问题10.对于矩阵的以下部分,验证(AB) -1 = B -1 A -1 。
(i)A = 和B =
解决方案:
To prove (AB)-1= B-1A-1
We take LHS
AB =
=
|AB| = 18 × 52 – 22 × 43
= 936 – 946 = -10
adj(AB) =
AB-1= adj(AB)/|AB| =
=
Now,
A =
|A| = 15 – 14 = 1
adj A =
Therefore, A-1 = adj A/|A| =
B =
|B| = 8 – 18 = -10
adj B =
Therefore, B-1= adj B/|B| =
Now, we take RHS
B-1A-1 =
=
=
LHS = RHS
Hence, Proved
(ii)A = 和B =
解决方案:
To prove (AB)-1 = B-1A-1
We take LHS
AB =
=
|AB| = 11 × 27 – 29 × 14
= 407 – 406 = 1
adj(AB) =
AB-1= adj(AB)/|AB| =
=
Now,
A =
|A| = 6 – 5 = 1
adj A=
Therefore, A-1 = adj A/|A| =
B =
|B| = 16 – 15 = 1
adj B =
Therefore, B-1= adj B/|B| =
Now, we take RHS
B-1A-1 =
=
LHS = RHS
Hence, Proved
问题11.让A = 和B = 。求(AB) -1 。
解决方案:
AB =
=
|AB| = 34 × 94 – 39 × 82 = -2
adj(AB) =
AB-1 = adj(AB)/|AB| =
=
问题12:给定A = ,计算A -1并显示2A -1 = 9I –A。
解决方案:
A =
|A| = 14 – 12 = 2
adj A =
Therefore, A-1 = adj A/|A| =
To show 2A-1 = 9I – A.
LHS = 2 × (1/2)
=
Now we take RHS
= 9I – A
= –
=
LHS = RHS
Hence Proved
问题13:如果A = ,则表明A – 3I = 2(I + 3A -1 )。
解决方案:
Here, A =
|A| = 4 – 10 = -6
adj A =
Therefore, A-1 = adj A/|A| =
To show, A – 3I = 2(I + 3A-1)
Now we take LHS
= A – 3I
= – 3
=
Now we take RHS
= 2I + 6A-1
= 2 + 6 × (1/6)
=
LHS = RHS
Hence Proved
问题14:求矩阵A的逆并显示aA -1 =(a 2 + bc + 1)I – aA。
解决方案:
Here, A =
|A| = (a + abc)/a – 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 – aA.
LHS = aA-1
= a
=
RHS = (a2 + bc + 1)I – aA
= – a
= –
=
LHS = RHS
Hence Proved
问题15:给定A = ,B -1 = ,计算(AB) -1 。
解决方案:
We know (AB)-1 = B-1A-1
Here, A =
|A| = 5(3 – 4) + 4(4 – 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
=
=
问题16:令F(α)= 和G(β)= , 显示
(i)[F(α)] -1 = F(-α)
解决方案:
We have F(α) =
|F(α)| = cos2α + sin2α = 1
Therefore, inverse of F(α) exists
Cofactors of F(α) are:
C11 = cosα C12 = -sinα C13 = 0
C21 = sinα C22 = cosα C23 = 0
C31 = 0 C32 = 0 C33 = 1
Adj F(α) =
=
=
[F(α)]-1 = 1/|F(α)|. adj F(α)
Hence, [F(α)]-1 = 1/1
=
Now, F(-α) =
=
So, [F(α)]-1 = F(-α)
Hence, Proved
(ii)[G(β)] -1 = G(-β)
解决方案:
We have G(β) =
|G(β)| = cos2β + sin2β = 1
Therefore, inverse of G(β) exists
Cofactors of G(β) are:
C11 = cosβ C12 = 0 C13 = sinβ
C21 = 0 C22 = 1 C23 = 0
C31 = -sinβ C32 = 0 C33 = sinβ
Adj G(β) =
=
=
[G(β)]-1 = 1/|G(β)|. adj G(β)
Hence, [G(β)]-1 = 1/1
=
Now, G(-β) =
=
So, [G(β)]-1 = G(-β)
Hence, Proved
(iii)[F(α)G(β)] -1 = F(-α)G(-β)
解决方案:
We already know that S[G(β)]-1 = G(-β)
[F(α)]-1 = F(-α)
Taking LHS = [F(α)G(β)]-1
= [F(α)]-1[G(β)]-1
= F(-α)G(-β) = RHS
Hence, Proved
问题17:如果A = ,验证A 2 – 4A + I = O,其中I = 和O = ,因此,找到A -1 。
解决方案:
Here, A =
A2 =
=
4A = 4
=
A2 – 4A + I = O
= – +
=
Hence, =
Now, A2 – 4A + I = O
A2 – 4A = -I
Multiplying both side by A-1 both sides we get
A.A(A-1) – 4AA-1 = -IA-1
AI – 4I = -A-1
A-1 = 4I – AI
= –
=
问题18:证明A = 满足方程A 2 + 4A – 42I =O。因此,找到A -1 。
解决方案:
Here, A =
A2 =
=
=
4A = 4
=
A2 + 4A – 42I = + –
=
Hence,
Now, A2 + 4A – 42I = 0
⇒ A-1A.A + 4A-1.A – 42A-1I = 0
⇒ IA + 4I – 42A-1 = 0
⇒ A-1 = 1/42 [A + 4I]
⇒ A-1 =
问题19:如果A = ,表明A 2 – 5A + 7I =O。因此找到A -1 。
解决方案:
Here, A =
A2 =
=
Now, A2 – 5A + 7I = + 5 + 7
=
=
Now, A2 – 5A + 7I = O
Multiplying by A-1 both sides
⇒ A-1AA + 5AA – 1 + 7IA-1 = 0
⇒ A-1 = 1/7[5I – A]
⇒ A-1 =
⇒ A-1 =
问题20.如果A = ,求x和y使得A 2 – xA + yI =O。因此,求A -1 。
解决方案:
Here, A =
A2 =
=
Now, A2 – xA + yI = O
⇒ – +
=
⇒ 22 – 4x + y = 0
⇒ 4x – y = 22 ………(i)
or
18 – 2x = 0
⇒ x = 9
Putting x = 9 in eq (i)
⇒ y = 14
A2 – 9A + 14I = 0
⇒ 9A = A2 + 14I
⇒ 9A-1A = A-1AA + 14A-1
⇒ 9I = IA + 14A-1
⇒ A-1 = 1/14[9I – A] = 1/14()
⇒ A-1=
问题21:如果A = ,找到λ的值,使A 2 =λA– 2I。因此,找到A -1 。
解决方案:
Here, A =
A2 =
=
If A2 = λA – 2I
λA = A2 + 2I
⇒ λ =
⇒ λ =
⇒ λ = 1
Now, A2 = λA – 2I
Multiplying both side A-1
⇒ A-1AA = A-1A – 2A-1I
⇒ A = I – 2A-1
⇒ 2A-1 = I – A =
A-1 =
问题22:证明A = 满足方程x 2 – 3x – 7 =0。因此,找到A -1 。
解决方案:
Here, A =
A2 =
Now, A2 – 3A – 7=
=
We have, A2 – 3A – 7 = 0
⇒ A-1AA – 3A-1A – 7A-1 = 0
⇒ A-3I – 7A-1 = 0
⇒ 7A-1 = A – 3I
⇒ 7A-1 = –
A-1 =
问题23:证明A = 满足方程x 2 – 12x + 1 =0。因此,找到A -1 。
解决方案:
Here, A =
A2 =
=
Now, A2 – 12A + I = –
=
We have, A2 – 12A + I = 0
⇒ A – 12I + A-1 = 0
⇒ A-1 = 12I – A
⇒ A-1 =
⇒ A-1 =
问题24.对于矩阵A = 证明A 3 – 6A 2 + 5A + 11I 3 =O。因此,找到A -1 。
解决方案:
Here, A =
A2 =
=
A3 =
=
A3 – 6A2 + 5A + 11I
= – 6
=
=
=
We have, A3 – 6A2 + 5A + 11I = O.
⇒ A-1(AAA) – 6A-1(AA) + 5A-1A + 11IA-1 = 0
⇒ A2 – 6A + 5I = -11A-1
⇒ -11A-1 = (A2 – 6A + 5I)
=
=
=
=
Therefore, A-1 =