660 Worked-Out Solutions to Exercises: Chapters 11 to 19
- Here are the equations from solution to Prob. 3, which we hope we have solved as a
linear system:
4 y−z=− 2
5 x− (3/2)y+ 8 z= 1
x+y+z= 1Let’s check the first equation with the values from the solution to Prob. 7:4 y−z=− 2
4 × (−20/11)− (−58/11)=− 2
−80/11+ 58/11 =− 2
−22/11=− 2
− 2 =− 2Checking in the second equation:5 x− (3/2)y+ 8 z= 1
5 × 89/11 − [(3/2) × (−20/11)]+ 8 × (−58/11)= 1
445/11− (−30/11)− 464/11 = 1
445/11+ 30/11 − 464/11 = 1
11/11= 1
1 = 1Checking in the third equation:x+y+z= 1
89/11+ (−20/11)+ (−58/11)= 1
89/11− 20/11 − 58/11 = 1
11/11= 1
1 = 1We can now be confident that the solution to the three-by-three linear system isx= 89/11
y=−20/11
z=−58/11- Here’s the three-by-three linear system we have been told to describe as a matrix:
x+y+z= 1
x+y+z= 2
x+y+z= 3