100%(1)1 out of 1 people found this document helpful
This preview shows page 17 - 20 out of 51 pages.
There are no solutions.14. The system reduces tox+2y=-2z=2-→x=-2-2yz= 2.Lety=t.xyz=-2-2tt215. The system reduces tox=4y=2z=1.16. The system reduces tox1+ 2x2+ 3x3+5x5=6x4+2x5=7-→x1= 6-2x2-3x3-5x5x4= 7-2x5.Letx2=r, x3=s, andx5=t.17
Chapter 1ISM:Linear Algebrax1x2x3x4x5=6-2r-3s-5trs7-2tt17. The system reduces tox1=-82214340x2=85918680x3=4695434x4=-459434x5=699434.18. a. No, since the third column contains two leading ones.b. Yesc. No, since the third row contains a leading one, but the second row does not.d. Yes19.0000and100020. Four, namely0000,1k00,0100,1001(kis an arbitrary constant.)21. Four, namely000000,1k0000,010000,100100(kis an arbitrary constant.)22. Seven, namely000000,1ab000,01c000,001000,10d01e,1f0001,010001.Here,a, b, . . . , fare arbitrary constants.23. We need to show that the matrix has the three properties listed on page 16.18
ISM:Linear AlgebraSection 1.2Property a holds by Step 2 of the Gauss-Jordan algorithm (page 17).Property b holds by Step 3 of the Gauss-Jordan algorithm.Property c holds by Steps 1 and 4 of the algorithm.24. Yes; each elementary row operation is reversible, that is, it can be “undone.” For example,the operation of row swapping can be undone by swapping the same rows again.Theoperation of dividing a row by a scalar can be reversed by multiplying the same row bythe same scalar.25. Yes; ifAis transformed intoBby a sequence of elementary row operations, then we canrecoverAfromBby applying the inverse operations in the reversed order (compare withExercise 24).26. Yes, by Exercise 25, since rref(A) is obtained fromAby a sequence of elementary rowoperations.27. No; whatever elementary row operations you apply to123456789, you cannot make thelast column equal to zero.28. Suppose (c1, c2, . . . , cn) is a solution of the systema11x1+a12x2+· · ·+a1nxn=b1a21x1+a22x2+· · ·+a2nxn=b2. . . . . . . . ..To keep the notation simple, suppose we addktimes the first equation to the second;then the second equation of the new system will be (a21+ka11)x1+· · ·+(a2n+ka1n)xn=b2+kb1.We have to verify that (c1, c2, . . . , cn) is a solution of this new equation. Indeed, (a21+ka11)c1+· · ·+ (a2n+ka1n)cn=a21c1+· · ·+a2ncn+k(a11c1+· · ·+a1ncn) =b2+kb1.We have shown that any solution of the “old” system is also a solution of the “new.” Tosee that, conversely, any solution of the new system is also a solution of the old system,note that elementary row operations are reversible (compare with Exercise 24); we canobtain the old system by subtractingktimes the first equation from the second equationof the new system.