ML Aggarwal Solution Class 10 Chapter 8 Matrices Exercise Test
Test
Question 1
Find the values of a and below
Sol :
[a+3b2+20−6]=[2a+13b0b2−5b]
comparing the corresponding elements
a + 3 = 2a + 1
⇒ 2a – a =3 – 1
⇒ a = 2
b² + 2 = 3b
⇒ b² – 3b + 2 = 0
⇒ b² – b – 2b + 2 = 0
⇒ b (b – 1) – 2 (b – 1) = 0
⇒ (b – 1) (b – 2) = 0.
Either b – 1 = 0, then b = 1
or b – 2 = 0, then b = 2
Hence a = 2, b = 2 or 1
Question 2
Find a, b, c and d if 3[abcd]=[4a+bc+d3]+[a6−12d]
Sol :
Given :
3[abcd]=[4a+bc+d3]+[a6−12d]
⇒[3a3b3c3d]=[4+aa+b+6c+d−13+2d]
Comparing the corresponding elements:
3a=4+a
⇒3a−a=4
⇒2a=4
∴a=2
3b=a+b+6⇒3b−b=2+6
⇒2b=8
∴b=4
3d=3+2d
⇒3d−2d=3
∴d=3
3c=c+d-1
⇒3c−c=3−1
2c=2
⇒c=1
Hence a=2, b=4, c=1, d=3
Question 3
Find X if Y=[3214] and 2X+Y=[10−32]
Sol :
Given :
2X+Y=[10−32]
⇒2X=2X+Y=[10−32]−Y
⇒2X=[10−32]−[3214]
=[1−30−2−3−12−4]=[−2−2−4−2]
X=12[−2−2−4−2]=[−1−1−2−1]
Question 4
Determine the matrices A and B when
A+2B=[126−3]...(i)
2A−B=[2−12−1]...(ii)
Multiplying (i) by 1 and (ii) by 2
Question 5
=[64−517]
Now A(4B−3C)=[12−34][64−517]
=[1×6+2(−5)1×4+2×17−3×6+4×(−5)−3×4+4×17]
=[6−104+34−18−20−12+68]=[−438−3856]
Question 6
If A=[1410],B=[213−1] and C=[2305] compute (AB)C = (CB)A ?
(AB) C=[[1410]×[213−1]][2305]
=[2+121−42+01+0][2305]
=[14−321][2305]
=[28+042−154+06+5]=[2827411]
(CB) A=[[2305][213−1]][1410]
=[4+92−30+150−5][1410]
=[13−115−5][1410]=[13−152+015−560+0]
=[12521060]
It is clear from above that
(AB)≠(CB)A
Question 7
If A=[3205] and B=[1012] find the each of the following and state it
they are equal:
(i) (A+B)(A−B)
(ii) A2−B2
Sol :
Given :
A=[3205]
B=[1012]
(i) (A+B)(A-B)
={[3205]+[1012]}×{[3205]−[1012]}
=[3+12+00+15+2]×[3−12−00−15−2]
=[4217]×[22−13]
=[8−28+62−72+21]=[614−523]
(ii) A2−B2
=[3205]×[3205]−[1012]×[1012]
=[9+06+100+00+25]−[1+00+01+20+4]
=[916025]−[1034]=[9−116−00−325−4]
=[816−321]
We see that (A+B)(A−B)≠A2−B2
Question 8
If A=[3−5−42] find A2−5A−141
Where I is unit matrix of order 2 x 2
Sol :
Given
∴A2−5 A−14I=[29−25−2024]
−[15⋅−25−2010]−14[1001]
=[29−25−2024]−[15−20−2010]−[140014]
=[29−15−14−25+25−0−20+20+024−10−14]=[0000]
Question 9
If A=[33pq] and A2=0 find p and q
Sol :
Given :
A=[33pq]
A2=A×A=[33pq][33pq]
=[9+3p9+3q3p+pq3p+q2]
But A2=0
∴[9+3p9+3q3p+pq3p+q2]=[0000]
Comparing the corresponding elements
9+3p=0
⇒3p=−9⇒p=−3
9+3q=0
⇒3q=−9⇒q=−3
Hence p=-3, q=-3
Question 10
If A=[3525xy] and A2=1, find x, y
Sol :
Given :
A=[3525xy]
A2=A×A=[3525xy][3525xy]
=[925+25x625+25y35x+xy25x+y2]
But A2=I=[1001]
[925+25x625+25y35x+xy25x+y2]=[1001]
Comparing the corresponding elements,
925+25x=1
⇒25x=1−925=1625
x=1625×52=85
625+25y=0
⇒25y=−625
y=−625×52=−35
Hence x=85,y=−35
Question 11
If [−1001][abcd]=[100−1] find a, b, c and d
Sol :
[−1001][abcd]=[100−1]
⇒[−a+0−b+00+c0+d]=[100−1]
⇒[−a−bcd]=[100−1]
Comparing the corresponding elements
−a=1⇒a=−1
−b=0⇒b=0
c=0 and d=−1
Hence a=-1, b=0, c=0, d=-1
Question 12
Find a and b if
Sol :
Given :
[a−bb−4b+4a−2][2002]=[−2−2140]
⇒[2a−2b+00+2a−2b2b+8+00+2a−4]=[−2−2140]
⇒[2a−2b2a−2b2b+82a−4]=[−2−2140]
Comparing the corresponding elements
$2 a-4=0
⇒2a=4⇒a=2
2 a-2 b=-2
⇒2×2−2b=−2
⇒4−2b=−2
⇒−2b=−2−4=−6
⇒b=3
Hence a=2, b=3
Question 13
If A=[sec60∘cos90∘−3tanA5∘sin90∘] and B=[0cos45∘−23sin90∘]
Find (i) 2A – 3B (ii) A² (iii) BA
Sol :
Given
A=[sec60∘cos90∘−3tan45∘sin90∘]=[20−31]
(∵sec60∘)=2,cos90∘=0,tan45∘=1sin90∘=1)
B=[0cot45∘−23sin90∘]=[01−23]
(∵cot45∘=1)
(i) 2A-3B
=2[20−31]−3[01−23]
=[40−62]−[03−69]
=[4−00−3−6+62−9]=[4−30−7]
(ii) A2=A×A=[20−31][20−31]
=[4+00+0−6−30+1]=[40−91]
(iii) BA=[01−23][20−31]
=[0−30+1−4−90+3]=[−31−133]
Comments
Post a Comment