Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.1 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.

Find the adjoint of the following:

Solution:

Question 2.

Find the inverse (if it exists) of the following.

Solution:

\(\begin{bmatrix} -2 & 4 \\ 1 & -3 \end{bmatrix}\)

|A| = 6 – 4 = 2 ≠ 0

∴ A^{-1} exists. A is non singular.

|A| = 2(8-7)-3(6-3)+1(21-12)

= 2 – 9 + 9 = 2 ≠ 0. A^{-1} exists.

Question 3.

Solution:

\(\left[\begin{array}{ccc}

\cos \alpha & 0 & \sin \alpha \\

0 & 1 & 0 \\

-\sin \alpha & 0 & \cos \alpha

\end{array}\right]\)

|F(α)| = cos α(cos α – 0) – 0 + sin α(0 + sin α)

= cos²α + sin²α = 1

|f(α)| = 1 ≠ 0. [F(α)]^{-1} exists.

[∵ cos (-θ) = cos θ ; sin(-θ) = -sin θ]

from (1) and (2) we have

[F(α)]^{-1} = F(-α)

Question 4.

If A = \(\begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}\), show that A² – 3A – 7I_{2} = O_{2}. hence find A^{-1}

Solution:

∴ A² -3A – 7I_{2} = O_{2}

Post multiply this equation by A^{-1
}A^{2}A^{-1 }– 3A A^{-1 }– 7I_{2 }A^{-1} = 0

A – 3I – 7A^{-1} = 0

A – 3I = 7 A^{-1}

A^{-1} = \(\frac {1}{7}\) (A – 3I)

Question 5.

Solution:

Question 6.

If A = \(\begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix}\) verify that A(adj A) = (adj A) A = \(\left| A \right|\)I_{2}.

Solution:

(1), (2) and (3) ⇒ A (adj A) = (adj A)A = |A| I_{2}.

Question 7.

Solution:

Question 8.

Solution:

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

= 48 – 80 + 48 = 16

Question 9.

Solution:

|adj A| = 0 + 2(36 – 18) + 0 = 2(18) = 36

Question 10.

Solution:

Question 11.

Solution:

Hence proved

Question 12.

Solution:

Question 13.

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}

Question 14.

Solution:

Hence proved.

Question 15.

Decrypt the received encoded message [2 -3] [20 4] with the encryption matrix \(\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\) 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:

So the sequence of decoded row matrics is [8 5] [12 16]

The receiver reads the message as “HELP”.