Find adjoint of each of the matrices in Exercises 1 and 2.
Question 1. [Tex]\begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}[/Tex]
A11 = 4; A12 = -3; A21 = -2; A22 = 1
adj A = [Tex]\begin{bmatrix} A_{11} &A_{21} \\ A_{12} & A_{22} \end{bmatrix}[/Tex]
adj A = [Tex]\begin{bmatrix} 4 &-2 \\ -3 & 1 \end{bmatrix}[/Tex]
Question 2. [Tex]\begin{bmatrix} 1 & -1 &2 \\ 2&3 &5 \\ -2& 0 &1 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1 & -1 &2 \\ 2&3 &5 \\ -2& 0 &1 \end{bmatrix}[/Tex]
A11 = [Tex]\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix}[/Tex]
A11 = 3 – 0 = 3
A12 = [Tex]-\begin{vmatrix} 2&5 \\ -2& 1 \end{vmatrix}[/Tex]
A12 = -(2 + 10) = -12
A13 = [Tex]\begin{vmatrix} 2&3\\ -2& 0 \end{vmatrix}[/Tex]
A13 = 0 + 6 = 6
A21 = [Tex]-\begin{vmatrix} -1&2\\ 0& 1 \end{vmatrix}[/Tex]
A21 = -(-1 – 0) = 1
A22 = [Tex]\begin{vmatrix} 1&2\\ -2& 1 \end{vmatrix}[/Tex]
A22 = 1 + 4 = 5
A23 = [Tex]-\begin{vmatrix} 1&-1\\ -2&0 \end{vmatrix}[/Tex]
A23 = -(0 – 2) = 2
A31 = [Tex]\begin{vmatrix} -1&2\\ 3&5 \end{vmatrix}[/Tex]
A31 = -5 – 6 = -11
A32 = [Tex]-\begin{vmatrix} 1&2\\ 2&5 \end{vmatrix}[/Tex]
A32 = -(5 – 4) = -1
A33 = [Tex]\begin{vmatrix} 1&-1\\ 2&3 \end{vmatrix}[/Tex]
A33 = 3 + 2 = 5
adj A = [Tex]\begin{bmatrix} A_{11} & A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}& A_{23} &A_{33} \end{bmatrix}[/Tex]
adj A = [Tex]\begin{bmatrix} 3& 1 &-11 \\ -12&5 &-1 \\ 6& 2 &5 \end{bmatrix}[/Tex]
Verify A(adj A) = (adj A)A = |A| I in exercises 3 and 4.
Question 3. [Tex]\begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}[/Tex]
Solution:
|A| = -12 -(-12)
|A| = -12 + 12 = 0
so, |A|*I = 0 * [Tex]\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}[/Tex]
|A|*I =[Tex] \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} [/Tex] ……… (1)
Now, for adjoint of A
A11 = -6
A12 = 4
A21 = -3
A22 = 2
adj A = [Tex]\begin{bmatrix} A_{11} &A_{21} \\ A_{12} & A_{22} \end{bmatrix}[/Tex]
adj A = [Tex]\begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix}[/Tex]
Now, A(adj A) = [Tex]\begin{bmatrix} 2&3 \\ -4&-6 \end{bmatrix} \begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix}[/Tex]
A(adj A) = [Tex]\begin{bmatrix} -12+12&-6+6 \\ 24-24&12-12 \end{bmatrix} [/Tex][Tex]\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]
A(adj A) = [Tex]\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} [/Tex]…………(2)
Now, (adj A)A = [Tex]\begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix} \begin{bmatrix} 2&3 \\ -4&-6 \end{bmatrix}[/Tex]
(adj A)A = [Tex]\begin{bmatrix} -12+12&-18+18 \\ 8-8&12-12 \end{bmatrix}[/Tex]
(adj A)A = [Tex]\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} [/Tex] …………….(3)
From eq(1), (2), and (3), you can see that A(adj A) = (adj A)A = |A|I
Question 4. [Tex]\begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}[/Tex]
|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0)
|A| = 11
|A| * I = [Tex]11\begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}[/Tex]
|A| * I = [Tex]\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]
Now, for adjoint of A
A11 = 0
A12 = -(9 + 2) = -11
A13 = 0
A21 = -(-3 – 0) = 3
A22 = 3 – 2 = 1
A23 = -(0 + 1) = -1
A31 = 2 – 0 = 2
A32 = -(-2 – 6) = 8
A33 = 0 + 3 = 3
adj A = [Tex]\begin{bmatrix} 0& 3 &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix}[/Tex]
Now,
A(adjA) = [Tex]\begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0& 3 &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix}[/Tex]
A(adj A) = [Tex]\begin{bmatrix} 0+11+0& 3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 &6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{bmatrix}[/Tex]
A(adj A) = [Tex]\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]
Also,
(adj A).A = [Tex]\begin{bmatrix} 0& 3 &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}[/Tex]
(adj A).A = [Tex]\begin{bmatrix} 0+9+2& 0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 &-22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{bmatrix}[/Tex]
(adj A).A = [Tex]\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]
From above, you can see,
A(adj A) = (adj A)A = I
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
Question 5. [Tex]\begin{bmatrix} 2 & -2\\ 4 & 3 \end{bmatrix}[/Tex]
Solution:
|A| = 6 – (-8) = 14
|A| ≠0, So inverse exists.
A11 = 3
A12 = 2
A21 = -4
A22 = 2
adj A = [Tex]\begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{1}{14}\begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}[/Tex]
Question 6. [Tex]\begin{bmatrix} -1 & 5\\ -3 & 2 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} -1 &5 \\ -3& 2 \end{bmatrix}[/Tex]
|A| = -2 + 15 = 13 ≠0
Hence, inverse exists.
A11 = 2
A12 = 3
A21 = -5
A22 = -1
adj A = [Tex]\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{1}{13}\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}[/Tex]
Question 7. [Tex]\begin{bmatrix} 1& 2 &3 \\ 0 & 2 &4 \\ 0 & 0 & 5 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1& 2 &3 \\ 0 & 2 &4 \\ 0 & 0 & 5 \end{bmatrix}[/Tex]
|A| = 1(10 – 0) – 2(0 – 0) + 3(0 – 0) = 10
For adj A
A11 = 10 – 0 = 0
A12 = -(0 – 0) = 0
A13 = 0 – 0 = 0
A21 = -(10 – 0) = -10
A22 = 5 – 0 = 5
A23 = -(0 – 0) = 0
A31 = 8 – 6 = 2
A32 = -(4 – 0) = -4
A33 = 2 – 0 = 2
adj A = [Tex]\begin{bmatrix} 10& -10 &2 \\ 0 & 5 &-4 \\ 0 & 0 & 2 \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{1}{10}\begin{bmatrix} 10& -10 &2 \\ 0 & 5 &-4 \\ 0 & 0 & 2 \end{bmatrix}[/Tex]
Question 8. [Tex]\begin{bmatrix} 1& 0 &0 \\ 3 & 3 &0 \\ 5 & 2 & -1 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1& 0 &0 \\ 3 & 3 &0 \\ 5 & 2 & -1 \end{bmatrix}[/Tex]
|A| = 1(-3 – 0) – 0 + 0 = -3 ≠0
Hence, inverse exists.
For adj A
A11 = -3 – 0 = -3
A12 = -(-3 – 0) = 3
A13 = 6 – 15 = -9
A21 = -(0 – 0) = 0
A22 = -1 – 0 = -1
A23 = -(2 – 0) = -2
A31 = 0 – 0 = 0
A32 = -(0 – 0) = 0
A33 = 3 – 0 = 3
adj A = [Tex]\begin{bmatrix} -3& 0 &0 \\ 3 & -1 &0 \\ -9 & -2 & 3 \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{-1}{3}\begin{bmatrix} -3& 0 &0 \\ 3 & -1 &0 \\ -9 & -2 & 3 \end{bmatrix}[/Tex]
Question 9. [Tex]\begin{bmatrix} 2& 1 &3 \\ 4 & -1 &0 \\ -7 & 2 & 1 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 2& 1 &3 \\ 4 & -1 &0 \\ -7 & 2 & 1 \end{bmatrix}[/Tex]
|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -3 ≠0
Hence, inverse exists.
For adj A
A11 = -1 – 0 = -1
A12 = -(4 – 0) = -4
A13 = 8 – 7 = 1
A21 = -(1 – 6) = 5
A22 = 2 + 21 = 23
A23 = -(4 + 7) = -11
A31 = 0 + 3 = 3
A32 = -(0 – 12) = 12
A33 = -2 – 4 = -6
adj A = [Tex]\begin{bmatrix} -1& 5 &3 \\ -4 & 23 &12 \\ 1 & -11 & -6 \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{-1}{3}\begin{bmatrix} -1& 5 &3 \\ -4 & 23 &12 \\ 1 & -11 & -6 \end{bmatrix}[/Tex]
Question 10. [Tex]\begin{bmatrix} 1& -1 &2 \\ 0 & 2 &-3 \\ 3 & -2 & 4 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1& -1 &2 \\ 0 & 2 &-3 \\ 3 & -2 & 4 \end{bmatrix}[/Tex]
|A| = 1(8 – 6) – 0 + 3(3 – 4) = -1
Now for adj A
A11 = 8 – 6 =2
A12 = -(0 + 9) = -9
A13 = 0 – 6 = -6
A21 = -(-4 + 4) =0
A22 = 4 – 6 = -2
A23 = -(-2 + 3) = -1
A31 = 3 – 4 = -1
A32 = -(-3 – 0) = 3
A33 = 2 – 0 = 2
adj A = [Tex]\begin{bmatrix} 2& 0 &-1 \\ -9 & -2 &3 \\ -6 & -1 & 2 \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]-\begin{bmatrix} 2& 0 &-1 \\ -9 & -2 &3 \\ -6 & -1 & 2 \end{bmatrix}[/Tex]
A-1 = [Tex]\begin{bmatrix} -2& 0 &1 \\ 9 & 2 &-3 \\ 6 & 1 & -2 \end{bmatrix}[/Tex]
Question 11. [Tex]\begin{bmatrix} 1& 0 &0 \\ 0 & cos\alpha &sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} 1& 0 &0 \\ 0 & cos\alpha &sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}[/Tex]
|A| = 1(-cos2α – sin2α) = -1
Now,
A11 = -cos2α – sin2α = -1
A12 = 0
A13 = 0
A21 = 0
A22 = -cosα
A23 = -sinα
A31 = 0
A32 = -sinα
A33 = cosα
adj A = [Tex]\begin{bmatrix} -1& 0 &0 \\ 0 & -cos\alpha &-sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]-\begin{bmatrix} -1& 0 &0 \\ 0 & -cos\alpha &-sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}[/Tex]
A-1 = [Tex]\begin{bmatrix} 1& 0 &0 \\ 0 & cos\alpha &sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}[/Tex]
Question 12. Let A = [Tex]\begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}[/Tex], verify that (AB) – 1 = B – 1A – 1
Solution:
A = [Tex]\begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}[/Tex]
|A| = 15 – 14 = 1
A11 = 5
A12 = -2
A21 = -7
A22 = 3
A-1 = (adj A)/|A|
A-1 = [Tex]\begin{bmatrix} 5 &-7 \\ -2& 3 \end{bmatrix}[/Tex]
B = [Tex]\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}[/Tex]
|B| = 54 – 56 = -2
adj B = [Tex]\begin{bmatrix} 9 &-8 \\ -7& 6 \end{bmatrix}[/Tex]
B-1 = (adj B)/|B|
B-1 = [Tex]\frac{-1}{2}\begin{bmatrix} 9 &-8 \\ -7& 6 \end{bmatrix}[/Tex]
B-1 = [Tex]\begin{bmatrix} \frac{-9}{2} &4 \\ \frac{7}{2}& -3 \end{bmatrix}[/Tex]
Now,
B-1A-1 = [Tex]\begin{bmatrix} \frac{-9}{2} &4 \\ \frac{7}{2}& -3 \end{bmatrix} \begin{bmatrix} 5 &-7 \\ -2& 3 \end{bmatrix}[/Tex]
B-1A-1 = [Tex]\begin{bmatrix} \frac{-45}{2}-8 &\frac{63}{2}+12 \\ \frac{35}{2}+6& \frac{-49}{2}-9 \end{bmatrix}[/Tex]
B-1A-1 = [Tex]\begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \frac{47}{2}& \frac{-67}{2} \end{bmatrix}[/Tex]
Now, AB = [Tex]\begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix} \begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}[/Tex]
AB = [Tex]\begin{bmatrix} 18+49 &24+63 \\ 12+35& 16+45 \end{bmatrix}[/Tex]
AB = [Tex]\begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}[/Tex]
|AB| = 67 * 61 – 87 * 47 = -2
adj (AB) = [Tex]\begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}[/Tex]
(AB)-1 = (adj AB)/|AB|
(AB)-1 = [Tex]\begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}[/Tex]
(AB)-1= [Tex]\begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \frac{47}{2}& \frac{-67}{2} \end{bmatrix}[/Tex]
From above, you can see that (AB)-1 = B-1A-1.
Hence, it is proved.
Question 13. A = [Tex]\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}[/Tex], show that A2 – 5A + 7I = O. Hence find A-1.
Solution:
A = [Tex]\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} \begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 9-1 &3+2\\ -3-2& -1+4 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}[/Tex]
So, A2 – 5A + 7I
= [Tex]\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix} [/Tex]- 5[Tex]\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix} [/Tex]+ 7[Tex]\begin{bmatrix} 1 &0\\ 0&1 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} [/Tex]+ [Tex]\begin{bmatrix} 7 &0 \\ 0& 7 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 0 &0\\ 0&0 \end{bmatrix}[/Tex]
= O
Hence, A2 – 5A + 7I = O
It can be written as
A.A – 5A = -7I
Multiplying by A-1 in both sides
A.A(A-1) – 5AA-1 = 7IA-1
A(AA-1) – 5I = -7A-1
AI – 5I = -7A-1
A-1 = -(A – 5I)/7
A-1 =1/7([Tex]\begin{bmatrix} 5 &0\\ 0&5 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 3 &1\\ -1&2 \end{bmatrix}[/Tex])
A-1 = [Tex]\frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}[/Tex]
Question 14. For the matrix A =[Tex]\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}[/Tex], find the numbers a and b such that A2 + aA + bI = O.
Solution:
A = [Tex]\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} \begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 9+2 &6+2 \\ 3+1& 2+1 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}[/Tex]
Now,
A2 – aA + bI = O
Multiplying by A-1 in both sides
(AA)A-1 + aAA-1 + bIA-1 = O
A(AA-1) + aI + b(IA-1) = O
AI + aI + bA-1 = O
A + aI = -bA-1
A-1 = -(A + aI)/b
Now,
A-1 = (adj A)/|A|
A-1 = [Tex]\begin{bmatrix} 1 &-2 \\ -1& 3 \end{bmatrix}[/Tex]
Now,
[Tex]\begin{bmatrix} 1 &-2 \\ -1& 3 \end{bmatrix} = -1/b[\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + \begin{bmatrix} a &0 \\ 0& a \end{bmatrix}][/Tex]
= -1/b[Tex]\begin{bmatrix} 3+a &2 \\ 1& 1+a \end{bmatrix} [/Tex]
= [Tex]\begin{bmatrix} \frac{-3-a}{b} &\frac{-2}{b} \\ \frac{-1}{b}& \frac{-1-a}{b} \end{bmatrix}[/Tex]
On comparing elements you will get
-1/b = -1
b = 1
(-3 – a)/b = 1
-3 – a = 1
a = -4
Hence, a = -4 and b = 1
Question 15. A = [Tex]\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}[/Tex], show that A3 – 6A2 + 5A + 11I = O. Hence find A-1
Solution:
A = [Tex]\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 1+1+2& 1+2-1 &1-3+3 \\ 1+2-6 & 1+4+3 &1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix}[/Tex]
A3 = A2.A
A3 = [Tex]\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1& 1 &1 \\ 1 &2 &-3 \\ 2 &-1 & 3 \end{bmatrix}[/Tex]
A3 = [Tex]\begin{bmatrix} 4+2+2& 4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 &-3-24-42 \\ 7-3+28 &7-6-14 & 7+9+42 \end{bmatrix}[/Tex]
A3 = [Tex]\begin{bmatrix} 8& 7 &1 \\ -23 &27 &-69 \\ 32&-13 & 58 \end{bmatrix}[/Tex]
A3 – 6A2 + 5A + 11I
[Tex]\begin{bmatrix} 8& 7 &1 \\ -23 &27 &-69 \\ 32&-13 & 58 \end{bmatrix} [/Tex]- 6[Tex]\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix} [/Tex]+ 5[Tex]\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}[/Tex] + 11[Tex]\begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 8& 7 &1 \\ -23 &27 &-69 \\ 32&-13 & 58 \end{bmatrix} [/Tex]- [Tex]\begin{bmatrix} 24& 12 &6 \\ -18&48 &-84 \\ 42&-18 & 84 \end{bmatrix}[/Tex] + [Tex]\begin{bmatrix} 5& 5 &5 \\ 5&10 &-15 \\ 10&-5 & 15 \end{bmatrix}[/Tex] + [Tex]\begin{bmatrix} 11& 0 &0 \\ 0&11 &0 \\ 0&0 & 11 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 0& 0 &0 \\ 0&0 &0 \\ 0&0 &0 \end{bmatrix}[/Tex]
= O
Hence, A3 – 6A2 + 5A + 11I = O
Now,
A3 – 6A2 + 5A + 11I = O
(AAA)A-1 – 6(AA)A-1 + 5(AA-1) + 11IA-1 = O
AA(AA-1) – 6A(AA-1) + 5(AA-1) = -11(IA-1)
A2 – 6A + 5I = -11 A-1
A-1 = -1/11(A2 – 6A + 5I) ………….(1)
Now, A2 – 6A + 5I
=[Tex]\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix} [/Tex] – 6[Tex]\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}[/Tex] + 5[Tex]\begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}[/Tex]
=[Tex]\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 6& 6 &6 \\ 6&12 &-18 \\ 12&-6 & 18 \end{bmatrix}[/Tex] + [Tex]\begin{bmatrix} 5& 0 &0 \\ 0&5 &0 \\ 0&0 &5 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 3& -4 &-5 \\ -9&1 &4 \\ -5&3 &1 \end{bmatrix}[/Tex]
From eq(1) you have
A-1 = -1/11 [Tex]\begin{bmatrix} 3& -4 &-5 \\ -9&1 &4 \\ -5&3 &1 \end{bmatrix}[/Tex]
Question 16. A = [Tex]\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}[/Tex], verify that A3 – 6A2 + 9A – 4I = O and hence fin A-1.
Solution:
A = [Tex]\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix} \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}[/Tex]
A2 = [Tex]\begin{bmatrix} 4+1+1& -2-2-1 &2+1+2 \\ -2-2-1&1+4+1 &-1-2-2 \\ 2+1+2&-1-2-2 &1+1+4 \end{bmatrix}[/Tex]
A2= [Tex]\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix}[/Tex]
A3 = A2.A
A3 = [Tex]\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix} \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}[/Tex]
A3 = [Tex]\begin{bmatrix} 12+5+5& -6-10-5 &6+5+10 \\ -10-6-5&5+12+5 &-5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}[/Tex]
A3 = [Tex]\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21 \\ 21&-21 &22 \end{bmatrix}[/Tex]
Now,
A3 – 6A2 + 9A – 4I
[Tex]\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21 \\ 21&-21 &22 \end{bmatrix}[/Tex] – 6[Tex]\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix}[/Tex] + 9[Tex]\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix} [/Tex]- 4[Tex]\begin{bmatrix} 1& 0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}[/Tex]
=[Tex]\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21 \\ 21&-21 &22 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 36& -30 &30 \\ -30&36 &-30 \\ 30&-30 &36 \end{bmatrix}[/Tex] + [Tex]\begin{bmatrix} 18& -9 &9 \\ -9&18 &-9 \\ 9&-9 &18 \end{bmatrix} [/Tex]- [Tex]\begin{bmatrix} 4& 0 &0 \\ 0&4 &0 \\ 0&0 &4 \end{bmatrix}[/Tex]
=[Tex]\begin{bmatrix} 40& -30 &30 \\ -30&40 &-30 \\ 30&-30 &40 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 40& -30 &30 \\ -30&40 &-30 \\ 30&-30 &40 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 0& 0 &0 \\ 0&0 &0 \\ 0&0 &0 \end{bmatrix}[/Tex]
= O
So, A3 – 6A2 + 9A – 4I = O
Now,
A3 – 6A2 + 9A – 4I = O
Multiplying by A-1 in both sides
(AAA)A-1 – 6(AA)A-1 + 9AA-1 – 4IA-1 = O
AA(AA-1) – 6A(AA-1) + 9(AA-1) = 4(IA-1)
AAI – 6AI +9I = 4A-1
A2 – 6A + 9I = 4A-1
A-1 = 1/4(A2 – 6A + 9I) ……….(1)
A2 – 6A + 9I
= [Tex]\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix}[/Tex] – 6[Tex]\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}[/Tex] + 9[Tex]\begin{bmatrix} 1& 0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 3& 1 &-1 \\ 1&3 &1 \\ -1&1 &3 \end{bmatrix}[/Tex]
From eq(1), you have
A-1 = [Tex]\frac{1}{4}\begin{bmatrix} 3& 1 &-1 \\ 1&3 &1 \\ -1&1 &3 \end{bmatrix}[/Tex]
Question 17. Let A be a non-singular matrix of order 3 * 3. Then |adj A| is equal to
(A) |A| (B) |A|2 (C) |A|3 (D) 3|A|
Solution:
You know,
(adj A)A = |A|I
(adj A)A = [Tex]\begin{bmatrix} |A| &0 &0 \\ 0 & |A| &0 \\ 0 &0 &|A| \end{bmatrix}[/Tex]
|(adj A)A| = [Tex]\begin{vmatrix} |A| &0 &0 \\ 0 & |A| &0\\ 0 &0 &|A| \end{vmatrix}[/Tex]
|(adj A)A| = |A|3[Tex]\begin{vmatrix} 1 &0 &0 \\ 0 & 1 &0\\ 0 &0 &1 \end{vmatrix}[/Tex]
|(adj A)A| = |A|3 I
|adj A| = |A|3
Hence, option B is correct.
Question 18. If A is an invertible matrix of order 2, then det(A-1) is equal to
(A) det(A) (B) 1/(det A) (C) 1 (D) 0
Solution:
Since A is an invertible matrix then A-1 exists.
And A-1 = (adj A)/|A|
Suppose a 2 order matrix is A = [Tex]\begin{bmatrix} a &b \\ c &d \end{bmatrix}[/Tex]
Then |A| = ad – bc
and adj A = [Tex]\begin{bmatrix} d &-b \\ -c &a \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{1}{|A|}\begin{bmatrix} d &-b \\ -c &a \end{bmatrix}[/Tex]
A-1 = [Tex]\begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \frac{c}{|A|}&\frac{a}{|A|} \end{bmatrix}[/Tex]
|A-1| = [Tex]\begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix}[/Tex]
|A-1| = [Tex]\frac{1}{|A|^{2}}\begin{vmatrix} d &-b \\ -c &a \end{vmatrix}[/Tex]
|A-1| = (ad – bc)/|A|2
|A-1| = |A|/|A|2
|A-1| = 1/|A|
det(A-1) = 1/(det A)
Hence, option B is correct.
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