 -7 -11 - + - 24 -13 -14 + - + 6/15 2/15 -1/15 A-1 = 21/15 7/15 -11/15 -24/15 -13/15 14/15
• 10. Identity Matrix 3 1 1 6/15 2/15 -1/15AA-1= 2 -4 -3 21/15 7/15 -11/15 7 -2 0 -24/15 -13/15 14/15 1 0 0 = 0 1 0 0 0 1
• 11. Remember:• The first thing we should do is to identify the correct determinant and finding the inverse and identity of the matrix given was done in order to prove whether the determinant used wasn’t wrong.
• 12. ERRORSCriss-Cross Multiplication Row Determinant Column Determinant
• 13. Criss-Cross 3 1 1 3 1 2 -4 -3 2 4 7 -2 0 7 -2 = -21 - 4 + 28 – 18 = -15
• 14. Criss-Cross 7 -2 0 7 -2 3 1 1 3 1 2 -4 -3 2 -4 = -21 - 4 + 28 – 18 = -15
• 15. Criss-Cross 7 -2 0 7 -2 2 -4 -3 2 -4 3 1 1 3 1 = -28 + 18 + 21 + 4 = 15 ERROR
• 16. Criss-Cross 3 1 1 3 1 7 -2 0 7 -2 2 -4 -3 2 -4 = 18 - 28 + 4 + 21 = 15 ERROR
• 17. Criss-Cross 2 -4 -3 2 -4 3 1 1 3 1 7 -2 0 7 -2 = -28 +18 + 21 – 4 = 15 ERROR
• 18. Criss-Cross 2 -4 -3 2 4 7 -2 0 7 -2 3 1 1 3 1 = -4 - 21 - 18 + 28 = -15
• 19. Column 3 1 1 2 -4 -3 7 -2 0 = 1(24) + 3(-13) + 0 = -15 = 1(21) + 4(-7) - 2(-11) = 15 ERROR = 3(-6) – 2(2) + 7(1) = -15
• 20. Column 3 1 1 7 -2 0 2 -4 -3 = 1(-24) – 0 – 3(-13) = 15 ERROR = 1(-21) + 2(-11) - 4(-7) = -15 = 3(6) – 7(1) + 2(2) = 15 ERROR
• 21. ROW3 1 12 -4 -37 -2 0 = 7(1) + 2(-11) + 0(-14) = -15 = 2(2) + 4(-7) - 3(-13) = 15 ERROR = 3(-6) – 1(21) + 1(24) = -15
• 22. ROW3 1 17 -2 02 -4 -3 = 2(2) + 4(-7) – 3(-13) = 15 ERROR = 7(1) + 2(-11) - O(-14) = -15 = 3(6) – 1(-21) + 1(-24) = 15 ERROR
• 23. Tip in finding the error:If the determinant you’ve found using criss-cross multiplication in matrix given is correct, the error in row and column was found in the middle row and column but if the determinant you’ve found using criss-cross multiplication in the given matrix is the error, the error in row and column was found in the first and last row and column.•