12th Maths Guide Solution Chapter 1 Application Of Matrices and Determinants Exercise 1.1

Here are the Class 12th maths guide solutions of Chapter 1 Application of Matrices and Determinants Exercise 1.1, according to the Samacheer Kalvi syllabus for Tamil Nadu state board students.

Maths Questions and Answer Solution in PDF for 12th class Tamil Nadu state board students, Samacheer Kalvi Syllabus

Below are the questions and answer notes in PDF for the Tamil Nadu State Board 12th class students according to the Samacheer Kalvi syllabus.

Question No. 1: Find the adjoint of the following.

Application Of Matrices and Determinants Exercise 1.1

Solution:-

Exe 1.1 2
Exe 1.1 3
Exe 1.1 4

Question No 2: Find the inverse (if it exists) of the following.

Exe 1.1 5

Solution:

Exe 1.1 6
Exe 1.1 7 i
Exe 1.1 7 ii
Exe 1.1 8

|A| = 2 (8-7) -3 (6-3)+1 (21-12)
= 2 – 9 + 9 = 2 ≠ 0. A-1  exists.

Exe 1.1 8 ii

Question No 3:

Exe 1.1 9 i

Solution:-

Exe 1.1 9 ii

|F(α)| = cos α(cos α – 0) – 0 + sin α(0 + sin α)
= cos²α + sin²α = 1
|f(α)| = 1 ≠ 0. [F(α)]-1 exists.

Exe 1.1 9 iii
Exe 1.1 9 iv

[cos (-θ) = cos θ ; sin(-θ) = -sin θ]
from (1) and (2) we have
[F(α)]-1 = F(-α)

Question No 4:

Exe 1.1 10 i 1

Solution:-

Exe 1.1 10 ii

A² – 3A – 7I2  = O2
Post multiply this equation by A-1
A2 A-1 – 3A A-1 – 7IA-1 = 0
A – 3I – 7A-1 = 0
A – 3I = 7 A-1
A-1 = 17 (A – 3I)

Exe 1.1 10 iii

Question No 5:

Exe 1.1 11 i

Solution:

Exe 1.1 11 ii
Exe 1.1 11 iii
Exe 1.1 11 iv

Question No 6:

Exe 1.1 12

Solution:-

Exe 1.1 13

(1) (2) and (3) ⇒ A (adj A) = (adj A)A = |A| I2.

Question No 7:

Exe 1.1 14 i

Solution:

Exe 1.1 14 ii
Exe 1.1 14 iii
Exe 1.1 14 iv

Question No 8:

Exe 1.1 15 i

Solution:

Exe 1.1 15 ii

|adj (A)| = 2 (24 – 0) + 4 (- 6 – 14) + 2(0 + 24)
= 48 – 80 + 48 = 16

Exe 1.1 15iii

Question No 9:

Exe 1.1 16i

Solution:-

Exe 1.1 16ii

Question No 10:

Exe 1.1 17

Question No 11:

Exe 1.1 18ii
Exe 1.1 18ii 1

Hence Proved.

Question No 12:

Exe 1.1 19i

Solution:-

Exe 1.1 19ii
Exe 1.1 19iii

Question No 13:

Exe 1.1 20i

Solution:-

Given A × B × C
⇒ A-1 A × B B-1 = A-1 C B-1
I × I = A-1 C B-1
⇒ X = A-1 CB-1
let us find A-1 and B-1

Exe 1.1 20ii
Exe 1.1 20iii

Question No 14:

Exe 1.1 21i

Solution:-

Exe 1.1 21ii
Exe 1.1 21iii

Hence Proved.

Question No 15:

Exe 1.1 22i

Solution:-

Exe 1.1 22ii

So the sequence of decoded row matrics is [8 5] [12 16]
The receiver reads the message as “HELP”.

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