Forms of linear equations #3 - Questions and Answers

Question 1 Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row echelon form. Solve the system by back substitution. Assume that the variables are named 𝑥 1 , 𝑥 2 , 𝑥 3 , … x 1 ​ ,x 2 ​ ,x 3 ​ ,… from left to right. Answer 1 The linear system corresponding to the augmented matrix is: 𝑥 1 + 5 𝑥 3 + 3 𝑥 4 = 2 𝑥 2 − 2 𝑥 3 + 4 𝑥 4 = − 7 𝑥 3 + 𝑥 4 = 3 x 1 ​ +5x 3 ​ +3x 4 ​ =2x 2 ​ −2x 3 ​ +4x 4 ​ =−7x 3 ​ +x 4 ​ =3 Solve for the leading variables: 𝑥 3 + 𝑥 4 = 3 x 3 ​ +x 4 ​ =3 𝑥 3 = 3 − 𝑥 4 x 3 ​ =3−x 4 ​ Substitute 𝑥 3 = 3 − 𝑥 4 x 3 ​ =3−x 4 ​ into the second equation: 𝑥 2 − 2 ( 3 − 𝑥 4 ) + 4 𝑥 4 = − 7 x 2 ​ −2(3−x 4 ​ )+4x 4 ​ =−7 𝑥 2 − 6 + 2 𝑥 4 + 4 𝑥 4 = − 7 x 2 ​ −6+2x 4 ​ +4x 4 ​ =−7 𝑥 2 + 6 𝑥 4 = − 1 x 2 ​ +6x 4 ​ =−1 𝑥 2 = − 1 − 6 𝑥 4 x 2 ​ =−1−6x 4 ​ Substitute 𝑥 3 = 3 − 𝑥 4 x 3 ​ =3−x 4 ​ into the first equation: 𝑥 1 + 5 ( 3 − 𝑥 4 ) + 3 𝑥 4 = 2 x 1 ​ +5(3−x 4 ​ )+3x 4 ​ =2 𝑥 1 + 15 − 5 𝑥 4 + 3 𝑥 4 = 2 x 1 ​ +15−5x 4 ​ +3x 4 ​ =2 𝑥 1 − 2 𝑥 4 = − 13 x 1 ​ −2x 4 ​ =−13 𝑥 1 = − 13 + 2 𝑥 4 x 1 ​ =−13+2x 4 ​ Assign arbitrary value to free variable 𝑥 4 x 4 ​ , say 𝑥 4 = 𝑡 x 4 ​ =t: The solution is described by the parametric equations: 𝑥 1 = − 13 + 2 𝑡 , 𝑥 2 = − 1 − 6 𝑡 , 𝑥 3 = 3 − 𝑡 , 𝑥 4 = 𝑡 x 1 ​ =−13+2t,x 2 ​ =−1−6t,x 3 ​ =3−t,x 4 ​ =t Answer 2 Given the row echelon form: [ 1 0 5 3 2 0 1 − 2 4 − 7 0 0 1 1 3 ] ​ 1 0 0 ​ 0 1 0 ​ 5 −2 1 ​ 3 4 1 ​ 2 −7 3 ​ ​ From the third row: 𝑥 3 + 𝑥 4 = 3 , 𝑥 3 = 3 − 𝑥 4 x 3 ​ +x 4 ​ =3,x 3 ​ =3−x 4 ​ From the second row: 𝑥 2 − 2 𝑥 3 + 4 𝑥 4 = − 7 x 2 ​ −2x 3 ​ +4x 4 ​ =−7 Substitute 𝑥 3 = 3 − 𝑥 4 x 3 ​ =3−x 4 ​ : 𝑥 2 − 2 ( 3 − 𝑥 4 ) + 4 𝑥 4 = − 7 x 2 ​ −2(3−x 4 ​ )+4x 4 ​ =−7 𝑥 2 + 6 𝑥 4 = − 1 , 𝑥 2 = − 1 − 6 𝑥 4 x 2 ​ +6x 4 ​ =−1,x 2 ​ =−1−6x 4 ​ From the first row: 𝑥 1 + 5 𝑥 3 + 3 𝑥 4 = 2 x 1 ​ +5x 3 ​ +3x 4 ​ =2 Substitute 𝑥 3 = 3 − 𝑥 4 x 3 ​ =3−x 4 ​ : 𝑥 1 + 15 − 5 𝑥 4 + 3 𝑥 4 = 2 x 1 ​ +15−5x 4 ​ +3x 4 ​ =2 𝑥 1 − 2 𝑥 4 = − 13 , 𝑥 1 = − 13 + 2 𝑥 4 x 1 ​ −2x 4 ​ =−13,x 1 ​ =−13+2x 4 ​ Final Solution: 𝑥 1 = − 13 + 2 𝑡 , 𝑥 2 = − 1 − 6 𝑡 , 𝑥 3 = 3 − 𝑡 , 𝑥 4 = 𝑡 x 1 ​ =−13+2t,x 2 ​ =−1−6t,x 3 ​ =3−t,x 4 ​ =t Answer 3 The system of linear equations is solved step by step using the back substitution process, as detailed in the solution above. Question 2 The last row of the reduced row echelon form corresponds to the equation 0 𝑥 1 + 0 𝑥 2 + 0 𝑥 3 = 1 0x 1 ​ +0x 2 ​ +0x 3 ​ =1. There are no values of 𝑥 1 , 𝑥 2 , x 1 ​ ,x 2 ​ , and 𝑥 3 x 3 ​ that satisfy this equation. Therefore, the system corresponding to the matrix has no solutions. Question 3 Write the equation in standard form of a linear equation: 𝑑 𝑦 𝑑 𝑥 = 𝑦 + 𝑦 𝑥 dx dy ​ = x y ​ +y ​ Answer The solution is written below: 1 𝑦 ⋅ 𝑑 𝑦 𝑑 𝑥 = 1 𝑥 ⋅ ( 1 + 𝑦 ) y ​ 1 ​ ⋅ dx dy ​ = x 1 ​ ⋅(1+ y ​ ) Question 4 The reduced row echelon form of the augmented matrix of a system of linear equations is given. Determine if the system has one solution, no solution, or infinitely many solutions: [ 101 ∣ 011 ∣ 000 ] [101∣011∣000] Answer There are two columns before the vertical bar, so there are two variables. Each variable is represented by a leading 1, so there are no parameters needed. There are no impossible equations, such as 0 = 1 0=1. The system is consistent and has exactly one solution. Question 5 The reduced row echelon form of the augmented matrix is given: [ 1 0 0 − 1 0 1 0 4 0 0 1 3 0 0 0 0 ] ​ 1 0 0 0 ​ 0 1 0 0 ​ 0 0 1 0 ​ −1 4 3 0 ​ ​ Answer The last row represents an equation that is always true ( 0 = 0 0=0). The system is consistent and has exactly one solution. (No parameters) Interpreting row by row gives: 𝑥 1 = − 1 , 𝑥 2 = 4 , 𝑥 3 = 3 x 1 ​ =−1,x 2 ​ =4,x 3 ​ =3