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S Chand CLASS 10 Chapter 8 MATRICES Exercise 8 B

 Exercise 8 B


Question 1

 A=[32],B=[21],C[13]

Ans: 
i)
X=A+B+C=[32]+[21]+[13]=[32+1213]=[22]

(ii)
X=A+BC
=[32]+[21][13]
=[32121+3]=[04]

(iii) XA=BX=A+B
X=[32]+[21]=[3221]=[11]

(iv) A+X=B+CX=B+CA
so, X=[21]+[13][32]
=[2+13132]=[46]

(v) 2X=AC
=[32][13]
=[312+3]
=[25]
So X=12[25]
=[152]

Question 2

Ans: B=[2345],C=[1344]
BC=[2345][1344]
=[213+34+454]
=[1001]

Question 3

Ans: A=[2031],B=[0123]
2A3B=2[2031]3[0123]
=[4062][0369]
=[40036(6)29]
=[436+67]
=[4307]

Question 4

A=[1423],B=[4132]

(i) 2A+B=2[1423]+[4132]
=[2846]+[4132]
=[24814362]
2[2714]
 
(ii)  C+B=[0000]
=[0000]B
[0000][4131]
=[4132]

Question 5

Ans: 
P=[4628],Q=[231y]
 So P+2Q=[4628]+2[2311]
=[4628]+[4622]
=[4+466228+2]
=[8006]

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=52x=51=4x=42=2
and 8+y=0y=8
So, x=2,y=8

Question 7

Ans: [a342]+[2b12][112c]
=[5073]
[a+213+b14+1+222c]
[5073]
Comparing the correspending elements
b+2=0b=2c=3c=3So,a=4,b=2,c=3

Question 8

Ans: [2120]+2A=[3543]
2A=[3543][2120]
=[325+14230]=[5623]
A=12[5623]=[523132]

Question 9

Ans : 2[3x01]+3[1342]
=[27158]
[62x02]+[393y6]
=[27158]
[62x+90+3y2+6]
=[27158]
[92x+93y8]
Comparing the corresponding elements 2x+9=72x+792x=16 x=162=8
3y=15y=153=5
z=9
x=8,y=5,z=9

Question 10

Ans : M=[2012] and N=[2012]
M+2N=[2012]+2[2012]
=[2012]+[4024]
=[2+40+0122+4]
=[6016]

Question 11

(a) 
A=[3003],B=[1301],C=[1001]
 So AB+C=[3003][1301]+[1001]
[31+103+000+031+1]
=[3303]

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

Question 12

P=[3125],Q=[1640],R=[4123]

(i)  2p+3φR=2[3125]+3[1640][4123]
=[62410]+[318120][4123]
=[6342+18+1412210+03]
=[717107]

(ii) 4P2Q+3R=4[3125]2[1640]+3[4123]
=[124820][21280]+[12369]
=[122+1241238+8+6200+9]
=[2192229]

Question 13

Ans: [cos2θ0cot2θ1]+[sin2θ1cosec2θ0]+[0j10]
=[cos2θ+sin2θ+0cot2θcosec2θ1]
=[1+00111]{sin2θ+cos2θ=1cot2θcosec2θ=1}
=[1001]

Question 14

Ans: A=[xyzw],B=[xyzw],C=[0y2z0]
Now L.H.S. =(A+B)+C
={[xy2w1]+[xyzw]}+[0y2z0]
=[x+xyyz2w+w]+[0y2z0]
=[2x+00+y0+2z2w+0]=[2xy2z.2w]
R.H.S =A+(B+C)
=[xyzω]+{[xyzω]+[0y2z0]}
=[xyzb]+[x+0y+yz+2zw+0]
=[2xy2z2w] =LHS=RHS

Question 15

Ans: A=[8152],B=[46212]

Now 2A + 3B +4C is a null matrix
2[8152]+3[46212]+4C=[0000]
[162104]+[1218636]+4C=[0000]
[16122+181064+36]+4C=[0000]
[420440]+4C=[0000].
4C=[0000][420440]
4C=[0402004090]=[420490]
C=[15110]

Question 16

P=[6246],Q=[5320]
Now 3P - 2Q + 3X =O 
So, 3X = - 3P+2Q = 2Q - 3p 
=3[5320]3[6246]
=[10640][1861218]
=[10186+64120+18]=[8128+18]
Sox=13[812818]
=[834836]

Question 17

x+y=[97211]
xy=[5643]
=[95762+4113]=[41614]
X=12[41614]
[21237]

Subtracting ,We get 
2y=[97211][5643]
=[9+57+624113]=[141328]
y=12[141328]=[713134]

Question 18

Ans: 2A+B=[3427]...........(i)
A2B=[4311]...........(ii)
2A4B=2[4311]=[8622]..........(iii)
Multiplying (ii) by 2
Subtracting (iii) from (i)
5B=[3427][8622]
=[38462272]=[51005]
B=15[51005]=[1201]
Now 2A=[3427]B
=[427][1201]
=[3+14+22071]=[4226]
A=12[4226]=[2113]
A=[2113] and B=[1201]















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