Ch. 7 – Matrices and Systems of Equations

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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 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!

REF it! Ex 1: Solve this system of equations. 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!

REF it! (cont’d): Solve this system of equations: Step 3: Third equation must be led by z Must get rid of x, so add -2•(first equation) and second equation Must get rid of y, so add 2nd and 3rd equations Back-substitute to get y = -1 and x = 1, so we get… …(1, -1, 2)

REF it! Ex 2: Solve this system of equations. Add -2(E2) and E3 and replace for E2… Divide 1st equation by 2… Add -4(E1) and E3 and replace for E3…

REF it! Ex 2: Cont’d. Add 5(E2) and 3(E3) and replace for E3… Divide E2 and E3 down to REF…

REF it! Ex 2: Cont’d. Now back-substitute… Answer: ( ½ , -3/2 , 1 )

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! Add E1 and E2 and replace for E2… Write as a 3x4 matrix… Add -2(E1) and E3 and replace for E3…

Ex 3: cont’d. Add -4(E2) and E3 and replace for E3… Divide E3 by -1 to get…

Ex 3: Cont’d. Now back-substitute… Answer: ( 69 , 101 , 46 )

Ex 4: Solve this system of equations. Add -2(E1) and E2 and replace for E2… Write as a 3x4 array… Add E1 and E3 and replace for E3…

Ex 4: Solve this system of equations. We’re left with 0 = 0 in the 3rd equation. Since this equation is true, we have an infinite number of solutions… Add 2(E2) and E3 and replace for E3…

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 Answer: ( 5/2 – ½a , 4a – 1 , a )