Algebra

1. MATRIX
ALGEBRA

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.
•