424 VECTORS AND MATRICES [APP. A
A.7. Findx,y,z,t, where 3
[
xy
zt]
=[
x 6
− 12 t]
+[
4 x+y
z+t 3]
.First write each side as a single matrix:
[
3 x 3 y
3 z 3 t]
=[
x+ 4 x+y+ 6
z+t− 12 t+ 3]Set corresponding entries equal to each other to obtain the system of four equations.3 x=x+ 4 , 3 y=x+y+ 6 , 3 z=z+t= 1 , 3 t= 2 t+ 3or
2 x= 4 , 2 y= 6 +x, 2 z=t− 1 ,t= 3
The solution isx=2,y=4,z=1,t=3.A.8. Prove Theorem A.1(v):k(A+B)=kA+kB.
LetA=[aij]andB=[bij]. Then theijentry ofA+Bisaij+bij. Hencek(aij+bij)is theijentry ofk(A+B).
On the other hand, theijentries ofkAandkBarekaijandkbij, respectively. Thuskaij+kbijis theijentry ofkA+kB.
However, for scalars,k(aij+bij)=kaij+kbij. Thusk(A+B)andkA+kBhave the sameijentries. Therefore,
k(A+B)=kA+kB.MATRIX MULTIPLICATION AND TRANSPOSE
A.9. Calculate: (a)[ 3 ,− 2 , 5 ]⎡
⎣6
1
− 4⎤
⎦; (b)[ 2 ,− 1 , 7 , 4 ]⎡
⎢
⎢
⎣5
− 3
− 6
9⎤
⎥
⎥
⎦.Multiply corresponding entries and then add:(a)[ 3 ,− 2 , 5 ]⎡
⎣6
1
− 4⎤
⎦= 18 − 2 − 20 =−4. (b)[ 2 ,− 1 , 7 , 4 ]⎡
⎢
⎢⎣5
− 3
− 6
9⎤
⎥
⎥⎦= 10 + 3 − 42 + 36 = 7.A.10. LetA=[
13
2 − 1]
andB=[
20 − 4
3 − 26]
.Find: (a)AB;(b)BA.(a) SinceAis 2×2 andBis 2×3, the productABis defined and is a 2×3 matrix. To obtain the first row ofAB,
multiply the first row [1, 3] ofAby the columns[
2
3]
,[
0
− 2]
,[
− 4
6]
ofB, respectively:[
13
2 − 1][
2 0 − 4
3 − 26]
=[
1 ( 6 )+ 3 ( 3 ) 1 ( 0 )+ 3 (− 2 ) 1 (− 4 )+ 3 ( 6 )]=[
2 + 90 − 6 − 4 + 18]
=[
11 − 614]To obtain the entries in the second row ofAB, multiply the second row[ 2 ,− 1 ]ofAby the columns ofB,
respectively:
[
13
2 − 1][
2 0 − 4
3 − 26]
=[
11 − 614
4 − 30 + 2 − 8 − 6]ThusAB=[
11 − 614
12 − 14](b) Note thatBis 2×3 andAis 2×2. Since the inner numbers, 3 and 2, are not equal, the productBAis not defined.