You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.1

Question 1.

Construct an m × n matrix A = [a_{ij}], where a_{ij} is given by

Solution:

(i) a_{ij} = \(\frac{(i-2 j)^{2}}{2}\)

Here m = 2, n = 3

So we have to construct a matrix of order 2 × 3

(ii) Here m = 3 and n = 4

So we have to construct a matrix order 3 × 4

The general form of a matrix of order 3 × 4 will be

Question 2.

Find the values of p, q, r and s if

Solution:

When two matrices (of the same order) are equal then their corresponding entries are equal.

p^{2} – 1 = 1 ………… (1)

– 31 – q^{3} = – 4 ……….. (2)

r + 1 = \(\frac{3}{2}\) ……… (3)

s – 1 = – π ……….. (4)

(1) ⇒ p^{2} – 1 ⇒ p^{2} = 1 + 1 = 2

p = ±√2

(2) ⇒ – 31 – q^{3} – 4

31 + q^{3} = 4

q^{3} = 4 – 31 = – 27

q^{3} = (-3)^{3}

q = – 3

(3) ⇒ r + 1 = \(\frac{3}{2}\)

r = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\)

(4) ⇒ s – 1 = – π

s = 1 – π

∴ The required values are

p = ±√2,

q = 3,

r = \(\frac{1}{2}\),

s = 1 – π

Question 3.

Determine the value of x + y if

Solution:

\(\left[ \begin{matrix} 2x\quad +\quad y & 4x \\ 5x\quad -\quad 7 & 4x \end{matrix} \right] \) = \(\left[ \begin{matrix} 7 & 7y\quad -\quad 13 \\ y & x\quad +\quad 6 \end{matrix} \right]\)

Equating the corresponding entries

2x + y = 7 ………. (1)

4x = 7y – 13

4x – 7y = – 13 ………. (2)

5x – 7 = y

5x – y = 7 ………. (3)

4x = x + 6

4x – x = 6

3x = 6

x = \(\frac{6}{3}\) = 2 ………. (4)

Substituting for x in equation (1)

(1) ⇒ 2 × 2 + y = 7

y = 7 – 4 = 3

The required values are x = 2 and y = 3

x + y = 2 + 3 = 5

x + y = 5

Question 4.

Determine the matrices A and B if they satisfy

Solution:

Question 5.

If A = \(\left[\begin{array}{ll}{\mathbf{1}} & {\boldsymbol{a}} \\ {\mathbf{0}} & {\mathbf{1}}\end{array}\right]\), then compute A^{4}

Solution:

Question 6.

Consider the matrix A_{α} = \(\left[\begin{array}{cc}{\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha}\end{array}\right]\)

(i) Show that A_{α}A_{β} = A_{α + β}.

(ii) Find all possible real values of satisfying the condition A_{α} + A^{T}_{α} = 1.

Solution:

General solution is α = 2nπ + \(\frac{\pi}{3}\), n ∈ Z

Question 7.

If A = \(\left[\begin{array}{rr}{4} & {2} \\ {-1} & {x}\end{array}\right]\) such that (A – 2I) (A – 3I) = 0, find the value of x.

Solution:

Question 8.

If A = \(\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{0}} \\ {\mathbf{0}} & {\mathbf{1}} & {\mathbf{0}} \\ {\boldsymbol{a}} & {\boldsymbol{b}} & {-\mathbf{1}}\end{array}\right]\), show that A^{2} is a unit matrix.

Solution:

Question 9.

If A = and A^{3} – 6A^{2} + 7A + KI = 0, find the value of k.

Solution:

Question 10.

Give your own examples of matrices satisfying the following conditions in each case:

(i) A and B such that AB ≠ BA.

(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.

(iii) A and B such that AB = 0 and BA ≠ 0.

Solution:

Question 11.

Show that f(x) f(y) = f(x + y), where f(x) =

Solution:

Question 12.

If A is a square matrix such that A^{2} = A, find the value of 7A – (I + A)^{3}.

Solution:

Given A is a square matrix such that A^{2} = A.

(I + A)^{3} = (I + A) (I + A) (I + A)

= (I . I + I . A + A . I + A . A) (I + A)

= (I + A + A + A^{2}) (I + A)

= (I + 2A + A ) (I + A)

[Given A^{2} = A]

= (I + 3A) (I + A)

= I . I + I . A + 3A . I + 3A . A

= I + A + 3A + 3A^{2}

= I + 4A + 3A

[Given A^{2} = A]

(I + A)^{3} = I + 7A

∴ 7A – (I + A)^{2} = 7A – (I + 7A)

= 7A – I – 7A

7A – (I + A)^{2} = – I

Question 13.

Verify the property A (B + C) = AB + AC, when the matrices A, B, and C are given by

Solution:

Question 14.

Find the matrix A which satisfies the matrix relation

Solution:

Question 15.

(i) (A + B)^{T} = A^{T} + B^{T} = B^{T} + A^{T}

(ii) (A – B)^{T} = A^{T} – B^{T}

(iii) (B^{T})^{T} = B.

Solution:

Question 16.

If A is a 3 × 4 matrix and B is a matrix such that both ATB and BAT are defined, what is the order of the matrix B?

Solution:

Given Order of A = 3 × 4

∴ Order of A^{T} = 4 × 3

Given that A^{T}B is defined.

∴ Number of columns of A^{T} = Number of rows of B

Number of rows of B = 3

Also given BA^{T} is defined.

∴ Number of columns of B = Number of rows of A^{T}

Number of columns of B = 4

∴ Order of B = 3 × 4

Question 17.

Express the following matrices is the sum of a symmetric matrix and a skew-symmetric matrix:

Solution:

Question 18.

Find the matrix A such that

Solution:

Question 19.

If A = is a matrix such that AA^{T} = 9I, find the values of x and y.

Solution:

Question 20.

Solution:

Question 21.

Construct the matrix A = [a_{ij}]_{3×3}, where a_{ij} = i- j. State whether A is symmetric or skew- symmetric.

Solution:

Given A is a matrix of order 3 × 3

Here A^{T} = -A

⇒ A is skew-symmetric

Question 22.

Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.

Solution:

Let A and B be two symmetric matrices

⇒ A^{T} = A and B^{T} = B …………….. (1)

Given that AB = BA (2)

To prove AB is symmetric:

Now (AB)^{T} = B^{T}A^{T} = BA

(from(1)) But (AB)^{T} = AB by ………….. (2)

⇒ AB is symmetric.

Conversely, let AB be a symmetric matrix.

⇒ (AB)^{T} = AB

i.e. B^{T}A^{T} = AB

i.e. BA = AB (from (1))

⇒ AB is symmetric

Question 23.

If A and B are symmetric matrices of the same order, prove that

(i) AB + BA is a symmetric matrix.

(li) AB – BA is a skew-symmetric matrix.

Solution:

Given A and B are symmetric matrices of the same order.

∴ A^{T} = A , B^{T} = B

(i) AB + BA is a symmetric matrix

(AB + BA)^{T} = (AB )^{T} + (BA)^{T}

= B^{T}A^{T} + A^{T}B^{T}

= BA + AB

= AB + BA

∴ AB + BA is a symmetric matrix.

(ii) AB – BA is a skew – symmetric matrix

(AB – BA)^{T} = (AB )^{T} (BA )^{T}

= B^{T}A^{T} – A^{T}B^{T}

= BA – AB

(AB – BA)^{T} = – (AB – BA)

∴ AB – BA is a skew symmetric matrix.

Question 24.

A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds.

The pack-I contains 100 gm of cashew nuts, 100 gm of raisins, and 50 gm of almonds. Pack-II contains 200 gm of cashew nuts, 100 gm of raisins, and 100 gm of almonds. Pack-III contains 250 gm of cashew nuts, 250 gm of raisins, and 150 gm of almonds. The cost of 50 gm of cashew nuts is ₹ 50, 50 gm of raisins is ₹ 10, and 50 gm of almonds is₹ 60. What is the cost of each gift pack?

Solution:

### Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.1 Additional Problems

Question 1.

Prove that (i) AB ≠ BA

(ii) A(BC) = (AB) C

(iii) A(B + C) = AB + AC

(iv) AI = IA = A

Solution:

Question 2.

If A = \(\left[\begin{array}{ll}{2} & {3} \\ {4} & {5}\end{array}\right]\) find A^{2} – 7A – 2I.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

If A = \(\left[\begin{array}{rr}{3} & {-5} \\ {-4} & {2}\end{array}\right]\), show that A^{2} – 5A – 14I = 0 where I is the unit matrix of order 2.

Solution:

Question 6.

If A = \(\left[\begin{array}{rr}{3} & {-2} \\ {4} & {-2}\end{array}\right]\), find k so that A^{2} = kA – 2I

Solution:

Question 7.

If A = \(\left[\begin{array}{lll}{1} & {2} & {2} \\ {2} & {1} & {2} \\ {2} & {2} & {1}\end{array}\right]\), show that A^{2} – 4A – 5I = 0

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution: