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Ch. 7 – Matrices and Systems of Equations 7.3 – Multivariable Linear Systems Systems of 3 equations Here is an example of a system of 3 linear equations: Can do elimination to solve, but must do it a lot x 2 y 3z 9 x 3 y 4 2 x 5 y 5 z 17 Here is an example of the same system of 3 linear equations in Row-Echelon Form (REF): REF equations are in a stair-step pattern, leading coefficient is 1 It’s much easier to solve! Just plug in 2 for z to find y, then plug that in to find x! We want multivariable equations to look like this! x 2 y 3z 9 y 3z 5 z2 REF it! Ex 1: Solve this system of equations. x 2 y 3z 9 x 3 y 4 2 x 5 y 5 z 17 To solve, we will do several eliminations to get the equations in REF! To do that, we need our stair-step pattern and leading coefficients of 1! ADD and REPLACE! Step 1: First equation must be led by x - CHECK! Step 2: Second equation must be led by y Must get rid of x, so add equations 1 and 2 The new equation is y + 3z = 5, so replace it for the second equation to get… Now 2nd equation is led by y! x 2 y 3z 9 y 3z 5 2 x 5 y 5 z 17 x 2 y 3z 9 y 3z 5 REF it! (cont’d): Solve this system of equations: 2 x 5 y 5 z 17 Step 3: Third equation must be led by z Must get rid of x, so add -2•(first equation) and second equation 2 x 4 y 6 z 18 2 x 5 y 5 z 17 y z 1 x 2 y 3z 9 y 3z 5 y z 1 Must get rid of y, so add 2nd and 3rd equations y 3z 5 y z 1 z2 2z 4 Back-substitute to get y = -1 and x = 1, so we get… …(1, -1, 2) x 2 y 3z 9 y 3z 5 z2 REF it! Ex 2: Solve this system of equations. 2 x 4 y z 4 2 x 4 y 6 z 13 x 2 y 0.5 z 2 2 x 4 y 6 z 13 4x 2 y z 6 4x 2 y z 6 Divide 1st equation by 2… 4 x 8 y 12 z 26 4x 2 y z 6 6 y 11z 20 Add -2(E2) and E3 and replace for E2… x 2 y 0.5 z 2 6 y 11z 20 4x 2 y z 6 Add -4(E1) and E3 and replace for E3… REF it! Ex 2: Cont’d. 4 x 8 y 2 z 8 4x 2 y z 6 10 y z 14 x 2 y 0.5 z 2 6 y 11z 20 10 y z 14 Add 5(E2) and 3(E3) and replace for E3… 30 y 55 z 100 30 y 3 z 42 58z 58 x 2 y 0.5 z 2 6 y 11z 20 58 z 58 Divide E2 and E3 down to REF… REF it! x 2 y 0.5 z 2 11 10 y z 6 3 z 1 Ex 2: Cont’d. Now back-substitute… 11 10 y 6 3 3 y 2 Answer: ( ½ , -3/2 , 1 ) 3 x 2 0.5(1) 2 2 1 x 2 Ex 3: Solve this system of equations. You can always rewrite the system as an augmented matrix (3x4) of coefficients – it will save you lead! Just get 1’s in the main diagonal! x 2 y 3z 5 x 3 y 5z 4 2 x 3z 0 Write as a 3x4 matrix… 1 2 3 1 3 5 4 5 0 1 2 9 1 2 3 1 2 3 0 5 4 3 0 5 Add E1 and E2 and replace for E2… 1 2 3 0 2 2 9 3 0 1 0 5 Add -2(E1) and E3 and replace for E3… Ex 3: cont’d. 2 4 6 10 0 3 2 0 0 4 9 10 1 2 3 0 0 2 9 9 10 1 4 5 Add -4(E2) and E3 and replace for E3… 0 4 8 0 9 10 4 36 0 0 1 46 1 2 3 0 0 2 9 1 46 1 0 5 Divide E3 by -1 to get… Ex 3: Cont’d. 3 0 0 2 9 1 46 1 0 5 x 2 y 3z 5 y 2z 9 z 46 Now back-substitute… y 2(46) 9 1 2 y 101 Answer: ( 69 , 101 , 46 ) x 2(101) 3(46) 5 x 69 Ex 4: Solve this system of equations. 2 x y 3z 4 4 x 2 z 10 2 x 3 y 13z 8 Write as a 3x4 array… 4 2 6 8 4 0 2 10 0 2 8 2 2 4 1 0 3 2 4 10 2 3 13 8 Add -2(E1) and E2 and replace for E2… 2 0 1 2 3 8 2 3 13 8 4 2 Add E1 and E3 and replace for E3… Ex 4: Solve this system of equations. 2 1 3 4 2 3 13 8 0 4 16 4 2 3 1 4 0 2 8 2 0 4 16 4 Add 2(E2) and E3 and replace for E3… 0 4 16 0 16 4 0 4 0 0 4 0 2 1 0 2 0 0 3 4 8 0 2 0 We’re left with 0 = 0 in the 3rd equation. Since this equation is true, we have an infinite number of solutions… Ex (cont’d): If the answer is infinite solutions, we must find a generic triple that solves the system: Step 1: Let z = a. Step 2: Back-substitute to get expressions for x and y in terms of a za 2 y 8(a) 2 2 x y 3z 4 2 y 8z 2 y 4a 1 2 x (4a 1) 3(a) 4 Answer: ( 5/2 – ½a , 4a – 1 , a ) 5 1 x a 2 2 0z 0