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

**12th Maths Exercise 1.1 Answers Question 1.**

Find the adjoint of the following:

Solution:

**Exercise 1.1 Class 12 Maths State Board Question 2.**

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

Solution:

For a matrix A, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}(\mathrm{adj} \mathrm{A})\). Where |A| ≠ 0. If |A| = 0 then A is called a singular matrix and so \(\mathrm{A}^{-1}\) does not exist.

**12th Maths Exercise 1.1 Question 3.**

If F(α) = \(\left[\begin{array}{ccc}{\cos \alpha} & {0} & {\sin \alpha} \\ {0} & {1} & {0} \\ {-\sin \alpha} & {0} & {\cos \alpha}\end{array}\right]\) show that \([\mathrm{F}(\alpha)]^{-1}=\mathrm{F}(-\alpha)\)

Solution:

Let A = F (α)

So \([\mathrm{F}(\alpha)]^{-1}=\mathrm{A}^{-1}\)

Now

**12th Maths Chapter 1 Exercise 1.1 Question 4.**

If A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\) show that A^{2} – 3A – 7I_{2} = O_{2}. Hence find A^{-1}.

Solution:

A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\)

To Find A-1

Now we have proved that A^{2} – 3A – 7I_{2} = O_{2}

Post multiply by A^{-1} we get

A – 3I – 7A^{-1} = O_{2}

**12th Maths 1.1 Exercise Question 5.**

If \(\mathbf{A}=\frac{1}{9}\left[\begin{array}{ccc}{-8} & {1} & {4} \\ {4} & {4} & {7} \\ {1} & {-8} & {4}\end{array}\right]\) prove that A^{-1} = A^{T}

Solution:

**12 Maths Exercise 1.1 Question 6.**

If \(\mathbf{A}=\left[\begin{array}{rr}{8} & {-4} \\ {-5} & {3}\end{array}\right]\), verify that A(adj A) = (adj A)A = |A| I_{2}

Solution:

**12th Maths 1st Chapter Exercise 1.1 Question 7.**

If \(\mathbf{A}=\left[\begin{array}{ll}{3} & {2} \\ {7} & {5}\end{array}\right]\), and \(\mathbf{B}=\left[\begin{array}{cc}{-1} & {-3} \\ {5} & {2}\end{array}\right]\) verify that (AB)^{-1} = B^{-1} A^{-1}.

Solution:

**12th Maths Application Of Matrices And Determinants Question 8.**

If adj (A) = \(\left[\begin{array}{ccc}{2} & {-4} & {2} \\ {-3} & {12} & {-7} \\ {-2} & {0} & {2}\end{array}\right]\) find A

Solution:

**12th Maths Ex 1.1 Question 9.**

If adj(A) = \(\left[\begin{array}{ccc}{0} & {-2} & {0} \\ {6} & {2} & {-6} \\ {-3} & {0} & {6}\end{array}\right]\) find A^{-1}

Solution:

**12th Exercise 1.1 Question 10.**

Find adj(adj(A)) if adj A = \(\left[\begin{array}{ccc}{1} & {0} & {1} \\ {0} & {2} & {0} \\ {-1} & {0} & {1}\end{array}\right]\)

Solution:

**12th Maths Exercise 1.1 Answers In Tamil Medium Question 11.**

Solution:

**12th Maths 1st Chapter Question 12.**

Find the matrix A for which A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)

Solution:

Given A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)

Let \(\mathrm{B}=\left(\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right) \text { and } \mathrm{C}=\left(\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right)\)

Given AB = C, To find A

Now AB = C

Post multiply by B^{-1} on both sides

ABB^{-1} = CB^{-1} (i.e) A (BB^{-1}) = CB^{-1}

⇒ A(I) = CB^{-1} (i.e) A = CB^{-1}

To find B^{-1}:

**12th Maths 1.1 Question 13.**

Given \(\mathbf{A}=\left[\begin{array}{cc}{1} & {-1} \\ {2} & {0}\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}{3} & {-2} \\ {1} & {1}\end{array}\right] \text { and } \mathbf{C}\left[\begin{array}{ll}{1} & {1} \\ {2} & {2}\end{array}\right]\), find a matrix X such that AXB = C.

Solution:

A × B = C

Pre multiply by A^{-1} and post multiply by B^{-1} we get

A^{-1} A × BB^{-1} = A^{-1}CB^{-1} (i.e) X = A^{-1}CB^{-1}

**12 Maths Samacheer Kalvi Solutions Question 14.**

Solution:

**12th Maths Exercise 1.1 5th Sum Question 15.**

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

Let the encoding matrix be \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\)

So the sequence of decoded matrices is [8 5], [12 16].

Thus the receivers read this message as HELP.

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

**12th Maths Chapter 1 Question 1.**

Using elementary transformations find the inverse of the following matrix

Solution:

**12th Maths Guide Question 2.**

Using elementary transformations find the inverse of the matrix

Solution:

**12th Maths Solutions Samacheer Kalvi Question 3.**

Using elementary transformation find the inverse of the matrix

Solution:

**12 Maths Chapter 1 Exercise 1.1 Question 4.**

Using elementary transformations find the inverse of the matrix

Solution:

**12th Maths Exercise 1.1 Solutions Question 5.**

Using elementary transformation, find the inverse of the following matrix

Solution:

**Samacheer Kalvi 12th Maths Guide Question 6.**

Solution:

**Samacheer Kalvi 12 Maths Solutions Question 7.**

Solution:

**12 Maths Solutions Samacheer Kalvi Question 8.**

Solution:

Question 9.

Solution:

Question 10.

Solution: