wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

(i) If A=1-202 130-21, find A−1. Using A−1, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii) A=3-422 351 01, find A−1 and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii) A=1-202130-21 and B=72-6-21-3-42 5, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

Open in App
Solution

(i) Here, A=1-202130-21 A=1 1+6+22-0+0-4-0 =7+4+0 =11Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+113-21=7, C12=-11+22301=-2, C13=-11+3 210-2=-4C21=-12+1-20-21=2, C22=-12+21001=1, C23=-12+31-20-2=2C31=-13+1-2013=-6, C32=-13+21023=-3, C33=-13+31-221=5adj A=7-2-4212-6-35T = 72-6-21-3-425A-1=1Aadj A=11172-6-21-3-425or, AX=Bwhere, A=1-202130-21,X=xyz and B=1087Now, X=A-1BX=11172-6-21-3-4251087X=11170+16-42-20+8-21-40+16+35xyz=11144-3311 x=4, y=-3 and z=1

(ii) Here, A=3-42235101 A=3 3-0+42-5+20-3 =9-12-6 =-9Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+13501=3, C12=-11+22511=3, C13=-11+3 2310=-3C21=-12+1-4201=4, C22=-12+23211=1, C23=-12+33-410=-4C31=-13+1-4235=-26, C32=-13+23225=-11, C33=-13+33-423=17adj A=33-341-4-26-1117T = 34-2631-11-3-417A-1=1Aadj A=1-934-2631-11-3-417AX=BHere, A=3-42235101, X=xyz and B=-172X=A-1BX=1-934-2631-11-3-417-172X=1-9-3+28-52-3+7-223-28+34xyz=1-9-27-189 x=3, y=2 and z=-1

(iii) Here, A=1-202130-21 and B=72-6-21-3-425AB=1-202130-21 72-6-21-3-425AB=7+4+02-2+0-6+6+014-2-124+1+6-12-3+150+4-40-2+20+6+5 =110001100011AB=11100010001AB=11I3111AB=I3111BA=I3A-1=111BA-1=11172-6-21-3-425X=A-1BX=11172-6-21-3-4251087xyz=11170+16-42-20+9-21-40+16+35xyz=11144-3311 x=4, y=-3 and z=1

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Angle Bisectors
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon