428 VECTORS AND MATRICES [APP. A
A.22. Reduce the matrixA=⎡
⎣1 −23 12
114 − 13
259 − 28⎤
⎦to row canonical form.First reduceAto echelon form by applying the operations “Add−R 1 toR 2 ” and “Add− 2 R 1 toR 3 ,” and then
the operation “Add− 3 R 2 toR 3 .” These operations yieldA∼⎡
⎣1 −23 12
031 − 21
093 − 44⎤
⎦∼⎡
⎣1 −23 12
031 − 21
00021⎤
⎦Now use back-substitution on the echelon matrix to obtain the row canonical form ofA. Specifically, first multiply
R 3 by^12 to obtain the pivota 34 =1, and then apply the operations “Add 2R 3 toR 2 ” and “Add−R 3 toRt.” These
operations yieldA∼⎡
⎣1 −23 12
031 − 21(^000112)
⎤
⎦∼
⎡
⎣
1 − (^23032)
0 3102
0 001^12
⎤
⎦
Now multiplyR 2 by^13 making the pivota 22 =1, and then apply the operation “Add 2R 2 toR 1 .” We obtain
A∼
⎡
⎢
⎣
1 − (^23032)
(^0113023)
(^000112)
⎤
⎥
⎦∼
⎡
⎢
⎣
(^101130176)
(^0113023)
00 01^12
⎤
⎥
⎦
Sincea 11 =1, the last matrix is the desired row canonical form ofA.
A.23. Solve each system using its augmented matrixM:
(a)
x+ y− 2 z+ 4 t= 5
2 x+ 2 y− 3 z+ t= 4
3 x+ 3 y− 4 z− 2 t= 3
(b)
x− 2 y+ 4 z= 2
2 x− 3 y+ 5 z= 3
3 x− 4 y+ 6 z= 7
(a) Reduce its augmented matrixMto echelon form and then to row canonical form:
M=
⎡
⎣
11 − 245
22 − 314
33 − 4 − 23
⎤
⎦∼
⎡
⎣
11 − 245
00 1− 7 − 6
00 21412
⎤
⎦∼
[
110 − 10 − 7
001 − 7 − 6
]
(The third row of the second matrix is deleted since it is a multiple of the second row and will result in a zero row.)
Write down the system corresponding to the row canonical form ofMand then transfer the free variables to
the other side to obtain the free variable form of the solution:
x+y− 10 t=− 7
z− 7 t=− 6 and then
x=− 7 −y+ 10 t
z=− 6 + 7 t
Herexandzare the basic variables andyandtare the free variables.
Theparametricform of the solution can be obtained by setting the free variables equal toparameters, sayy=
a andt = b. This process yieldsx =− 7 −a+ 10 b, y = a,z =− 6 + 7 b, t = b or
u=(− 7 −a+ 10 b, a,− 6 + 7 b,b)(whichis another form of the solution).
Aparticular solutioncan be obtained by assigning any values to the free variables (or parameters) and
solving for the basic variables using either form of the general solution. For example, settingy= 2 ,t=3, we
obtainx= 21 ,z=15. Thus the following is a particular solution of the system:
x= 21 ,y= 2 ,z= 15 ,t=3oru=( 21 , 2 , 15 , 3 )
(b) First row reduce its augmented matrixMto echelon form:
M=
⎡
⎣
1 − 242
2 − 353
3 − 467
⎤
⎦∼
⎡
⎣
1 − 242
01 − 3 − 1
02 − 61
⎤
⎦∼
⎡
⎣
1 − 242
01 − 3 − 1
0003
⎤
⎦
In echelon form, the third row corresponds to the degenerate equation 0x+ 0 y+ 0 z=3.
Thusthesystem has no solution. (Note that the echelon form indicates whether or not the system has a solution.)