If A is a non-singular matrix of order 3, then we get
Question 1 :
|adj A| = 2(24 - 0) + 4(-6 - 14) + 2(0 + 24)
= 2(24) + 4(-20) + 2(24)
= 48 - 80 + 48
= 96 - 80
= 16
√adj A = √16 = 4
By applying these values in the formula, we get
Question 2 :
|adj A| = 0(12 + 0) + 2(36 - 18) + 0(0 + 6)
= 0 + 2(18) + 0
|adj A| = 36
√adj A = √36 = 6
Question 3 :
Solution :
Question 4 :
Solution :
|A| = 1 + tan2 x
|A| = sec2 x
Hence proved.
Question 4 :
Solution :
5a - b = 14 -------(1)
3a - 2b = 7 -----(2)
(1)x2==> 10a - 2b = 28
3a - 2b = 7
(-) (+) (-)
-----------------
7a = 21
a = 21/7 = 3
By applying the value of a in (1), we get
5(3) - b = 14
-b = 14 - 15
-b = -1
b = 1
5c - d = 7 -------(3)
3c - 2d = 7 -----(4)
(3) x 2 ==> 10c - 2d = 14
3c - 2d = 7
(-) (+) (-)
-----------------
7c = 7
c = 7/7 = 1
By applying the value of c in (3), we get
5(1) - d = 7
5 - d = 7
-d = 7 - 5
d = -2
Question 5 :
Solution :
3(a-c)+(b-d) = 1
3a - 3c + b - d = 1 ---(1)
-2(a-c)+(b-d) = 1
-2a+2c + b - d = 1------(2)
6a + 2b = 2 ----(3)
-4a + 2b = 2 -----(4)
(3)/2 ==> 3a + b = 1
(4)/2 ==> -2a + b = 1
By subtracting the above equations, we get
a = 0
By applying the value of a in (3), we get
3(0) + b = 1
b = 1
From (1),
3a - 3c + b - d = 1 ---(1)
-3c + 1 - d = 1
-3c - d = 1 - 1
-3c - d = 0 --(A)
From (2),
-2a + 2c + b - d = 1------(2)
2c + 1 - d = 1
2c - d = 1 - 1
2c - d = 0--(B)
By solving (A) and (B), we get
(A) - (B)
-3c - 2c = 0
c = 0
By applying the value of c in (A), we get
d = 0
Question 6 :
Solution :
A-1 = (1/|A|) adj A
|A| = 0(0-1) - 1(0-1) + 1(1-0)
= 0 + 1 + 1
|A| = 2
Hence proved.
Question 7 :
Decrypt the received encoded message [2 -3] [20 4] with the encryption matrix
and the decryption matrix as its inverse, where the system of codes are described by the numbers 1-26 to the letters A− Z respectively, and the number 0 to a blank space.
Solution :
Hence the required word is "HELP".
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