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

**11th Maths Matrices And Determinants Solutions 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

**11th Maths Exercise 7.1 Answers Question 2.**

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

Solution:

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

⇒ p^{2} – 1 = 1

⇒ p^{2} = 1 + 1 = 2

p = ± \(\sqrt{2}\)

-31 – q^{3} = -4

-q^{3} = -4 + 31 = 27

q^{3} = -27 = (-3)^{3}

⇒ q = -3

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

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

s – 1 = π

⇒ s = – π + 1 (i.e.,) s = 1 – π

So, p = ± \(\sqrt{2}\), q = -3, r = 1/2 and s = 1 – π

**11th Maths 7.1 Exercise Question 3.**

Determine the value of x + y if

Solution:

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

⇒ 2x + y = 7 ………….. (1)

4x = 7y – 13 ………….. (2)

5x – 7 = y …………… (3)

4x = x + 6 ……………. (4)

from (4) 4x – x = 6

3x = 6 ⇒ x = \(\frac{6}{3}\) = 2

Substituting x = 2 in (1), we get

2(2) + y = 7 ⇒ 4 + y = 7 ⇒ y = 7 – 4 = 3

So x = 2 and y = 3

∴ x + y = 2 + 3 = 5

**11th Maths Exercise 7.1 Question 4.**

Determine the matrices A and B if they satisfy

Solution:

**11th Maths Matrix Solutions Question 5.**

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

Solution:

**11th Maths Volume 2 Exercise 7.1 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

**11th Maths Exercise 7.1 Tamil Medium 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:

**11th Maths Matrices And Determinants Pdf 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:

**Matrices And Determinants Class 11 State Board Solutions Question 9.**

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

Solution:

**11th Maths Matrix And Determinants 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:

**11th Maths Volume 2 Exercise 7.1 Answers Question 11.**

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

Solution:

**11 Matric Maths Solutions Question 12.**

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

Solution:

Given A^{2} = A

So 7A – (I + A)^{3} = 7A – (I + 3A + 3A^{2} + A^{3}]

= 7A – I – 3A – 3 A^{2} – A^{3}

Given A^{2} = A

7A – I – 3A – 3A – A^{3} = -I + A – A^{3}

= -I + A – (A^{2} × A)

= -I + A – (A × A) = -I + A – A^{2}

= -I + A – A = -I

So the value of 7A – (I + A)^{3} = -I.

**Class 11th Maths Exercise 7.1 Question 13.**

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

Solution:

**11th Maths Determinants Solutions Question 14.**

Find the matrix A which satisfies the matrix relation

Solution:

**Matrices And Determinants Class 11 Solutions Pdf 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:

**Class 11 Matric Maths Solutions 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?

Sol.

A is a matrix of order 3 × 4

So AT will be a matrix of order 4 × 3

AT B will be defined when B is a matrix of order 3 × n

BA^{T} will be defirted when B is of order m × 4

from (1) and (2) we see that B should be a matrix of order 3 × 4

**Class 11 Maths Ex 7.1 Solutions Question 17.**

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

Solution:

**11th Matrices And Determinants Question 18.**

Find the matrix A such that

Solution:

**Class 11th Maths Chapter 7 Exercise 7.1 Question 19.**

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

Solution:

**11th Maths Book 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 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

⇒ – A^{T} = A and B^{T} = B

(i) To prove AB + BA is a symmetric matrix.

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

= BA + AB = AB + BA

i.e. (AB + BA)^{T} = AB + BA

⇒ (AB + BA) is a symmetric matrix.

(ii) To prove AB – BA is a skew symmetric matrix.

Proof: (AB – BA)^{T} = (AB)^{T} – (BA)^{T} = B^{T}A^{T} – A^{T}B^{T} = BA – AB

i.e. (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.

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: