1. Use elementary operations to find the solution set to the given linear systems.
Given:7x1+3x211x1+5x2=6=12[71135∣∣612]Work:
Multiply top row by 3, and the bottom row by 2:
[2122910∣∣1824]
Swap the rows:
[2221109∣∣2418]
Subtract second row from first row and divide second row by 3:
[1713∣∣66]
Subtract 7* Row 1 from Row 2:
[101−4∣∣6−36]
Divide row 2 by -4:
[1011∣∣69]Answer:x2=9, and x1+9=6, so x1=−3
3. Determine whether or not the following matrices are in echelon or reduced echelon form.
100010001∣∣∣−1−30
Matrix 1 is in reduced echelon form
010301−103∣∣∣041
Matrix 2 is not in reduced or echelon form
1000−11000210∣∣∣∣4−150
Matrix 3 is in reduced echelon form
4. Use Gaussian elimination to find the solution set to the following linear systems.
Given:8x1+6x29x1+7x2=−4,=2.[8967∣∣−42]Work:R1=R2−R1, R2=R1:
[1816∣∣6−4]R2=(R2−8R1)/−2:
[1011∣∣626]Answer:x2=26 and x1+26=6, so x1=−20
Given:2x1+3x24x1+7x33x2−x3=−1,=4,=0.24030307−1∣∣∣−140Work:R1=R1−R3:
24000317−1∣∣∣−140R2=R2−2R120000315−1∣∣∣−160
Swap R2 and R3:
2000301−15∣∣∣−106
Divide all the shits:
1000100.5−1/31∣∣∣−0.506/5Answer:x3=6/5x2−(1/3)(6/5)=0 so x2=2/5x1+(1/5)=−0.5 so x1=−7/10
5. Use Gauss–Jordan elimination to find the solution set to the following linear systems.
Given:2x1+3x2−x34x1+5x2−3x3−x1+3x2+5x3=−2=−2=−824−1353−1−35∣∣∣−2−2−8Work:R1=R1+R3:
14−16534−35∣∣∣−10−2−8R2=R2+4R3:
10−161734175∣∣∣−10−26−8R2=R2/17:
10−1613415∣∣∣−10−26/17−8R3=(R3+R1)/9:
100611411∣∣∣−10−26/17−2Answer:
The matrix is inconsistent because −26/17 can’t be equal to −2