Inverse 3x3 Matrices
Slides in this set (2)
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- How to find the inverse of a 3x3 matrix (e.g. A = [1 2 1][0 5 4][2 -2 0] ):
- 1. Find the determinant of the matrix ( |A| = 14 )
- 2. Transpose the matrix , which essentially means reflect in diagonal from top left to bottom right (A^T = [1 0 2][2 5 -2][1 4 0] )
- 3. From A^T , calculate the matrix of minor determinants. To do this, delete the row and its corresponding column and then find the minor determinant. For example for A^T , the top left element for the matrix of minor determinants (MOM) is found by: (5)(0) - (-2)(4) = 8. (MOM = [8 2 3][-8 -2 4][-10 -6 5] .
- 4. From the MOM , we find the matrix of co-factors (MOC). To do this, multiply the matrix by [+ - +][- + -][+ - +]. (MOC = [8 -2 3][8 -2 -4][-10 6 5] )
- 5. Multiply the MOC by 1/|A| for the inverse. (A^-1 = 1/14[8 -2 3][8 -2 -4][-10 6 5] ). To check this is correct, AA^-1 should equal I .
- -We can find this on our calculators to by defining a matrix and then putting it to the power of -1
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- 1a. A = [a b 0][0 a b][b 0 a] . Given that A^-1 exists , show a ≠ -b. b). Find A^-1
- a). |A| = a|a b||0 a| - b|0 b||b a| + 0 , a(a^2 - 0) - b(0 - b^2) = a^3 + b^3 , a^3 + b^3 ≠ 0 , a^3 ≠ -b^3 , a ≠ -b as required
- b). A^T = [a 0 b][b a 0][0 b a] , MOM: [a^2 ab b^2][-b^2 a^2 ab][-ab -b^2 a^2] , MOC: [a^2 -ab b^2][b^2 a^2 -ab][-ab b^2 a^2] , A^-1 = 1/a^3+b^3 [a^2 -ab b^2][b^2 a^2 - ab][-ab b^2 a^2]
- 2. Show how these 3 simultaneous can be written using matrices: 2x+4y-z = -9 , x+4z = 24 , x-2y+3z = 25
- We can say: [2 4 -1][1 0 4][1 -2 3] x [x][y][z] = [-9][24][25] , and lets say [2 4 -1][1 0 4][1 -2 3] = A , we need to find A^-1
- A^-1 = 1/22 [8 -10 16][1 7 -9][-2 8 -4]
- We now say: [x][y][z] = 1/22 [8 -10 16][1 7 -9][-2 8 -4] x [-9][24][25]
- This cancels to [x][y][z] = [4][-3][5] . x=4 , y=-3 , z=5