S Chand CLASS 10 Chapter 8 MATRICES Exercise 8 B

 Exercise 8 B


Question 1

 ⇒A=[−32],B=[−2−1],C[1−3]

Ans: 
i)
⇒X=A+B+C=[32]+[−2−1]+[1−3]=[−3−2+12−1−3]=[2−2]

(ii)
⇒X=A+B−C
=[32]+[−2−1]−[1−3]
=[3−2−12−1+3]=[04]

(iii) ⇒X−A=B⇒X=A+B
X=[32]+[−2−1]=[3−22−1]=[11]

(iv) ⇒A+X=B+C⇒X=B+C−A
so, X=[−2−1]+[1−3]−[32]
=[−2+1−3−1−3−2]=[−4−6]

(v) ⇒2X=A−C
=[32]−[1−3]
=[3−12+3]
=[25]
So X=12[25]
=[152]

Question 2

Ans: B=[2−3−45],C=[1−3−44]
B−C=[2−3−45]−[1−3−44]
=[2−1−3+3−4+45−4]
=[1001]

Question 3

Ans: ⇒A=[2031],B=[01−23]
2A−3B=2[20−31]−3[01−23]
=[40−62]−[03−69]
=[4−00−3−6−(−6)2−9]
=[4−3−6+6−7]
=[4−30−7]

Question 4

⇒A=[1423],B=[−4−1−3−2]

(i) ⇒2A+B=2[1423]+[−4−1−3−2]
=[2846]+[−4−1−3−2]
=[2−48−14−36−2]
2[−2714]
 
(ii)  ⇒C+B=[0000]
=[0000]−B
[0000]−[−4−1−3−1]
=[4132]

Question 5

Ans: 
⇒P=[462−8],Q=[2−3−1y]
 So P+2Q=[462−8]+2[2−3−11]
=[462−8]+[4−6−22]
=[4+46−62−2−8+2]
=[800−6]

Question 6

Ans: 
⇒2[345u]+[1y01]=[70105]
⇒[68102x]+[1y01]=[70105]
⇒[6+18+y1002x+1]=[70105]
⇒[78+y102x+1]
=[70105]
Comparing the corresponding elements, 
2x+1=5⇒2x=5−1=4⇒x=42=2
and 8+y=0⇒y=−8
So, x=2,y=−8

Question 7

Ans: ⇒[a342]+[2b1−2]−[11−2c]
=[5073]
⇒[a+2−13+b−14+1+22−2−c]
⇒[5073]
Comparing the correspending elements
b+2=0⇒b=−2−c=3⇒c=−3So,a=4,b=−2,c=−3

Question 8

Ans: ⇒[2−120]+2A=[−3543]
⇒2A=[−3543]−[2−120]
=[−3−25+14−23−0]=[−5623]
A=12[−5623]=[−523132]

Question 9

Ans : ⇒2[3x01]+3[1342]
=[2−7158]
⇒[62x02]+[393y6]
=[2−7158]
⇒[62x+90+3y2+6]
=[2−7158]
⇒[92x+93y8]
Comparing the corresponding elements 2x+9=−7⇒2x+−7−9⇒2x=−16 ⇒x=−162=−8
3y=15⇒y=153=5
z=9
∴x=−8,y=5,z=9

Question 10

Ans : ⇒M=[2012] and N=[20−12]
M+2N=[2012]+2[20−12]
=[2012]+[40−24]
=[2+40+01−22+4]
=[60−16]

Question 11

(a) 
⇒A=[3003],B=[1301],C=[1001]
 So A−B+C=[3003]−[1301]+[1001]
⇒[3−1+10−3+00−0+03−1+1]
=[3−303]

(b)
= So , A and B are '2 × 2 Matrices 
A−B=A+B⇒−B=B
It is possible if B is a zero matrix

Question 12

⇒P=[−3−125],Q=[16−40],R=[4−12−3]

(i)  2p+3φ−R=2[−3−125]+3[16−40] - [4−123]
=[−6−2410]+[318−120]−[4−123]
=[−6−3−4−2+18+14−12−210+0−3]
=[−717−107]

(ii) ⇒4P−2Q+3R=4[−3−125]−2[16−40]+3[4−123]
=[−12−4820]−[212−80]+[12−369]
=[−12−2+12−4−12−38+8+620−0+9]
=[−2−192229]

Question 13

Ans: ⇒[cos2θ0cot2θ1]+[sin2θ1−cosec2θ0]+[0−j−10]
=[cos2θ+sin2θ+0cot2θ−cosec2θ−1]
=[1+001−11]{∵sin2θ+cos2θ=1cot2θ−cosec2θ=1}
=[1001]

Question 14

Ans: ⇒A=[xyzw],B=[x−y−zw],C=[0y2z0]
Now L.H.S. =(A+B)+C
={[xy2w1]+[x−y−zw]}+[0y2z0]
=[x+xy−yz−2w+w]+[0y2z0]
=[2x+00+y0+2z2w+0]=[2xy2z.2w]
R.H.S =A+(B+C)
=[xyzω]+{[x−y−zω]+[0y2z0]}
=[xyzb]+[x+0−y+y−z+2zw+0]
=[2xy2z2w] =L⋅H⋅S=R⋅H⋅S

Question 15

Ans: ⇒A=[8152],B=[−46−212]

Now 2A + 3B +4C is a null matrix
2[8152]+3[−46−212]+4C=[0000]
⇒[162104]+[−1218−636]+4C=[0000]
⇒[16−122+1810−64+36]+4C=[0000]
⇒[420440]+4C=[0000].
⇒4C=[0000]−[420440]
⇒4C=[0−40−200−40−90]=[−4−20−4−90]
⇒C=[−1−5−1−10]

Question 16

⇒P=[6−24−6],Q=[5320]
Now 3P - 2Q + 3X =O 
So, 3X = - 3P+2Q = 2Q - 3p 
=3[5320]−3[6−24−6]
=[10640]−[18−612−18]
=[10−186+64−120+18]=[−812−8+18]
Sox=13[−812−818]
=[−834−836]

Question 17

⇒x+y=[97211]
x−y=[−5−643]
=[9−57−62+411−3]=[41614]
∴X=12[41614]
∴[21237]

Subtracting ,We get 
2y=[97211]−[−5−643]
=[9+57+62−411−3]=[1413−28]
y=12[1413−28]=[713−134]

Question 18

Ans: ⇒2A+B=[3−427]...........(i)
A−2B=[4311]...........(ii)
2A−4B=2[4311]=[8622]..........(iii)
Multiplying (ii) by 2
Subtracting (iii) from (i)
5B=[3−427]−[8622]
=[3−84−62−27−2]=[−5−1005]
B=15[−5−1005]=[−1−201]
Now 2A=[3−427]−B
=[−−427]−[−1−201]
=[3+1−4+22−07−1]=[4−226]
A=12[4−226]=[2−113]
∴A=[2−113] and B=[−1−201]















Labels: , , , ,

No comments:

Post a Comment

Contact Form

Name

Email *

Message *