CHAPTER 5: MATRICES AND LINEAR ALGEBRA 144 Find the inverse of the matrix F = [ − − − ] Answer: Step 1: DETERMINANT (det) det() = | 1 −2 1 2 1 −1 3 −1 2 | = (1) | 1 −1 −1 2 | − (−2) | 2 −1 3 2 | + (1) | 2 1 3 −1 | = 1(2 − 1) + 2(4 + 3) + 1(−2 − 3) = 10 Step 2: MINOR (M) = [ | 1 −1 −1 2 | | 2 −1 3 2 | | 2 1 3 −1 | | −2 1 −1 2 | | 1 1 3 2 | | 1 −2 3 −1 | | −2 1 1 −1 | | 1 1 2 −1 | | 1 −2 2 1 | ] = [ 1 7 −5 −3 −1 5 1 −3 5 ] Step 3: COFACTOR (cof) Cofactor is the signed minor which means that the sign at the minor matrix will be changed by simply attach it with the “checkerboard pattern” as below: = ( ... ... ... ... ... ... ... ) () = [ 1 −7 −5 3 −1 −5 1 3 5 ] Example 14
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 145 Step 4: ADJOINT (adj) adj = () () = [ 1 3 1 −7 −1 3 −5 −5 5 ] Step 5: Inverse matrix (3x3) formula −1 = 1 det() ∙ () −1 = 1 det() ∙ () = 1 10 ∙ [ 1 3 1 −7 −1 3 −5 −5 5 ] = [ 1 10 ⁄ 3 10 ⁄ 1 10 ⁄ −7 10 ⁄ −1 10 ⁄ 3 10 ⁄ − 1 2 ⁄ −1 2 ⁄ 1 2 ⁄ ] Notes: If a square matrix A has an inverse, A is said to be non-singular or invertible. Otherwise, it is called singular or non-invertible. 1. Find the inverses (if they exist) of: = [ 0 2 3 4 ] ; = [ 1 4 3 12] i. −1 = Answer: i)[ −2 3 ⁄ 1 3 ⁄ 1 2 ⁄ 0 ] ii. −1 = Answer: ii)Undefined Exercise 8
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 146 5.2.1 Explain the procedure to solve linear equations 1. Isolate the terms of x, y and z on left hand sided (LHS) by using the matrix form (3 x 1 order). 2. Isolate the coefficients and put aside with x, y and z term in matrix form (3 x 3 order) called matrix A. The operation of these two matrix is multiple sign. 3. Group the answer of the equations in the matrix form (3 x 1 order) at the right hand side (RHS). 4. Find the determinants value of the matrix A correctly. 5. Choose the solution by using Cramer Rule or Inverse Matrix Method. 6. Check the solution and if it is correct, both sides of the equation will be equal. 2. Find the inverse of the matrix A = [ 3 1 −1 2 −2 0 1 2 −1 ] Answer: [ 1 −1 2 ⁄ −1 1 −1 −1 3 −5 2 ⁄ −4 ] 5.2 Explain system of linear equations
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 147 Use Cramer’s Rule to solve below equations. + 2 + 3 = 17 3 + 2 + = 11 − 5 + = −5 5.2.2 Solve systems of linear equations Solving systems of linear equations is a common problem encountered in many disciplines. Solving such problems is so important that the techniques for solving them (substitution, elimination) are learned early on in algebra studies. There are 4 types of methods that can be choose to solve systems of linear equations: a. Row Reduction Techniques b. Interpreting Solutions from Reduced Row Echelon Form c. Multiplication by Inverse of Coefficient Matrix d. Cramer's Rule (Source: https://brilliant.org) 5.2.3 Solve simultaneous equations with three variables using Cramer’s Rule Simultaneous equations means two or more equations that share variables. As for an example, two equations that share the variables x and y as write down as: x + y = 3 −3x + y = 6 For two equations with two unknown variables, the suitable methods as mention as before which are substitution or elimination methods. Meanwhile, to solve three equations that contains three variables, the Cramer Rule is more preferable compared to other methods. But, to get better understanding in Cramer’s rule, determinant subtopic must be mastered first. Cramer’s rule fail if the determinant of the coefficient array is zero, since you can’t divide by zero. Example 15
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 148 Answer: Step 1: Rearrange into matrix form [ 1 2 3 3 2 1 1 −5 1 ][ ] = [ 17 11 −5 ] Step 2: The coefficient matrix of this is system = [ 1 2 3 3 2 1 1 −5 1 ] Step 3: Find the determinant of the matrix A = [ 1 2 3 3 2 1 1 −5 1 ] det() = || = 1 [ 2 1 −5 1 ] − (2) [ 3 1 1 1 ] + (3) [ 3 2 1 −5 ] = 1(2 + 5) − 2(3 − 1) + 3(−15 − 2) = −48 Step 4: Find the determinant of the matrix = [ 17 2 3 11 2 1 −5 −5 1 ] | | = 17 [ 2 1 −5 1 ] − (2) [ 11 1 −5 1 ] + (3) [ 11 2 −5 −5 ] = 17(2 + 5) − 2(11 + 5) + 3(−55 + 10) = 119 − 32 − 135 = −48 Step 5: Find the determinant of the matrix = [ 1 17 3 3 11 1 1 −5 1 ] || = 1 [ 11 1 −5 1 ] − (17) [ 3 1 1 1 ] + (3) [ 3 11 1 −5 ] = 1(11 + 5) − 17(3 − 1) + 3(−15 − 11) = 16 − 34 − 48 = −96
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 149 Step 6: Find the determinant of the matrix = [ 1 2 17 3 2 11 1 −5 −5 ] | | = 1 [ 2 11 −5 −5 ] − (2) [ 3 11 1 −5 ] + (17) [ 3 2 1 −5 ] = 1(−10 + 55) − 2(−15 − 11) + 17(−15 − 2) = 45 + 52 − 289 = −192 Step 7: Solve for x,y and z = | | || = −48 −48 = 1 = || || = −96 −48 = 2 = | | || = −192 −48 = 4 1. Solve the simultaneous equations below by using Cramer’s rule. + + = 4 − = −3 + 2 − = 0 Answer: = −1 ; = 2 ; = 3 2. Solve the simultaneous equations below by using Cramer’s rule. 13 − 14 − 15 = 25 −12 + 13 + 15 = −22 11 − 7 − 14 = 15 Answer: = 2 ; = −1 ; = 1 Exercise 9
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 150 3. Solve the following simultaneous equation by using Cramer’s rule. 2 + − = 0 + = 4 + + = 0 Answer: = 8 3 ⁄ = 4 3 ⁄ = 4 3 ⁄ 4. Find the unknown variables by using Cramer’s rule. − + 3 + = 1 2 + 5 = 3 3 + − 2 = −2 Answer: = 7 3 ⁄ = −1 3 ⁄ = 13 3 ⁄
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