Class 12 RD Sharma Solutions β Chapter 7 Adjoint and Inverse of a Matrix β Exercise 7.1 | Set 2
Question 10. For the following parts of matrices verify that (AB)-1 = B-1A-1.
(i) A = and B =
Solution:
To prove (AB)-1= B-1A-1
We take LHS
AB =
=
|AB| = 18 Γ 52 β 22 Γ 43
= 936 β 946 = -10
adj(AB) =
AB-1= adj(AB)/|AB| =
=
Now,
A =
|A| = 15 β 14 = 1
adj A =
Therefore, A-1 = adj A/|A| =
B =
|B| = 8 β 18 = -10
adj B =
Therefore, B-1= adj B/|B| =
Now, we take RHS
B-1A-1 =
=
=
LHS = RHS
Hence, Proved
(ii) A = and B =
Solution:
To prove (AB)-1 = B-1A-1
We take LHS
AB =
=
|AB| = 11 Γ 27 β 29 Γ 14
= 407 β 406 = 1
adj(AB) =
AB-1= adj(AB)/|AB| =
=
Now,
A =
|A| = 6 β 5 = 1
adj A=
Therefore, A-1 = adj A/|A| =
B =
|B| = 16 β 15 = 1
adj B =
Therefore, B-1= adj B/|B| =
Now, we take RHS
B-1A-1 =
=
LHS = RHS
Hence, Proved
Question 11. Let A = and B = . Find (AB)-1.
Solution:
AB =
=
|AB| = 34 Γ 94 β 39 Γ 82 = -2
adj(AB) =
AB-1 = adj(AB)/|AB| =
=
Question 12. Given A = , Compute A-1 and show that 2A-1 = 9I β A.
Solution:
A =
|A| = 14 β 12 = 2
adj A =
Therefore, A-1 = adj A/|A| =
To show 2A-1 = 9I β A.
LHS = 2 Γ (1/2)
=
Now we take RHS
= 9I β A
= β
=
LHS = RHS
Hence Proved
Question 13. If A = , then show that A β 3I = 2(I + 3A-1).
Solution:
Here, A =
|A| = 4 β 10 = -6
adj A =
Therefore, A-1 = adj A/|A| =
To show, A β 3I = 2(I + 3A-1)
Now we take LHS
= A β 3I
= β 3
=
Now we take RHS
= 2I + 6A-1
= 2 + 6 Γ (1/6)
=
LHS = RHS
Hence Proved
Question 14. Find the inverse of the matrix A = and show that aA-1 = (a2 + bc + 1)I β aA.
Solution:
Here, A =
|A| = (a + abc)/a β bc = 1
Therefore, inverse of A exists
Cofactor of A are,
C11 = (1 + bc)/a C12 = -c
C21 = -b C22 = a
adj A =
=
=
A-1 = 1/|A|. adj A
= 1/1
=
To show that
aA-1 = (a2 + bc + 1)I β aA.
LHS = aA-1
= a
=
RHS = (a2 + bc + 1)I β aA
= β a
= β
=
LHS = RHS
Hence Proved
Question 15. Given A = , B-1 = , Compute (AB)-1.
Solution:
We know (AB)-1 = B-1A-1
Here, A =
|A| = 5(3 β 4) + 4(4 β 3) = -5 + 4 = -1
Co-factors of A are:
C11 = -1 C12 = 0 C13 = 1
C21 = 8 C22 = 1 C23 = -10
C31 = -12 C32 = -2 C33 = 15
adj A =
A-1 = 1/|A|. adj A
Hence, A-1 =
=
(AB)-1 = B-1A-1
=
=
Question 16. Let F(Ξ±) = and G(Ξ²) = , Show that
(i) [F(Ξ±)]-1 = F(-Ξ±)
Solution:
We have F(Ξ±) =
|F(Ξ±)| = cos2Ξ± + sin2Ξ± = 1
Therefore, inverse of F(Ξ±) exists
Cofactors of F(Ξ±) are:
C11 = cosΞ± C12 = -sinΞ± C13 = 0
C21 = sinΞ± C22 = cosΞ± C23 = 0
C31 = 0 C32 = 0 C33 = 1
Adj F(Ξ±) =
=
=
[F(Ξ±)]-1 = 1/|F(Ξ±)|. adj F(Ξ±)
Hence, [F(Ξ±)]-1 = 1/1
=
Now, F(-Ξ±) =
=
So, [F(Ξ±)]-1 = F(-Ξ±)
Hence, Proved
(ii) [G(Ξ²)]-1 = G(-Ξ²)
Solution:
We have G(Ξ²) =
|G(Ξ²)| = cos2Ξ² + sin2Ξ² = 1
Therefore, inverse of G(Ξ²) exists
Cofactors of G(Ξ²) are:
C11 = cosΞ² C12 = 0 C13 = sinΞ²
C21 = 0 C22 = 1 C23 = 0
C31 = -sinΞ² C32 = 0 C33 = sinΞ²
Adj G(Ξ²) =
=
=
[G(Ξ²)]-1 = 1/|G(Ξ²)|. adj G(Ξ²)
Hence, [G(Ξ²)]-1 = 1/1
=
Now, G(-Ξ²) =
=
So, [G(Ξ²)]-1 = G(-Ξ²)
Hence, Proved
(iii) [F(Ξ±)G(Ξ²)]-1 = F(-Ξ±)G(-Ξ²)
Solution:
We already know that S[G(Ξ²)]-1 = G(-Ξ²)
[F(Ξ±)]-1 = F(-Ξ±)
Taking LHS = [F(Ξ±)G(Ξ²)]-1
= [F(Ξ±)]-1[G(Ξ²)]-1
= F(-Ξ±)G(-Ξ²) = RHS
Hence, Proved
Question 17. If A = , Verify that A2 β 4A + I = O, where I = and O = , Hence, find A-1.
Solution:
Here, A =
A2 =
=
4A = 4
=
A2 β 4A + I = O
= β +
=
Hence, =
Now, A2 β 4A + I = O
A2 β 4A = -I
Multiplying both side by A-1 both sides we get
A.A(A-1) β 4AA-1 = -IA-1
AI β 4I = -A-1
A-1 = 4I β AI
= β
=
Question 18. Show that A = satisfies the equation A2 + 4A β 42I = O. Hence, Find A-1.
Solution:
Here, A =
A2 =
=
=
4A = 4
=
A2 + 4A β 42I = + β
=
Hence,
Now, A2 + 4A β 42I = 0
β A-1A.A + 4A-1.A β 42A-1I = 0
β IA + 4I β 42A-1 = 0
β A-1 = 1/42 [A + 4I]
β A-1 =
Question 19. If A = , show that A2 β 5A + 7I = O. Hence find A-1.
Solution:
Here, A =
A2 =
=
Now, A2 β 5A + 7I = + 5 + 7
=
=
Now, A2 β 5A + 7I = O
Multiplying by A-1 both sides
β A-1AA + 5AA β 1 + 7IA-1 = 0
β A-1 = 1/7[5I β A]
β A-1 =
β A-1 =
Question 20. If A = , find x and y such that A2 β xA + yI = O. Hence, evaluate A-1.
Solution:
Here, A =
A2 =
=
Now, A2 β xA + yI = O
β β +
=
β 22 β 4x + y = 0
β 4x β y = 22 β¦β¦β¦(i)
or
18 β 2x = 0
β x = 9
Putting x = 9 in eq (i)
β y = 14
A2 β 9A + 14I = 0
β 9A = A2 + 14I
β 9A-1A = A-1AA + 14A-1
β 9I = IA + 14A-1
β A-1 = 1/14[9I β A] = 1/14()
β A-1=
Question 21. If A = , find the value of Ξ» so that A2 = Ξ»A β 2I. Hence, find A-1.
Solution:
Here, A =
A2 =
=
If A2 = Ξ»A β 2I
Ξ»A = A2 + 2I
β Ξ» =
β Ξ» =
β Ξ» = 1
Now, A2 = Ξ»A β 2I
Multiplying both side A-1
β A-1AA = A-1A β 2A-1I
β A = I β 2A-1
β 2A-1 = I β A =
A-1 =
Question 22. Show that A = satisfies the equation x2 β 3x β 7 = 0. Thus, find A-1.
Solution:
Here, A =
A2 =
Now, A2 β 3A β 7=
=
We have, A2 β 3A β 7 = 0
β A-1AA β 3A-1A β 7A-1 = 0
β A-3I β 7A-1 = 0
β 7A-1 = A β 3I
β 7A-1 = β
A-1 =
Question 23. Show that A = satisfies the equation x2 β 12x + 1 = 0. Thus, find A-1.
Solution:
Here, A =
A2 =
=
Now, A2 β 12A + I = β
=
We have, A2 β 12A + I = 0
β A β 12I + A-1 = 0
β A-1 = 12I β A
β A-1 =
β A-1 =
Question 24. For the matrix A = show that A3 β 6A2 + 5A + 11I3 = O. Hence, find A-1.
Solution:
Here, A =
A2 =
=
A3 =
=
A3 β 6A2 + 5A + 11I
= β 6
=
=
=
We have, A3 β 6A2 + 5A + 11I = O.
β A-1(AAA) β 6A-1(AA) + 5A-1A + 11IA-1 = 0
β A2 β 6A + 5I = -11A-1
β -11A-1 = (A2 β 6A + 5I)
=
=
=
=
Therefore, A-1 =