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RD Sharma Class 12 Exercise 6.1 Adjoint and inverse of Matrix Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 6.1 Adjoint and inverse of Matrix Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 02:54 PM IST

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RD Sharma Class 12 Solutions Chapter6 Adjoint & Inverse of Matrix - Other Exercise

Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 (i)

Answer:
Adj(A)=[4−2−5−3]
Hint:
Here, we use basic concept of adjoint of matrix .
Given:
AdjA=[−3524]
Solution:
A=[−3524]
|A|=−3×4−5×2=−12−10=−22C11=4,C12=−2,C21=−5,C22=−3
Cij=[4−2−5−3]
Taking transpose,
Adj(A)=[4−5−2−3]Adj(A)×A=[4−5−2−3][−3524]=[−12+(−10)20−206−6−10−12]
=[−2200−22] (1)
|A|I=−22[1001]=[−2200−22] (2)
A(Adj(A))=[−3524][4−5−2−3]
=[−2200−22] (3)
From equation (1), (2) and (3)
A(Adj(A))=|A|I=Adj(A)×A

Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 (ii)

Answer:
Adj(A)=[d−b−ca]
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
A=[abcd]
Solution:
Let
A=[abcd]|A|=|abcd||A|=a×d−c×b|A|=ad−cb
Let’s find cofactor
C11=d,C12=−c,C21=−b,C22=aCij=[d−c−ba]
Let’s transpose Cij
Adj(A)=[d−b−ca]
Let’s prove this below
Adj(A)×A=|A|I=A×Adj(A)
Adj(A)×A=[d−b−ca]×[abcd]=[da+(−bc)db−bd−ca+ac−cb+ad]Adj(A)×A=[da−bc00da−bc] (1)
MHS
|A|I=ad−cb[1001]=[ad−cb00ad−cb]=[da−bc00ad−cb] (2)
RHS
A×Adj(A)=[abcd]×[d−b−ca]
=[ad−bc00da−bc] (3)
From equation (1), (2) and (3)
Adj(A)×A=|A|I=A×Adj(A)

Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 (iii)

Answer:
[cos⁡α−sin⁡α−sin⁡αcos⁡α]
Hint:
Here, we use basic concept of determinant.
Given:
A=[cos⁡αsin⁡αsin⁡αcos⁡α]
Solution:
Let’s find |A|
|A|=cos2⁡α−sin2⁡α=cos⁡2α
Let’s find cofactor
C11=cos⁡α,C12=−sin⁡α,C21=−sin⁡α,C22=cos⁡αCij=[cos⁡α−sin⁡α−sin⁡αcos⁡α]
Let’s transpose Cij
Adj(A)=[cos⁡α−sin⁡α−sin⁡αcos⁡α]
Let’s prove below
Adj(A)×A=|A|×A=A×Adj(A)Adj(A)×A=[cos⁡α−sin⁡α−sin⁡αcos⁡α]×[cos⁡αsin⁡αsin⁡αcos⁡α]
=[cos2⁡α−sin2⁡αcos⁡αsin⁡α−cos⁡αsin⁡α−sin⁡αcos⁡α+cos⁡αsin⁡αcos2⁡α−sin2⁡α]
=[cos⁡2α00cos⁡2α] (1)
|A|I=cos⁡2α×[1001]
=[cos⁡2α00cos⁡2α] (2)
A×Adj(A)=[cos⁡αsin⁡αsin⁡αcos⁡α]×[cos⁡α−sin⁡α−sin⁡αcos⁡α]=[cos2⁡α−sin2⁡α00cos2⁡α−sin2⁡α]
=[cos⁡2α00cos⁡2α] (3)
From equation (1), (2) and (3)
A×Adj(A)=|A|I=Adj(A)×A

Adjoint and Inverse of a Matrix exercise 6.1 question 1 (iv)

Answer:
Adj(A)=|1tan⁡α2−tan⁡α21|
Hint:
Here, we use basic concept of determinant.
Given:
A=|1tan⁡α2−tan⁡α21|
Solution:
Let’s find |A|
|A|=|1tan⁡α2−tan⁡α21|
|A|=1−[(−tan⁡α2)(tan⁡α2)]|A|=1+tan2⁡α2|A|=sec2⁡α2
Let’s find cofactor
C11=1,C12=tan⁡α2,C21=−tan⁡α2,C22=1
Cij=[1tan⁡α2−tan⁡α21]
Take transpose of Cij
Adj(A)=[1−tan⁡α2tan⁡α21]
Let’s prove below
Adj(A)×A=|A|I=A×Adj(A)Adj(A)×A=[1−tan⁡α2tan⁡α21]×[1tan⁡α2−tan⁡α21]
=[1+tan2⁡α2tan⁡α2−tan⁡α2tan⁡α2−tan⁡α21+tan2⁡α2]=[sec2⁡α200sec2⁡α2] (1)
|A|I=sec2⁡α2×[1001]=[sec2⁡α200sec2⁡α2] (2)
A×Adj(A)=[1tan⁡α2−tan⁡α21]×[1−tan⁡α2tan⁡α21]
=[1+tan2⁡α2001+tan2⁡α2]=[sec2⁡α200sec2⁡α2] (3)
From equation (1), (2) and 3
Adj(A)×A=|A|×I=A×Adj(A)
Hence proved

Adjoint and Inverse of a Matrix exercise 6.1 question 2 (i)

Answer:
Adj(A)=[−3222−3222−3]
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
A=[122212221]
Solution:
Let’s find |A|
|A|=1|1221|−2|2221|+2|2122|=1(1−4)−2(2−4)+2(4−2)=(−3)+4+4=5
Let’s find cofactors
C11=+(1−4)=−3C12=−(2−4)=2C13=+(4−2)=2C21=−(2−4)=2C22=+(1−4)=−3

C23=−(2−4)=2C31=+(4−2)=2C32=−(2−4)=2C33=+(1−4)=−3Cij=[−3222−3222−3]
Let’s transpose it
Adj(A)=[−3222−3222−3]
Let’s verify below
Adj(A)×A=|A|I=A×Adj(A)Adj(A)×A=[−3222−3222−3]×[122212221]
=[−3+4+40004−3+40004+4−3]
=[500050005] (1)
|A|I=5×[100010001]=[500050005] (2)
A×Adj(A)=[122212221]×[−3222−3222−3]=[−3+4+4000−3+4+4000−3+4+4]
=[500050005] (3)
From equation (1), (2) and (3)
A×Adj(A)=|A|I=Adj(A)×A

Adjoint and Inverse of a Matrix exercise 6.1 question 2 (ii)

Answer:
Adj(A)=[23−13−3695−3−1]
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
A=[125231−111]
Solution:
Let’s find |A|
|A|=1|3111|−2|21−11|+5|23−11|=1(3−1)−2(2+1)+5(2+3)=2−6+25=21
Let’s find cofactors
C11=+(3−1)=2C12=−(2+1)=−3C13=+(2+3)=5C21=−(2−5)=3C22=+(1+6)=6
C23=−(1+2)=−3C31=+(2−15)=−13C32=−(1−10)=9C33=+(3−4)=−1
Cij=[2−3536−3−139−1]
Let’s take transpose of Cij
Adj(A)=[23−13−3695−3−1]
Let’s verify below
Adj(A)×A=|A|I=A×Adj(A)Adj(A)×A=[23−13−3695−3−1]×[125231−111]=[210002100021] (1)
|A|I=21×[100010001]=[210002100021] (2)
A×Adj(A)=[125231−111]×[23−13−3695−31]=[210002100021] (3)
From equation (1), (2) and (3)
Adj(A)×A=|A|I=A×Adj(A)

Adjoint and Inverse of Matrices Excercise 6.1 Question 2 (iii).
Answer:

Adj(A)=[−2211−114−2216−88]
Hint:
Here, we use basic concept of determinant and adjoint of matrix.
Given:
A=|2−1342504−1|
Solution:
Let’s find |A|
|A|=|2−1342504−1|
|A|=2|254−1|−(−1)|450−1|+3|4204|=2(−2−20)+1(−4−0)+3(16−0)=−44−4+48=0|A|=0
Let’s find cofactor
C11=+(−2−20)=−22C12=−(−4+0)=4C13=+(16−0)=16C21=−(1−12)=11C22=+(−2−0)=−2C23=−(−8−0)=−8C31=+(−5−6)=−11C32=−(10−12)=2C33=+(4+4)=8
Cij=[−2241611−2−8−1128]
Let’s take transpose of Cij
Adj(A)=[−2211−114−2216−88]
Let’s prove this below
Adj(A)×A=|A|I=A×Adj(A)Adj(A)×A=[−2211−114−2216−88]×[2−1342504−1]
=[−44+44+0000−4+(−4)+800048−40−8]
=[000000000] (1)
|A|I=0×[100010001]=[000000000] (2)
A×Adj(A)=[2−1342504−1]×[−224−114−2216−88]=[000000000] (3)
So, from equation (1), (2) and (3)
Adj(A)×A=|A|I=A×Adj(A)

Adjoint and Inverse of Matrices Excercise 6.1 Question 2 (iv).

Answer:
Adj(A)=[3−11−157−54−22]
Hint:
Here, we use basic concept of determinant and adjoint of matrix.
Given:
A=[20−1510113]
Solution:
Let’s find |A|
|A|=|20−1510113|=2|1013|−0|5013|+(−1)|5111|=2(3)−0−1(5−1)=6−4=2|A|=2
Let’s find cofactor
C11=+(3−0)=3C12=−(15−0)=−15C13=+(5−1)=4C21=−(0+1)=−1C22=+(6+1)=7C23=−(−2−0)=−2C31=+(2−1)=1C32=−(0+5)=−5C33=+(2−0)=2
Cij=[3−154−17−21−52]
Let’s take the transpose of Cij
Adj(A)=[3−11−157−54−22]
Let’s prove,
Adj(A)×A=|A|I=A×Adj(A)Adj(A)×A=[3−11−157−54−22]×[20−1510113]=[6+(−5)+10000+7−5000−4+0+6]
=[200020002] (1)
|A|I=2×[100010001]=[200020002] (2)
A×Adj(A)=[20−1510113]×[3−11−157−54−22]=[200020002] (3)
From the equation (1), (2) and (3)
Adj(A)×A=|A|I=A×Adj(A)

Adjoint and Inverese of Matrices Exercise 6.1 Question 3

Answer:
Proved A(Adj(A))=0
Hint:
Here, we have to use advance method of finding adjoint of matrix.
Given:
A=[1−1123018210]
Solution:
We know that,
A×Adj(A)=|A|I Then,
|A|I=0 (1)
From the equation (1)
|A|I=0
|1−1123018210|×1=0 since, |I|=1
1|30210|−(−1)|201810|+1|23182|=0
0=1(30)+(20−0)+(4−54)0=30+20+4−540=54−540=0
So, A×Adj(A)=|A|I=0

Adjoint and Inverese of Matrices Exercise 6.1 Question 4

Answer:
Proved Adj(A)=A
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
A=[−4−3−3101443]
Solution:
Here, let’s find cofactor
C11=+(0−4)=−4C12=−(−1)=1C13=+(4−0)=4C21=−(−3)=3C22=+(0)=0C23=−(−4)=4C31=+(−3−0)=−3C32=−(−1)=1C33=+(3)=3
So,
Cij=[−414304−313]
Let’s find Adj(A)
Take the transpose of Cij
Adj(A)=[−43−3101443]
Hence we clearly see that Adj(A)=A

Adjoint and Inverese of Matrices Exercise 6.1 Question 5

Answer:
Adj(A)=3AT
Hint:
Here, we use basic concept of determinant and adjoint of matrix.
Given:
A=[−1−2−221−22−21]
Solution:
A=[−1−2−221−22−21]
Let’s find cofactors Cij=(−1)i+j
C11=+(1−4)=−3C12=−(2+4)=−6C13=+(−4−2)=−6C21=−(−2−4)=6C22=+(−1+4)=3C23=−(2+4)=−6C31=+(4+2)=6C32=−(2+4)=−6C33=+(−1+4)=3
Cij=[−3−6−663−66−63]
So,
Adj(A) = Transpose of Cij
Adj(A)=[−366−63−6−6−63] (1)
3AT=3×[−122−21−2−2−21]=[−366−63−6−6−63] (2)
Here, from equation (1) and (2)
Clearly see that,
3AT=Adj(A)
Hence, proved

Adjoint and Inverse Matrix Exercise 6.1 Question 6

Answer:
Adj(A)=[250002500025]
Hint:
Here, we use basic concept of determinant.
Given:
A=[1−2302−1−452]
Solution:
Let’s find cofactor of A
C11=9,C21=19,C31=−4C12=4,C22=14,C32=1C13=8,C23=3,C33=2
Cij=[94819143−412]
Adj(A) = Transpose of Cij
Adj(A)=[919−44141832]A×Adj(A)=[1−2302−1−452]×[919−44141832]
=[9−8+2419−28−9−4−2+60+8−80+28−30+2−2−36+20+16−76+70+616+5+4]=[250002500025]
Hence, A×Adj(A)=25I

Adjoint and Inverse Matrix Exercise 6.1 Question 7 (i)

Answer:
Adj(A)=[250002500025]
Hint:
Here, we use basic concept of determinant.
Given:
A=[1−2302−1−452]
Solution:
Let’s find cofactor of A
C11=9,C21=19,C31=−4C12=4,C22=14,C32=1C13=8,C23=3,C33=2
Cij=[94819143−412]
Adj(A) = Transpose of Cij
Adj(A)=[919−44141832]A×Adj(A)=[1−2302−1−452]×[919−44141832]
=[9−8+2419−28−9−4−2+60+8−80+28−30+2−2−36+20+16−76+70+616+5+4]=[250002500025]
Hence, A×Adj(A)=25I

Adjoint and Inverse Matrix Exercise 6.1 Question 7 (ii)

Answer:
A−1=[0110]
Hint:
Here, we use basic concept of inverse of matrix
Such that A−1=1|A|×Adj(A)
Given:
A=[0110]
Solution:
|A|=|0110|=0−1=−1
Let’s find Adj(A)
C11=0,C12=−1C21=−1,C22=0
So,
Cij=[0−1−10]
Adj(A) Is transpose of Cij
So,
Adj(A)=[0−1−10]
So,
A−1=1|A|×Adj(A)A−1=−11×[0−1−10]A−1=−1×[0−1−10]A−1=[0110]

Adjoint and Inverse Matrix Exercise 6.1 Question 7 (iii)

Answer:

Answer:
A−1=[1+bca−b−ca]
Hint:
Here, we use basic concept of inverse.
Given:
A=[abc1+bca]
Solution:
We know that
A−1=1|A|×Adj(A)|A|=|abc1+bca|=1+bc−bc=1
Let’s find Adj(A)
C11=1+bcaC12=−cC21=−bC22=aCij=[1+bca−c−ba]
Adj(A)=CijTAdj(A)=[1+bca−b−ca]
So, let’s put value in formula
A−1=11×[1+bca−b−ca]A−1=[1+bca−b−ca]

Adjoint and Inverse of a Matrix Exercise 6.1 Question 7 (i)

Answer:
A−1=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]
Hint:
Here, we use basic concept of inverse
Given:
A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]
Solution:
A−1=1|A|×Adj(A)
Let’s find |A|
|A|=cos2⁡θ+sin2⁡θ=1
So,
A−1=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]

Adjoint and Inverse of Matrices Excercise 6.1 Question 7 (iv)

Answer:
A−1=117[1−532]
Hint:
Here, we use basic concept of determinant and inverse of matrix.
Given:
A=[25−31]
Solution:
A−1=1|A|×Adj(A)
So let’s find |A|
|A|=|25−31|=2−[(5)×(−3)]=2+15=17
Let’s find Adj(A)
C11=1,C12=3C21=−5,C22=2
So,
Cij=[13−52]Adj(A)=CijTAdj(A)=[1−532]
Let’s put the values in formula
A−1=117[1−532]

Adjoint and Inverse of Matrices Excercise 6.1 Question 8 (i).

Answer:
A−1=118×[−51717−57−51]
Hint:
Here, we use basic concept of determinant and inverse of matrix.
Given:
A=[123231312]
Solution:
A−1=1|A|×Adj(A)|A|=1|3112|−2|2132|+3|2331|=1(6−1)−2(4−3)+3(2−9)|A|=5−2−21=−18
Let’s find Adj(A)
For that let’s find cofactor
C11=+(6−1)=5C12=−(4−3)=−1C13=+(2−9)=−7C21=−(4−3)=−1C22=+(2−9)=−7C23=−(1−6)=5C31=+(2−9)=−7C32=−(1−6)=5C33=+(3−4)=−1
Cij=[5−1−7−1−75−75−1]Adj(A)=Cj→T
Adj(A)=[5−1−7−1−75−75−1]A−1=1|A|×Adj(A)
A−1=−118×[5−1−7−1−75−75−1]A−1=118×[−51717−57−51]

Adjoint and Inverse of Matrices Excercise 6.1 Question 8 (ii)

Answer:
A−1=127[4173−1−11651−3]
Hint:
Here, we use basic concept of determinant and inverse of matrix
Given:
A=[1251−1−123−1]
Solution:
We know that
A−1=1|A|×Adj(A)
So let’s find |A|
|A|=1|−1−13−1|−2|1−12−1|+5|1−123|=1(1+3)−2(−1+2)+5(3+2)=4−2+25|A|=27
Hence A−1 exist
Cofactor of A are
C11=4,C21=17,C31=3C12=−1,C22=−11,C32=6C13=5,C23=1,C33=−3
 So Adj(A)=CijTAdj(A)=[4−1517−11136−3]T=[4173−1−11651−3]
A−1=1|A|×Adj(A)A−1=127[4173−1−11651−3]

Adjoint and Inverse of Matrices Excercise 6.1 Question 8 (iii)

Answer:
A−1=14[31−1131−113]
Hint:
Here, we use basic concept of determinant and inverse of matrix.
Given:
A=[2−11−12−11−12]
Solution:
A−1=1|A|×Adj(A)|A|=2|2−1−12|−(−1)|−1−112|+1|−121−1|=2(4−1)+1(−2+1)+1(1−2)=6−2=4≠0
Hence A−1 exist
Cofactor of A
C11=3,C21=1,C31=−1C12=3,C22=3,C32=1C13=−1,C23=1,C33=3
Adj(A)=CijTAdj(A)=[31−1131−113]T=[31−1131−113]A−1=1|A|×Adj(A)=14[31−1131−113]



Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (iv)

Answer:
A−1=[3−11−156−55−22] .
Hint:
Here, we use basic concept of determinant and inverse of matrix.
A−1=1|A|×Adj(A)
Given:
A=[10−1510013]
Solution:
|A|=2|1013|−0|5003|−1|5101|=2(3−0)−0−1(5)=6−5|A|=1
Hence A−1 exist
Cofactor of A are
C11=3,C12=−15,C13=5C21=−1,C22=6,C23=−2C31=1,C32=−5,C33=2
Adj(A)=CijTAdj(A)=[3−155−16−21−52]T=[3−11−156−55−22]
A−1=1|A|×Adj(A)A−1=11×[3−11−156−55−22]


Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (v)

Answer:
A−1=[01−14−343−34]
Hint:
Here, we use basic concept of determinant and inverse of matrix.
A−1=1|A|×Adj(A)
Given:
A=[01−14−343−34]
Solution:
|A|=0|−34−34|−1|4434|+(−1)|4−33−3|=0−1(16−12)−1(−12+9)=−4+3=−1
Hence A−1 exist
Cofactor of A are
C11=0,C12=−4,C13=−3C21=−1,C22=3,C23=−4C31=−3,C32=3,C33=−4Adj(A)=CijT
Adj(A)=[0−11−43−4−33−4]A−1=1|A|×Adj(A)
=−11[0−11−43−4−33−4]A−1=[01−14−343−34]

Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (vi)

Answer:
A−1=[−211114−12−34−100]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[00−1345−2−4−7]
Solution:
|A|=0|45−4−7|−0|35−2−7|−1|34−2−4|=0−0−1(−12+8)=4
Hence A−1 exist
Cofactor of A are
C11=−8,C21=4,C31=4C12=11,C22=−2,C32=−3C13=−4,C23=0,C33=0
Adj(A)=[811−44−204−30]T=[84411−2−3−400]A−1=1|A|×Adj(A)
A−1=14×[−84411−2−3−400]A−1=[−211114−12−34−100]

Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (vii)

Answer:
A−1=[1000cos⁡αsin⁡α0sin⁡α−cos⁡α]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[1000cos⁡αsin⁡α0sin⁡α−cos⁡α]
Solution:
Let’s find |A|
|A|=1|cos⁡αsin⁡αsin⁡α−cos⁡α|−0+0=−cos2⁡α−sin2⁡α|A|=−1
So A−1 exist
Cofactors of A are
C11=−1,C21=0,C31=0C12=0,C22=−cos⁡α,C32=−sin⁡αC13=0,C23=−sin⁡α,C33=cos⁡αAdj(A)=CijT
Adj(A)=[−1000−cos⁡α−sin⁡α0−sin⁡αcos⁡α]T=[−1000−cos⁡α−sin⁡α0−sin⁡αcos⁡α]A−1=1|A|×Adj(A)
=1−1[−1000−cos⁡α−sin⁡α0−sin⁡αcos⁡α]A−1=[1000cos⁡αsin⁡α0sin⁡α−cos⁡α]

Adjoint and Inverse of a Matrix exercise 6.1 question 9 (i)

Answer:
A−1=[7−3−3−110−101]
Hint:
Here, we use basic concept of determinant and inverse of matrix.
A−1=1|A|×Adj(A)
Given:
A=[133143134]
Solution:
Here, let’s find |A|
|A|=1|4334|−3|1314|+3|1413|=1(16−9)−3(4−3)+3(3−4)=7−3−3=1
Hence A−1 exist
Let’s find cofactor of A
C11=7,C21=−3,C31=−3C12=−1,C22=1,C32=0C13=−1,C23=0,C33=1Adj(A)=CijT
Adj(A)=[7−3−3−110−101]
A−1=1|A|×Adj(A)
A−1=11[7−3−3−110−101]=[7−3−3−110−101]A−1×A=[7−3−3−110−101]×[133143134]
=[7−3−321−12−921−9−12−1+1+0−3+4+0−3+3+0−1+0+1−3+0+3−3+0+4]A−1×A=[100010001]=I

Adjoint and Inverse of a Matrix exercise 6.1 question 9 (ii)

Answer:
A−1=12[11−1−3119−5−1]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[231341372]
Solution:
|A|=2|4172|−3|3132|+1|3437|=2(8−7)−3(6−3)+1(21−12)=2−9+9=2
Hence A−1 exist
Cofactor of A
C11=1,C21=1,C31=−1C12=−3,C22=1,C32=1C13=9,C23=−5,C33=−1
Adj(A)=CijTAdj(A)=[1−3911−5−11−1]T=[11−1−3119−5−1]
A−1=1|A|×Adj(A)A−1=12[11−1−3119−5−2]
A−1×A=12[11−1−3119−5−1][231341372]=12[2+3−33+4−71+1−2−6+3+3−9+4+7−3+1+218−15−327−20−79−5−2]=12[200020002]=[100010001]
Hence A−1×A=I

Adjoint and Inverse Matrix Exercise 6.1 Question 11

Answer:
[−4739241−17]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[3275],B=[6789]
Solution:
A=[3275]
Let’s find |A|
|A|=|3275|=15−14=1Adj(A)=[5−2−73]A−1=1|A|×Adj(A)=11[5−2−73]=[5−2−73]
B=[6789]|B|=54−56=−2Adj(B)=[9−7−86]B−1=1|B|×Adj(A)=1−2[9−7−86]
Now, we know that
(AB)−1=B−1A−1=1−2[9−7−86][5−2−73]=1−2[45+49−18−21−40−4216+18]
=−12[94−39−8234](AB)−1=[−4739241−17]

Adjoint and Inverse Matrix Exercise 6.1 Question 12

Answer:
Hence proved 2A−1=9I−A
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[2−3−47]
Solution:
Here let’s find |A|,Adj(A)&A−1
|A|=|2−3−47|=14−12=2Adj(A)=[7342]A−1=12[7342]
To show 2A−1=9I−A
2A−1=2×12[7342]=[7342] (1)
9I−A=[9009]−[2−3−47]=[7342] (2)
From equation (1) and (2)
Hence, 2A−1=9I−A

Adjoint and Inverse Matrix Exercise 6.1 Question 13

Answer:
Hence proved A−3I=2(I+3A−1)
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[4521]
Solution:
Let’s find |A|,Adj(A)&A−1
|A|=4−10=−6Adj(A)=[1−5−24]A−1=1|A|×Adj(A)=1−6[1−5−24]
To show A−3I=2(I+3A−1)
LHS
A−3I=[4521]−3[1001]=[152−2] (1)
RHS
2(I+3A−1)=2I+6A−1=2[1001]+66[−152−4]=[2002]+[−152−4]
=[152−2] (2)
Here from equation (1) and (2)
A−3I=2(I+3A−1)

Adjoint and Inverse of Matrices Excercise 6.1 Question 14

Answer:
A−1=[1+bca−b−ca]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[abc1+bca]
Solution:
|A|=a+abca−bc=a+abc−abca=aa=1
So, hence A−1 exist
Cofactor of A
C11=1+bca,C12=−cC21=−b,C22=aAdj(A)=[1+bca−c−ba]T=[1+bca−b−ca]
A−1=1|A|×Adj(A)=11[1+bca−b−ca]A−1=[1+bca−b−ca]
To show aA−1=(a2+bc+1)I−aA
LHS
aA−1=a[1+bca−b−ca]=[1+bc−ab−aca2]
RHS
(a2+bc+1)I−aA=[a2+bc+100a2+bc+1]−[a2abac1+bc]=[1+bc−ab−aca2]
LHS = RHS

Adjoint and Inverse of Matrices Excercise 6.1 Question 15

Answer:
[−219−27−218−25−329−42]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[504232121],B−1=[133143134]
Solution:
For (AB)−1 we know that
(AB)−1=B−1A−1
Here B−1 is also given so
Let’s fin A−1
|A|=−5+4=−1
Cofactor of A are
C11=−1,C21=8,C31=−12C12=0,C22=1,C32=−2C13=1,C23=−10,C33=15
Adj(A)=[−10181−10−12−215]T=[−18−1201−21−1015]A−1=−1[−18−1201−21−1015]=[1−8120−12−110−15]
(AB)−1=B−1A−1=[133143134][1−8120−12−110−15]=[1+0−3−8−3+3012+6−451+0−3−8−4+3012+8−451+0−4−8−3+4012+6−60]
(AB)−1=[−219−27−218−25−329−42]

Adjoint and Inverse of Matrices Excercise 6.1 Question 16 question (i)

Answer:
Hence proved [F(α)]−1=F(−α)
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
F(α)=[cos⁡α−sin⁡α0sin⁡αcos⁡α0001],G(β)=[cos⁡β0sin⁡β010−sin⁡β0cos⁡β]
Solution:
|F(α)|=|cos⁡α−sin⁡α0sin⁡αcos⁡α0001||F(α)|=cos2⁡α+sin2⁡α=1
Cofactor of A are
C11=cos⁡α,C21=sin⁡α,C31=0C12=−sin⁡α,C22=cos⁡α,C32=0C13=0,C23=0,C33=1Adj(F(α))=CijT
Adj(F(α))=[cos⁡α−sin⁡α0sin⁡αcos⁡α0001]TAdj(F(α))=[cos⁡αsin⁡α0−sin⁡αcos⁡α0001] (1)
[F(α)]−1=11[cos⁡αsin⁡α0−sin⁡αcos⁡α0001]F(−α)=[cos⁡(−α)sin⁡(−α)0−sin⁡(−α)cos⁡(−α)0001]
=[cos⁡αsin⁡α0−sin⁡αcos⁡α0001] (2)
From equation (1) and (2)
[F(α)]−1=F(−α)

Adjoint and Inverese of Matrices Exercise 6.1 Question 16 (ii)

Answer:
Hence proved [G(β)]−1=G(−β)
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
F(α)=[cos⁡α−sin⁡α0sin⁡αcos⁡α0001],G(β)=[cos⁡β0sin⁡β010−sin⁡β0cos⁡β]
Solution:
Let’s find |G(β)|
|G(β)|=cos2⁡β+sin2⁡β=1
Cofactor of A
C11=cos⁡β,C21=0,C31=sin⁡βC12=0,C22=1,C32=0C13=sin⁡β,C23=0,C33=cos⁡β
Adj(G(β))=[cos⁡β0sin⁡β010−sin⁡β0cos⁡β]T
Adj(G(β))=[cos⁡β0−sin⁡β010−sin⁡β0cos⁡β][G(β)]−1=1|G(β)|×Adj(G(β))
[G(β)]−1=11[cos⁡β0−sin⁡β011sin⁡β0cos⁡β]G(−β)=[cos⁡(−β)0sin⁡(−β)010sin⁡(−β)0cos⁡(−β)]=[cos⁡β0−sin⁡β010sin⁡β0cos⁡β]
Hence
[G(β)]−1=G(−β)

Adjoint and Inverese of Matrices Exercise 6.1 Question 17

Answer:
[2−3−12]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[2312]A2−4A+I=0I=[1001]&O=[0000]
Solution:
A2=[2312][2312]=[4+36+62+23+4]A2=[71247]
4A=4×[2312]=[81248]I=[1001]
A2−4A+I=[71247]−[81248]+[1001]=[7−8+112−12+04−4+07−8+1]=[0000]
Hence
A2−4A+I=0A×A−4A=−I
Let’s multiply A−1 both sideA×A(A−1)−4AA−1=−IA−1A−4I=−A−1A−1=4I−A=[4004]−[2312]=[2−3−12]



Adjoint and Inverse of Matrices Excercise 6.1 Question 18

Answer:
A−1=142[−4528]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[−8524]
Solution:
A2=[−8524][−8524]=[64+10−40+20−16+810+16]=[74−20−826]4A=4×[−8524]=[−3220816]42I=42×[1001]=[420042]
Now,
A2+4A−42I=[74−20−826]+[−3220816]−[420042]
=[74−74−20+20−2+842−42]=[0000]
A2+4A−42I=0
Multiplying by A−1 both sides
A×A×A−1+4A×A−1−42I×A−1=0A+4I−42A−1=042A−1=A+4I
So,
A−1=142[A+4I]A−1=142([−8524]+[4004])A−1=142[−4528]

Adjoint and Inverse of a Matrix exercise 6.1 question 19

Answer:
A−1=17[2−113]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[31−12] Show that A2−5A+7I=0
Solution:
A=[31−12]A2=[31−12][31−12]=[9−13+2−3−2−1+4]=[85−53]
Now,
A2−5A+7I=[85−53]−5×[31−12]+7[1001]=[8−15+75−5+0−5+5+03−10+7]=[0000]A2−5A+7I=0
Multiply by A−1 both sides
A×A×A−1−5A×A−1+7I×A−1=0A−5I+7A−1=0A−1=17[5I−A]
A−1=17[[5005]−[31−12]]A−1=17[2−113]

Adjoint and Inverse of Matrices Excercise 6.1 Question 20

Answer:
X=9andy=14 , A−1=114[5−3−24]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[4325]
Solution:
A2=[4325][4325]=[16+612+158+106+25]=[22271831]A2−xA+yI=[22371831]−x[4325]+[y00y]=[0000]
22−4x+y=04x−y=22x=9
So,
4x−y=224×9−y=2236−22=yy=14
So A2−9A+14I=0
Multiply by A−1 both sides,
A×A×A−1−9A×A−1+14I×A−1=0A−9I+14A−1=0A−1=114[9I−A]
A−1=114[9009]−[4325]A−1=114[5−3−24]

Adjoint and Inverse of Matrices Excercise 6.1 Question 21

Answer:
Λ =1, A−1=12[−22−43]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[3−24−2]
Solution:
A2=[3−24−2][3−24−2]=[9−8−6+412−8−8+4]=[1−24−4]
Now,
A2=λA−2IλA=A2+2IλA=[1−24−4]+[2002]=[3−24−2]
λ=[3−24−2]÷[3−24−2]λ=1 So, A2=A−2I
Multiply A−1 both side
A−1A×A=AA−1−2LA−12A−1=I−A=[1001]−[3−24−2]=[−22−43]A−1=12[−22−43]

Adjoint and Inverse of Matrices Excercise 6.1 Question 22

Answer:
A−1=17[23−1−5]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[53−1−2]
Solution:
A2=[53−1−2][53−1−2]=[25−315−6−5+2−3+4]=[229−31]
Now,
A2−3A−7I=0[229−31]−3[53−1−2]−7[1001][22−15−79−9−0−3+3−01+6−7]=[0000]A2−3A−7I=0
Multiply A−1 both side
A×A×A−1−3A×A−1−7LA−1=0A−3I−7A−1=07A−1=A−3I
=17[53−1−2]−3[1001]A−1=17[23−1−5]

Adjoint and Inverese of Matrices Exercise 6.1 Question 23

Answer:
A−1=[6−5−76]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[6576]
Solution:
A2−12A+I=0A2=[6576][6576]=[36+3530+3042+4235+36]=[71608471]
Now,
A2−12A+1=0[71608471]−12[6576]+[1001]=[71−72+160−60+084−89+01−72+1]=[0000]
Also,
A2−12A+1=0A−12I+A−1=0A−1−12I=A
A−1=[120012]−[6576]=[12−6−5−712−6]A−1=[6−5−76]

Adjoint and Inverese of Matrices Exercise 6.1 Question 24

Answer:
A−1=−111[3−4−5−914−531]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given
A=[11112−32−13]
Solution:
A3=A2×AA2=[11112−32−13][11112−32−13]=[1+1+21+2−11−3+31+2−61+4+31−6+92−1+62−2−32+3+9]A2=[421−38−147−314]
A2×A=[421−38−147−314][11112−3213]=[4+2+24+4−14−6+3−3+8−28−3+14+14−3−24−427−3+287−6−147+9+42]=[871−2327−6932−1358]
Now,
A3−6A2+5A+11I[871−2327−6932−1358]−6[421−38−147−314]+5[11112−32−13]+11[100010001]
[8−247−121−6−23+1827−48−69+8432−42−13+1858−84]+[5+1155510+11−1510−526]=[000000000]
A3−6A2+5A+11I=0
Now,
(A×A×A)A−1−6(A×A)A−1+5AA−1+11IA−1=0AA(A−1A)−6A(A−1A)+5A−1A=−11(A−1I)A−1=−111(A2−6A+5I)
Now,
A2−6A+5I
[421−38−147−314]−6[11112−32−13]+5[100010001]
=[421−38−147−314]−[666612−1812−618]+[500050005]=[−3−4−5−914−531]
Hence,
A−1=−111[3−4−5−914−531]

Adjoint and Inverese of Matrices Exercise 6.1 Question 25

Answer:
A−1=[−9−8−2872−5−4−1]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[10−2−2−12341]
Solution:
A3−A2−3A−I3=0
So,
A3=A2×A
A2=A×A=[10−2−2−12341][10−2−2−12341]=[1+0−60+0−8−2+0−2−2+2+60+1+84−2+23−8+30−4+4−6+8+1]
=[−5−8−4694−203]
A2×A=[−5−8−4692341]×[10−2−2−12341]=[−5+16−120−8+1610−16−46−18+120−9+16−12+18+4−2−0+90−0−124+0+3]
=[−1−8−1007107127]
Now,
A3−A2−3A−I
[−1−8−1007107127]−[−5−8−4694−203]−3[10−2−2−12341]−[100010001]
=[−1+5−8+8−10+40−67−910−47+212−07−3]+[−3−106−06−03−1−6−0−9−0−12+0−3−1]=[40−6−6−269124]+[−40662−6−9−12−4]=[000000000]
Thus,
A3−A2−3A−I now (AAA)×A−11−AAA−1−3AA−1−LA−1=0A2−A−3A−I=0A−1=(A2−A−3I)
=[−5−8−4694−203]−[10−2−2−12341]−[300030003]
[−5−1−3−8+0−4+26+2−09+1−34−2−2−3−0−43−1−3]=[−9−8−2872−5−4−1]
A−1=[−9−8−2872−5−4−1]

Adjoint and Inverse Matrix Exercise 6.1 Question 26
Answer:

A−1=14[31−1131−113]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[2−11−12−11−12]
Solution:
A3=A2×A
=[6−55−56−55−56][2−11−12−11−12]=[12+5+5−6+10−56+5+10−10−6−55+12+5−5−6−1010+5+6−5−10−65+5+12]
=[22−2121−2122−2121−2122]
Now,
A3−6A2+9A−4I
[22−2121−2122−2121−2122]−6[6−55−56−55−56]+9[2−11−12−11−12]−4[100010001]
=[22−36+18−4−21+30−921−30+9−21+30−922−36+18−4−21+30−921−30+9−0−21+30−922−36+18−4]=[000000000]
Thus,
A3−6A2+9A−4IA2−6A+9I=4A−1A−1=14[A2−6A+9I]
A2−6A+9I=[6−55−56−55−56]−6[2−11−12−11−12]+9[100010001]
=[6−12+9−5+6+05−6+0−5+6+06−12+9−5+6+05−6+0−5−6+06−12+3]=[31−1131−113]
Hence,
A−1=14[31−1131−113]

Adjoint and Inverse Matrix Exercise 6.1 Question 27

Answer:
A−1=19[−84114−8474]=AT
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=19[−8141471−84] Find AT=A−1
Solution:
AT=A−1LHSAT=19[8144471−84]T=19[−84114−8474]
RHSA−1=1|A|×Adj(A)
Let’s find
|A|&Adj(A)|A|=19[−8(16+56)−1(16−7)+4(−32−4)]|A|=−81
Cofactor of A
C11=72,C21=−36,C31=−9C12=−9,C22=−36,C32=72C13=−36,C23=−63,C33=−36Adj(A)=[72−9−36−36−36−63972−36]T=[72−36−9−9−3672−36−63−36]
A−1=1|A|×Adj(A)=1−9[72−36−9−9−367236−63−36]A−1=19[−84114−8474]=AT


Adjoint and Inverse Matrix Exercise 6.1 Question 28

Answer:
A−1=A3
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[3−342−340−11]
Solution:
A−1=A3
RHS
A−1=1|A|×Adj(A)
Let’s find |A|&Adj(A)
|A|=3+6−8=1
Cofactor of A
C11=1,C21=−1,C31=0C12=−2,C22=3,C32=−4C13=−2,C23=3,C33=−3
Adj(A)=[1−10−23−4−23−3]A−1=1|A|×Adj(A)=11[1−10−23−4−23−3] (1)
RHS
A3=A2×AA2=[3−342−340−11][3−342−340−11]=[9−6+0−9+9−412−12+46−6+0−6+9−48−12+40−2+00+3−10−4+1]
=[3−440−10223]
A3=A2×A=[3−440−10−22−3][3−342−340−11]
=[1−10−23−4−23−3] (2)
So, here equation (1) and (2)
LHS = RHS
A−1=A3

Adjoint and Inverse of Matrices Excercise 6.1 Question 29

Answer:
A−1=A2
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[−120−111010]
Solution:
|A|=−1|1110|−2|−1100|+0|A|=−1(−1)−0+0=1|A|=1
Cofactor of A
C11=−1,C21=0,C31=2C12=0,C22=0,C32=1C13=−1,C23=1,C33=1Adj(A)=[−10−1001211]T=[−102001−111]
A−1=1|A|×Adj(A)=11[−102001−111]=[−102001−111]
A2=A×A=[−120−111010][−120−111010]=[−1−2+0−2+2+00+2+01−1+1−2+1+1−1+1−00−1+00+1−00+1−0]=[−102001−111]
Hence A−1=A2

Adjoint and Inverse of Matrices Excercise 6.1 Question 30

Answer:
X=[−3−14417]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[5411],B=[1−213]
Solution:
AX=BX=A−1B|A|=1
Cofactor of A
C11=1,C21=−4C12=−1,C22=5Adj(A)=[1−1−45]T=[1−4−15]
A−1=1|A|×Adj(A)A−1=11[1−4−15]X=[1−4−15]×[1−213]=[−3−14417]

Adjoint and Inverse of a Matrix exercise 6.1 question 31

Answer:
X=[−7−576]
Hint:
Here, we use basic concept of determinant and inverse of matrix
Given:
A=[53−1−2],B=[14777]
Solution:
AX=BX=A−1B|A|=−7
Cofactor of A
C11=−2,C12=1C21=−3,C22=5Adj(A)=[−21−35]T=[−2−315]A−1=1|A|×Adj(A)
A−1=1−7[23−1−5]X=−17[23−1−5][14777]
=−17[28+2114+21−14−35−7−35]X=[−7−576]

Adjoint and Inverse of a Matrix exercise 6.1 question 32

Answer:
A−1=[−16324−5]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[3275],B=[−11−21],C=[2−104]
Solution:
Then the given equation becomes as,
AXB=CX=A−1CB−1|A|=15−14=1|B|=−1+2=1
A−1=1|A|×Adj(A)=11[5−2−73]
B−1=1|B|×Adj(B)=1[1−12−1]
X=A−1CB−1=1[5−2−73][2−104][1−12−1]=[10+0−5−8−14+07+12][1−12−1]
=[10−26−10+13−14+3814−19]X=[−16324−5]

Adjoint and Inverse of a Matrix exercise 6.1 question 33

Answer:
X=[9−14−1625]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[2153],B=[5332],C=[1001]
Solution:
Then the given equation becomes
A×B=IX=A−1B−1|A|=6−5=1|B|=10−9=1
A−1=1|A|×Adj(A)=11[3−1−52]
B−1=1|B|×Adj(B)=11[2−3−35]
X=A−1B−1=[3−1−52][2−3−35]
X=[6+3−14−1625]
X=[9−14−1625]

Adjoint and Inverse of a Matrix exercise 6.1 question 33
Edit Q



Adjoint and Inverse of a Matrix exercise 6.1 question 33

Answer:

Answer:
X=[9−14−1625]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[2153],B=[5332],C=[1001]
Solution:
Then the given equation becomes
A×B=IX=A−1B−1|A|=6−5=1|B|=10−9=1
A−1=1|A|×Adj(A)=11[3−1−52]
B−1=1|B|×Adj(B)=11[2−3−35]
X=A−1B−1=[3−1−52][2−3−35]
X=[6+3−14−1625]
X=[9−14−1625]

Adjoint and Inverese of Matrices Exercise 6 point 1 Question 34

Answer:
A−1=15[−3222−3222−3]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[122212121]
A2−4A−5I=0
Solution:
A=[122212221]
A2=[122212221][122212221]=[1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1]
A2=[988898889]
A2−4A−5I=0
[988898889]−4[122212221]+5[100010001]
=[9−4−58−8−08−8−08−8−09−4−58−8−08−8−08−8−09−4−5]=[000000000]
Also,
A2−4A−5I=0(AA)A−1−4AA−1−5IA−1=0A−4I−5A−1=0A−1=15(A−4I)
A−1=15[122212221]−4[100010001]
A−1=15[1−42−02−02−01−42−02−02−01−4]=15[−3222−3222−3]


Adjoint and Inverese of Matrices Exercise 6.1 Question 35

Answer:
|A×Adj(A)|=|A|n
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A is square matrix
Solution:
|A×Adj(A)|=|A|nLHS|A×Adj(A)|=|A|×|A|n−1 [Adj(A)=|A|n−1]
=|A|n+1−1=|A|n=RHS

Adjoint and Inverese of Matrices Exercise 6.1 Question 36

Answer:
(AB)−1=[9−35−210102]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A−1=[3−11−156−55−22],B=[12−2−1300−21]
Find (AB)−1
Solution:
A−1=[3−11−156−55−22]
So, we know that
(AB)−1=B−1A−1
So let’s find B−1
B−1=1|B|×Adj(B)|B|=1(3−0)−2(1−0)−2(2−0)|B|=3+2−4=1
Now cofactor of B
C11=3,C21=2,C31=6C12=1,C22=1,C32=2C13=2,C23=2,C33=5Adj(B)=[326112225]
Now,
B−1=[326112225]
(AB)−1=B−1A−1
=[326112225][3−11−156−55−22]=[9−30+30−3+12−123−10+123−15+10−1+6−41−5+46−30+25−2+12−102−10+10]
(AB)−1=[9−35−210102]

Adjoint and Inverse Matrix Exercise 6.1 Question 37

Answer:
[−9−8−2872−5−4−1]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[1−230−14−221]
Solution:
AT=[10−2−2−12341]
Let’s find |AT|
|AT|=(−1−8)−0−2(−8+3)=−9+10=1
Cofactor of AT
C11=−9,C12=8,C13=−5C21=−8,C22=7,C23=−4C31=−2,C32=2,C33=1
Adj(AT)=[−9−8−2872−5−4−1]
(AT)−1=11[−9−8−2872−5−4−1]
(AT)−1=[−9−8−2872−5−4−1]

Adjoint and Inverse Matrix Exercise 6.1 Question 38

Answer:
A×Adj(A)=27[100010001]
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=|−1−2−221−22−21|
Solution:
|A|=|−1−2−221−22−21||A|=−1(1−4)+2(2+4)−2(−4−2)=3+12+12|A|=27
Cofactor of A
C11=−3,C21=6,C31=6C12=−6,C22=3,C32=−6C13=−6,C23=−6,C33=3
Adj(A)=[−366−63−6663]
A×Adj(A)=[−1−2−221−22−21][−366−63−6663]=[270002700027]
A×Adj(A)=27[100010001]
A×Adj(A)=|A|I

Adjoint and Inverse Matrix Exercise 6.1 Question 39

Answer:
A−1=12(A2−3I)
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[011101110]
Solution:
|A|=0−1(0−1)+1(1−0)=0+1+1=2
Cofactor of A
C11=−1,C21=1,C31=1C12=1,C22=−1,C32=1C13=1,C23=1,C33=−1
Adj(A)=[−1111−1111−1]
A−1=12[−1111−1111−1]
A2−3I=[011101110][011101110]−3[100010001]
=[211121112]−[300030003]=[−1111−1111−1]
Hence A−1=12(A2−3I)

Adjoint and Inverese of Matrices Exercise 6.1 Question 10 (i)

Answer:
Proved (AB)−1=B−1A−1
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[3275],B=[4632]
Solution:
Let’s find |A|
|A|=|3275|=15−14=1
Adj(A)=[5−2−73]
A−1=1|A|×Adj(A)
A−1=11[5−2−73]
Then let’s find |B|Adj(B)&B−1
|B|=|4632|=8−18=−10
Adj(B)=[2−6−34]
B−1=1|B|×Adj(B)
B−1=1−10×[2−6−34]
Then find AB
A×B=[3275]×[4632]
=[12+618+428+1542+10]=[18224352]
Then let’s find |AB|,Adj(AB) and inverse of AB
|AB|=|18224352|=936−946=−10
Adj(AB)=[52−22−4318]
(AB)−1=1−10×[52−22−4318]=110[−522243−18] (1)
Now B−1A−1
=110[2−634][5−2−73]
=110[10+42−4−18−15−286+12]
=110[−522243−18] (2)
From equation (1) and (2)
(AB)−1=B−1A−1
Hence proved

Adjoint and Inverese of Matrices Exercise 6.1 Question 10 (ii)

Answer:
Proved (AB)−1=B−1A−1
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
A=[2153],B=[4534]
Solution:
Let’s find |A|Adj(A)&A−1
|A|=|2153|=6−5=1
Adj(A)=[3−1−52]
A−1=1|A|×Adj(A)=11[3−1−52]=[3−1−52]
Let’s find |B|Adj(B)&B−1
|B|=|4534|=16−15=1
Adj(B)=[4−5−34]
B−1=1|A|×Adj(A)=11[4−5−34]
Then find AB
A×B=[2153][4534]=[11142937]
Let’s find |AB|,Adj(A)&(AB)−1
|AB|=407−406=1
Adj(AB)=[37−14−2911]
(AB)−1=1|AB|×Adj(A)=[37−14−2911] (1)
Now
B−1A−1=[4−5−34][3−1−52]
B−1A−1=[37−14−2911] (2)
Hence proved from equation (1) and (2)(AB)−1=B−1A−1

Adjoint and Inverese of Matrices Exercise 6.1 Question 16 (iii)

Answer:
Hence proved [F(α)G(β)]−1=G(−β)F(−α)
Hint:
Here, we use basic concept of determinant and inverse of matrix
A−1=1|A|×Adj(A)
Given:
F(α)=[cos⁡α−sin⁡α0sin⁡αcos⁡α0001],G(β)=[cos⁡β0sin⁡β010−sin⁡β0cos⁡β]
Solution:
We have to show
[F(α)G(β)]−1=G(−β)F(−α)
We already know that,
[G(β)]−1=G(−β)
[F(α)]−1=F(−α)
LHS=[F(α)G(β)]−1
=[G(β)]−1[F(α)]−1
=G(−β)F(−α)
LHS=RHS
Hence proved

RD Sharma Class 12th Exercise 6.1 consists of the chapter, Adjoint and Inverse of Matrix. This particular exercise consists of 58 Level 1 sums that are very fundamental and direct. Students can efficiently complete them in a day without a fuss if they understand the chapter. To help students cover as many questions as possible, Career360 has provided RD Sharma Class 12th Exercise 6.1 material.

The sums in this chapter are divided into two parts, i.e., Level 1 and Level 2. These levels are based on difficulty and weightage. Level one questions usually require fundamental knowledge and can be completed quickly, whereas level two sums require some extra understanding and are more complex.

It has solutions for the entire RD Sharma book that students can utilize to complete their syllabus. As it complies with the CBSE syllabus, students can refer to it for their classes and compare their progress. This exercise can be quickly completed with the help of RD Sharma Class 12th Exercise 6.1 by Career360.

Matrix multiplication, Adjoint, and Inverse, and other algebra are discussed in this chapter. As the material from Career360 contains solutions for all the questions from the book, there is nothing apart from this that students need to follow. These solutions are created by experts that have specifically designed them to help students get a good grasp of the subject. This is an easier and more efficient way to complete the preparation.

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Similarly, this chapter contains hundreds of questions. As teachers can't explain every one of those questions, this is where Career360 comes to help with RD Sharma Class 12 Chapter 6 Exercise 6.1 material. It has all the questions and covers the entire syllabus. It is beneficial and convenient for students now to study from home.

For the convenience of students, this material is free of cost. They can visit the website and download the material of choice for free. Thousands of students have already started preparing this material. Students who haven’t tried it yet should definitely refer to it.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. How can I download this material?

Students can download RD Sharma Class 12th Exercise 6.1 from Career360’s website for free. Simply search the book name and the exercise and download it on any laptop or mobile.

2. How are NCERT books compared to RD Sharma?

NCERT books are suitable for a basic study and contain some essential questions as well. Nevertheless, RD Sharma solutions are best suited for maths as they have detailed material. To check the material for matrices, check Class 12 RD Sharma Chapter 6 Exercise 6.1 Solution.

3. What is a Matrix?

A matrix is a collection of values of tabulated rows and columns. It is a rectangular array wherein the values are independent of each other and have no direct relation. To learn more about matrices, refer to RD Sharma Class 12 Solutions Adjoint and inverse of Matrix Ex 6.1.

4. What is the Inverse of a Matrix?

The inverse of a matrix is a value that, when multiplied, gives an identity matrix in multiple forms. If M is a matrix, then its inverse is denoted by M-1, and its equation can be given as M x M-1 = I. To learn more about the inverse of matrices, refer, RD Sharma Class 12 Solutions Adjoint and inverse of Matrix Ex 6.1.

5. Explain Adjoint of a Matrix

The adjoint is obtained by finding the cofactors and then finding the transpose of that resultant matrix. To learn more about adjoints of matrices, check RD Sharma Class 12 Solutions Chapter 6 Ex 6.1.

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