1. Spring 2013 Solving a System of Linear Equations Matrix inverse Simultaneous equations Cramer’s rule Second-order Conditions Lecture 7

2. Spring 2013 1. Matrix Equation and Matrix Inversion Market equilibrium model Demand: Q + P = 10 Supply : Q – 4P = -3 In matrix form,

3. Spring 2013 General equation system (N equations, N unknowns)

4. Spring 2013 If we can find a matrix A -1 such that A -1 Ax = x, then we can solve the equations by observing that A -1 Ax = A -1 b x = A -1 b Existence of an inverse A -1 : A is a square matrix. A is not a null matrix. Rows and columns are linearly independent. A -1 A = AA -1 = I N.

5. Spring 2013 In the previous example, Ax = b Example 7.1: Consider the matrix

6. Spring 2013 2. Determinants 2.1 Determinants of Order 2 |A| = a 11 a 22 – a 12 a 21. |A| = 1·4 - 2·3 = -2. Example 7.2:

7. Spring 2013 |A| = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 - a 11 a 23 a 32 – a 12 a 21 a 33 – a 13 a 22 a 31 2.2 Determinants of Order 3

8. Spring 2013 a 11 a 12 a 21 a 22 a 31 a 32 + + +– – – |A| = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 - a 11 a 23 a 32 – a 12 a 21 a 33 – a 13 a 22 a 31

9. Spring 2013 Example 7.3: Find the determinants of the following square matrices.

10. Spring 2013 |M ij | = 2.3 Expansion by Cofactors 1. The minor |M ij | of a matrix is the determinant of the matrix obtained by deleting the i th row and j th column.

11. Spring 2013 |M 11 | = 4, |M 12 | = 3, |M 21 | = 2, |M 22 | = 1 |M 11 | = 5, |M 12 | = -17, |M 13 | = -14 Example 7.4: Find all the minors of the matrices A and B.

12. Spring 2013 2. The cofactor |C ij | of a matrix = (-1) i+j |M ij | Example 7.5: Find all the cofactors of the following matrices.

13. Spring 2013 3. The determinant is the expansion along any row or column. |A| = a i1 C i1 + a i2 C i2 +  + a ij C ij +  a in C in = a 1j C 1j + a 2j C 2j +  + a ij C ij +  a nj C nj

14. Spring 2013 1 st row: |A| = a 11 C 11 + a 12 C 12 = 1·4 + 2·(-3) = -2 2 nd row: |A| = a 21 C 21 + a 22 C 22 = 3·(-2) + 4·1 = -2 1 st column: |A| = a 11 C 11 + a 21 C 21 = 1·4 + 3·(-2) = -2 2nd column: |A| = a 12 C 12 + a 22 C 22 = 2·(-3) + 4·1 = -2 Example 7.6: Calculate the determinant of matrix A.

15. Spring 2013 Expansion along the 1st row |B| = 8·5 + 3·17 + 2·(-14) = 63. Expansion along the 3rd column

16. Spring 2013 2.4 Basic Rules for Determinatns Let A be an n  n matrix. Then: If all the elements in a row (or column) of A are zero, then |A| = 0. If two of the rows (or columns) of A are proportional, then |A| = 0.

17. Spring 2013 Let A and B be n  n matrices. Then: If all the elements in a single row (or column) of A are multiplied by a number , the determinant is multiplied by . The value of the determinant of A is unchanged if a multiple of one row (or column) is added to a different row (or column) of A. |AB| = |A|  |B|.

18. Spring 2013 3. Matrix Inversion Let A be n by n. If there is a n by n matrix X such that AX = XA = I n, then X is called the inverse of A and is denoted by A -1.

19. Spring 2013 Example 7.7: Consider the following three matrices: Show that A and X are inverses of each other and B has no inverse.

20. Spring 2013 Step 1: Check the singularity of A |A|  0  Carry on matrix inversion |A| = 0  Stop Step 2: Find the cofactor matrix of A 3.1 Finding the Inverse of a Matrix

21. Spring 2013 Step 3: Find the adjoint of A adj(A) = C T. Step 4: Compute the inverse A -1 = adj(A) / |A|. Example 7.8: Find the inverse of

22. Spring 2013 Step 1: |A| = 1x4 – 3x2 = -2 < 0 Step 2: M 11 = 4, M 12 = 3, M 21 = 2, M 22 = 1 C 11 = 4, C 12 = -3, C 21 = -2, M 22 = 1 Step 3:

23. Spring 2013 Step 4: Example 7.9: Show that the matrix A has an inverse and find the inverse.

24. Spring 2013 3.2 Solving Equations Matrix equation Ax = b  A -1 Ax = A -1 b  x = A -1 b.

25. Spring 2013 Example 7.10: Solve the following simultaneous equations. 3x 1 – 2x 2 = 11 2x 1 + x 2 = 12 Matrix equation 1. |A| = 3  1 – 2  (-2) = 7 > 0

26. Spring 2013

27. 4. Cramer’s Rule Matrix equation Let

28. Spring 2013 Cramer’s rule: If |A|  0, then for i = 1,...,N, In the previous example, x 1 * = A 1 / |A| = 35/7 = 5 x 2 * = A 2 / |A| = 14/7 = 2

29. Spring 2013 Example 7.11: Solve the following system for the unknowns x 1, x 2 and x 3 using Cramer’s rule. 2x 1 + 4x 2 – x 3 = 15 x 1 – 3x 2 + 2x 3 = -5 6x 1 + 5x 2 + x 3 = 28

30. Spring 2013 Example 7.12: Solve the following system of equations: x 1 + x 2 = 3 x 1 + x 3 = 2 x 2 + x 3 + x 4 = 6 x 2 + x 4 = 1

31. Spring 2013 3 Possible Cases of an Equation System Case 1: Unique solution |A|  0

32. Spring 2013 Case 2: No solution |A| = 0 and equations are inconsistent. Case 3: Multiple solutions |A| = 0 and equations are consistent.