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MATRIX
ALGEBRA
A systematic approach of the
elimination method for solving a
system of linear equations provides
another method of solution that
involves a simplified notation.
    3 ways in finding determinants:
    Criss-cross multiplication
    Row
    Column
DETERMINING
 THE ERROR
     OF
 3X3 MATRIX
The Given Matrix:

          3 1 1
    A=    2 -4 -3
          7 -2 0
3 1 1           3 1
       2 -4 -3         2 -4
       7 -2 0          7 -2
     (0 -21 -4) - (-28 +18+0 )
             = -15
Criss-cross multiplication
Cofactor:
 3= -4   -3
    -2    0   = -6

 1= 2    -3
    7    0    = 21

 1= 2 -4
    7 -2      = 24
Cofactor:
  2= 1    1
     -2   0    =2
  -4= 3   1
      7   0    = -7
  -3= 3    1
      7   -2   = -13
Cofactor:
 7= 1 1
    -4 -3    =1
 -2= 3    1
     2   -3 = -11
 0=   3 1
      2 -4   = -14
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
Identity Matrix
      3        1       1   6/15 2/15 -1/15
AA-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
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.
ERRORS
Criss-Cross Multiplication

    Row Determinant


  Column Determinant
Criss-Cross
 3    1   1        3    1
 2   -4   -3       2    4
 7   -2    0       7    -2

               = -21 - 4 + 28 – 18
               = -15
Criss-Cross
  7   -2    0      7   -2
  3    1   1       3    1
  2   -4   -3      2   -4

                = -21 - 4 + 28 – 18
                = -15
Criss-Cross
 7   -2    0      7   -2
 2   -4   -3      2   -4
 3    1    1      3    1

               = -28 + 18 + 21 + 4
               = 15 ERROR
Criss-Cross
  3    1   1       3    1
  7   -2    0      7   -2
  2   -4   -3      2   -4

                = 18 - 28 + 4 + 21
                = 15 ERROR
Criss-Cross
  2   -4   -3      2   -4
  3    1   1       3    1
  7   -2    0      7   -2

                = -28 +18 + 21 – 4
                = 15 ERROR
Criss-Cross
  2   -4   -3      2    4
  7   -2    0      7   -2
  3    1   1       3    1

                = -4 - 21 - 18 + 28
                = -15
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
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
ROW
3    1   1
2   -4   -3
7   -2    0
                   = 7(1) + 2(-11) + 0(-14)
                   = -15
           = 2(2) + 4(-7) - 3(-13)
           = 15 ERROR
 = 3(-6) – 1(21) + 1(24)
 = -15
ROW
3    1   1
7   -2    0
2   -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
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.
•

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Matrix algebra determining errors

  • 2. A systematic approach of the elimination method for solving a system of linear equations provides another method of solution that involves 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 ) = -15 Criss-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 -1/15 AA-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. ERRORS Criss-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. ROW 3 1 1 2 -4 -3 7 -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. ROW 3 1 1 7 -2 0 2 -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. •