Matematikos uždavinių sprendimas
1. Užduotis.
Apskaičiuokite: a) (A+2BT)C b) 3B+2C , kai
|A= | 5 2 |B= | |C= | |BT=|3 -2 |
| |-2 4 | |3 1 7 | |-1 -1 | |1 3 |
| |2 1 | |-2 3 5 | |2 | |7 5 |
| | | | | |-2 3 | | |
| | | | | |-2 | | |
a) ( A + 2BT ) C
| | 5 | 3 | 5 2 | 6 -4|5+6 |11 |
|A + 22BT |2 |-2 |-2 4 |2 6 |2+(-4) |-2 |
|= |-2 |+ 2 1 |+ |= |(-2)+2 4+6 |0 |
| |4 |3 = |2 1 |14 10 |= |10 |
| |2 |7 5 | | |2+14 1+10 |16 |
| |1 | | | | |11 |
| |11 | | |11*(-1)+(-2)*(-2) 11*(-1)+(-2)*3 |
|( A + 2BT ) |-2 |-1 -1 |=|11*2+(-2)*(-2) |
|C = |0 |2 | |0*(-1)+10*(-2) 0*(-1)+10*3 |
| |10 |-2 3 | |0*2+10*(-2) = |
| |16 |-2 | |16*(-1)+11*(-2) 16*(-1)+11*3 |
| |11 | | |16*2+11*(-2) |
| |(-11)+4 (-11)+(-6) | -7 -17 |
|= |22+4 |26 |
| |0+(-20) 0+30 |-20 30 |
| |0+(-20) = |-20 |
| |(-16)+(-22) (-16)+33 |-38 17 |
| |32+(-22) |10 |
b) 3B + 2C
|3| 3 1 |+2| -1 -1|=| 9 3 |+|-2 -2|=| 7 1 25 |
| |7 | |2 | |21 | |4 | |-10 15 11 |
| |-2 3 | |-2 3 | |-6 9 | |-4 6| | |
| |5 | |-2 | |15 | |-4 | | |
2. Užduotis.
Apskaičiuokite determinantą:
| 2 4 | |=| 2 0 |=|1*(-1)4| |
|2 4 | | |0 -2 | |+3 |2 0 -2 |
|-2 1 | | |-2 1 0| | |-2 1 -1|
|0 -1 |*2+ | |-1 | | |= |
|3 -6 -2|*(-2)+ | |3 -2 0 | | |3 -2 8 |
|2 | | |8 | | | |
|0 2 1| | |0 2 1 | | | |
|3 | | |3 | | | |
=(-1) * (16+0-8+6-0-4)=(-1) ** 10=-10
3. Užduotis.
Sudarykite atvirkštinę matricą , kai
4 0 1
A = -2 1 1
1 -2 2
det A = 8+0+4-1-0+8=19
|A11=(-1)1| 1 |=2+2=4 |
|+1 |1 | |
| |-2 | |
| |2 | |
| | | |
|A12=(-1)1|-2 |=-(-4-1)=5 |
|+2 |1 | |
| |1 2| |
| | | |
|A13=(-1)1|-2 |=4-1=3 |
|+3 |1 | |
| |1 | |
| |-2 | |
| | | |
|A21=(-1)2| 0 |=-(0+2)=-2 |
|+1 |1 | |
| |-2 | |
| |2 | |
| | | |
|A22=(-1)2|4 1|=8-1=7 |
|+2 |1 2| |
| | | |
|A23=(-1)2| 4 |=-(-8-0)=8 |
|+3 |0 | |
| |1 | |
| |-2 | |
| | | |
|A31=(-1)3|0 1|=0-1=-1 |
|+1 |1 1| |
| | | |
|A32=(-1)3| 4 |=-(4+2)=-6 |
|+2 |1 | |
| |-2 | |
| |1 | |
| | | |
|A33=(-1)3| 4 |=4+0=4 |
|+3 |0 | |
| |-2 | |
| |1 | |
|A-1 * A=|4 -2 | 4 0 |4*4+(-2)*(-2)+(-1)*1 4*0+(-2)*1+(-1)*(-2) |
| |-1 |1 |4*1+(-2)*1+(-1)*2 |
| |5 7 |-2 1 |5*4+7*(-2)+(-6)*1 5*0+7*1+(-6)*(-2) |
| |-6 * |1 = |5*1+7*1+(-6)*2 = |
| |3 8 |1 -2 |3*4+8*(-2)+4*1 3*0+8*1+4*(-2) |
| |4 |2 |3*1+8*1+4*2 |
| 16+4-1 0-2+2 |19 0 0 |
|4-2-2 |0 19 0 |
|= 20-14-6 0+7+12 |0 0 19 |
|5+7-12 = | |
|12-16+4 0+8-8 3+8+8 | |
|A-1=1/19|4 -2 -1| 4 0 1 |1/19|19 0 0|
| |5 7 -6|-2 1 1 | |0 19 0|
| |* |= | |0 0 19|
| |3 8 4|1 -2 2 | | |
4. Užduotis.
Išspręskite lygčių sistemą determinantų pagalba ir atvirkštinės matricos
metodu:
x + 2y + z = 4
3x – 5y + 3z = 1
2x + 7y – z = 8
a) Sprendimas atvirkštinės matricos metodu
A – pagrindinė matrica iš koeficientų prie nežinomųjų;
X – nežinomųjų matrica;
B – laisvųjų radikalų matrica.
|A= |1 2 1 |X= ||x |B= |4 |
| |3 -5 3 | |y | |1 |
| |2 7 -1 | |z | |8 |
det A= 5+12+21+10+6-21=33
|A11=(-1)1|-5 |=5-21=-16 |
|+1 |3 | |
| |7 | |
| |-1 | |
| | | |
|A12=(-1)1| 3 |=-(-3-6)=9 |
|+2 |3 | |
| |2 -1| |
| | | |
|A13=(-1)1|3 |=21+10=31 |
|+3 |-5 | |
| |2 | |
| |7 | |
| | | |
|A21=(-1)2| 2 |=-(-2-7)=9 |
|+1 |1 | |
| |7 | |
| |-1 | |
| | | |
|A22=(-1)2|1 |=(-1)-2=-3 |
|+2 |1 | |
| |2 | |
| |-1 | |
| | | |
|A23=(-1)2| 1 |=-(7-4)=-3 |
|+3 |2 | |
| |2 7| |
| | | |
|A31=(-1)3| 2 |=6+5=11 |
|+1 |1 | |
| |-5 | |
| |3 | |
| | | |
|A32=(-1)3| 1 |=-(3-3)=0 |
|+2 |1 | |
| |3 3| |
| | | |
|A33=(-1)3| 1 |=(-5)-6=-11 |
|+3 |2 | |
| |3 | |
| |-5 | |
x=A-1*B
| |-16 9 |4 |1/33 *|(-16)*4+9*1+11*8 |(-64)+9+88 |
|x=1/3|11 |1 | |9*4+(-3)*1+0*8 |36+(-3)+0 |
|3 |9 -3 |= | |= 1/33 * |= |
| |0 * |8 | |31*4+(-3)*1+(-11)*8 |124+(-3)+(-88)|
| |31 -3 | | | | |
| |-11 | | | | |
|=1/33|33 |1 |
| |33 |1 |
| |= |1 |
| |33 | |
| x + 2y + z = 4 | 1 + 2*1 + 1 = 4 |
|3x – 5y + 3z = 1 |3*1 – 5*1 + 3*1 = 1 |
|2x + 7y – z = 8 |2*1 + 7*1 – 1 = 8 |
b) Sprendimas determinantų pagalba
x + 2y + z = 4
3x – 5y + 3z = 1
2x + 7y – z = 8
|∆= |1 2 |= 5+12+21+10+6-21=33 |
| |1 | |
| |3 -5 | |
| |3 | |
| |2 7 | |
| |-1 | |
| | | |
|∆x= |4 2 |= 20+48+7+40+2-84=33 |
| |1 | |
| |1 -5 | |
| |3 | |
| |8 7 | |
| |-1 | |
| | | |
|∆y= |1 4 |= (-1)+24+24-2+12-24=33 |
| |1 | |
| |3 1 | |
| |3 | |
| |2 8 | |
| |-1 | |
| | | |
|∆z= |1 2 |= (-40)+4+84+40-48-7=33 |
| |4 | |
| |3 -5 | |
| |1 | |
| |2 7 | |
| |8 | |
|x=|∆x |=|33 |=1 |
| |∆ | |33 | |
|y=|∆y |=|33 |=1 |
| |∆ | |33 | |
|z=|∆z |=|33 |=1 |
| |∆ | |33 | |
| x + 2y + z = 4 | 1 + 2*1 + 1 = 4 |
|3x – 5y
+ 3z = 1 |3*1 – 5*1 + 3*1 = 1 |
|2x + 7y – z = 8 |2*1 + 7*1 – 1 = 8 |
5. Užduotis.
|Ekonominės sistemos technologinė | 0 |. Koks turi būti gamybos |
|matrica A = |0,5 |planas |
| |0,6 | |
| |0,4 | |
X = ( x1x2 )T , kad būtų patenkinta paklausa C = ( 50 200 )T ?
| | 0 |X = ( x1x2|X = |x1 |C = ( 50 |C = | 50 |
|A= |0,5 |)T | |x2 |200 )T | |200 |
| |0,6 | | | | | | |
| |0,4 | | | | | | |
( E – A ) * X = C
|0 |-| 0 |*|x1 |=| 50 |
|0 1 | |0,5 | |x2 | |200 |
| | |0,6 | | | | |
| | |0,4 | | | | |
x1 – 0,5 x2 = 50 / *0,6
-0,6 x1 + 0,6 x2 = 200
x1 – 0,5 xx2 = 50
0,3 x2 = 230
x1 – 0,5 x2 = 50
x2 = 230 / 0,3 = 766,7
x1 – 0,5 *766,7 = 50
x2 = 766,7
x1 – 383,4 = 50
x2 = 766,7
x1 = 50 ++ 383,4
x2 = 766,7
x1 = 433,4
x2 = 766,7
433,4
766,7
———————–
2 4 2 4
-2 1 0 -1
3 -6 -2 2
0 2 1 3
