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**Q.1. If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?****Ans.**

The possible orders that a matrix having 28 elements are {28 Ã— 1, 1 Ã— 28, 2 Ã— 14, 14 Ã— 2, 4 Ã— 7, 7 Ã— 4}. The possible orders of a matrix having 13 elements are {1 Ã— 13, 13 Ã— 1}.**Q.2. In the matrix A = write:****(i) The order of the matrix A ****(ii) The number of elements ****(iii) Write elements a _{23}, a_{31}, a_{12}**

So, a

Q.3. Construct a

Let

Hence, the matrix A =

Hence, the matrix A =

Q.4. Construct a 3 Ã— 2 matrix whose elements are given by a

Given that a

a

a

a

Hence, the matrix

Given that A = B

â‡’

Equating the corresponding elements, we get

a + 4 = 2a + 2, 3b = b

â‡’ 2a - a = 2, b

âˆ´ a = 2

but here 2 is common.

Hence, the value of a = 2 and b = 2.

The order of matrix A = 2 Ã— 2 and the order of matrix B = 2 Ã— 3 Addition of matrices is only possible when they have same order. So, A + B is not possible.

Given that

Equating the corresponding elements, we get

5 + a = 0 â‡’ a = â€“ 5, 2 + b = 0 â‡’ b = â€“ 2, â€“ 2 + c = 0 â‡’ c = 2

12 + d = 0 â‡’ d = â€“ 12, e = 0, 1 + f = 0 â‡’ f = â€“ 1

Hence, the matrix,

The given equation can be written as

Equating the corresponding elements we get

12x = 48, 3x + 8 = 20, x

âˆ´ x = 48/12 = 4, 3x = 20 â€“ 8 = 12, â‡’ x

âˆ´ x = 4, â‡’ x

Hence, the non-zero values of x is 4.

Given that

A + B =

â‡’ A + B =

â‡’

âˆ´ (A + B) Ã— (A - B) =

Now, R.H.S. = A

= A Ã— A - B Ã— B

Hence,

Hence, (A + B) . (A â€“ B) â‰ A

Given that,

â‡’

â‡’

â‡’ [2x + 16 + 10x + 12 + x^{2} + 4x] = 0; â‡’ x^{2} + 16x + 28 = 0

â‡’ x^{2} + 14x + 2x + 28 = 0; â‡’ x(x + 14) + 2(x + 14) = 0

â‡’ (x + 2) (x + 14) = 0; x + 2 = 0 or x + 14 = 0

âˆ´ x = â€“ 2 or x = â€“ 14

Hence, the values of x are â€“ 2 and â€“ 14.**Q.11. Show thatsatisfies the equation A ^{2} â€“ 3A â€“ 7I = O and hence find A^{â€“1}.**

Given that

A

A

We are given A

â‡’ A

[Pre-multiplying both sides by A

â‡’ A

â‡’ I Ã— A â€“ 3I â€“ 7A

â‡’ A â€“ 3I â€“ 7A

â‡’ â€“7A

Hence,

Let

Equating the corresponding elements, we get,

â€“ 6a â€“ 3c + 10b + 5d = 1 ...(1)

â€“9a â€“ 6c + 15b + 10d = 0 ...(2)

4a + 2c â€“ 6b â€“ 3d = 0 ...(3)

6a + 4c â€“ 9b â€“ 6d = 1 ...(4)

Multiplying eq. (1) by 2 and subtracting eq. (2), we get,

â€“ 3a + 5b = 2 ...(5)

Now, multiplying eq. (3) by 2 and subtracting eq. (4), we get

2a â€“ 3b = â€“ 1 ...(6)

Solving eq. (5) and (6) i.e.,

â€“ 3a + 5b = 2

2a â€“ 3b = â€“ 1

2 x (- 3a + 5b = 2) â‡’ 6a + 10b = 4

3 x (2a - 3b = -1) â‡’ 6a - 9b = - 3

Adding b = 1

Putting the value of b in eq. (6), we get,

2a â€“ 3 x 1 = - 1

â‡’ 2a - 3 = - 1 â‡’ 2a = 3 - 1 â‡’ 2a = 2

âˆ´ a = 1

Now, putting the values of a and b in equations (1) and (3)

â€“6 x 1 - 3c + 10 x 1 + 5d = 1

â‡’ - 6 - 3c + 10 + 5d = 1

â‡’ â€“ 3c + 5d + 4 = 1 â‡’ 3c + 5d = - 3 ...(7)

and from eq. (3)

4 x 1 + 2c - 6 x 1 - 3d = 0; â‡’ 4 + 2c - 6 - 3d = 0

â‡’ - 2 + 2c - 3d = 0 â‡’ 2c - 3d = 2 ...(8)

Solving eq. (7) and (8) we get,

Adding

Putting the value of d in eq. (8) we get,

2c â€“ 3 x 0 = 2 â‡’ 2c= 2 â‡’ c = 1

Hence,

Order ofis 3 Ã— 1 and order ofis 3 x 3. So, the order of matrix A must be 1 x 3.

Equating the corresponding elements to get the values of a, b and c

4a = â€“ 4, 4b = 8, 4c = 4

âˆ´ a = â€“ 1 âˆ´ b = 2 âˆ´ c = 1

Hence, matrix A = [- 1 2 1]

Here,

âˆ´

Here, number of columns of first i.e., 3 is not equal to the number of rows of second matrix i.e., 2.

So, B

Hence, (BA)

Hence,

Let

Hence,

Q.17. Given Is (AB)â€² = Bâ€²Aâ€²?

Here,

L.H.S.

Hence, L.H.S. = R.H.S.

Q.18. Solve for x and y:

Given that:

Comparing the corresponding elements of both sides, we get,

2x + 3y â€“ 8 = 0 â‡’ 2x + 3y = 8 ...(1)

x + 5y â€“ 11 = 0 â‡’ x + 5y = 11 ...(2)

Multiplying eq. (1) by 1 and eq. (2) by 2, and then on subtracting, we get,

âˆ´ y = 2

Putting y = 2 in eq. (2) we get,

x + 5 x 2 = 11 â‡’ x + 10 = 11

âˆ´ x = 11 â€“ 10 = 1

Hence, the values of x and y are 1 and 2 respectively.

Q.19. If X and Y are 2 Ã— 2 matrices, then solve the following matrix equations for X and Y

Given that:

Multiplying eq. (1) by 3 and eq. (2) by 2, we get,

On subtracting eq. (4) from eq. (3) we get

Now, putting the value of Y in equation (1) we get,

Hence,

Given that:

Let,

AC = BC (Given)

â‡’ [3Î± + 5Î²] = [7Î± + 3Î²]

â‡’ 3Î± + 5Î² = 7Î± + 3Î²

â‡’ 3Î± â€“ 7Î± = 3Î² â€“ 5Î²

â‡’ â€“ 4Î± = â€“ 2Î²

âˆ´

So, let Î± = K and Î² = 2K, K is some real number.

Hence, matrix C =

Let,

Hence, AB = AC for matrix A is non-zero and B â‰ C.

**(i) (AB) C = A (BC) ****(ii) A (B + C) = AB + AC.****Ans.**

Given that **(i) **To verify: (AB)C = A(BC)

L.H.S. = R.H.S.

So, (AB)C = A(BC)**(ii) **To verify: A(B + C) = AB + AC

L.H.S. B + C

L.H.S.A(B + C)

R.H.S. AB

AC

R.H.S. AB + AC

L.H.S. = R.H.S.

Hence, A(B + C) = AB + AC

Q.23. If**Ans.**

Given that:

Now

Hence, PQ = QP.**Q.24. If :find A.****Ans.**

Given that:

L.H.S.

â‡’

Hence, matrix A = [â€“ 4]**Q.25. If A = [ 2 1] , B =verify that A (B + C) = (AB + AC).****Ans.**

Given that:

L.H.S.

R.H.S.

L.H.S. = R.H.S.

Hence, A(B + C) = (AB + AC) is verified.**Q.26. If ****then verify that A ^{2} + A = A (A + I), where I is 3 Ã— 3 unit **

Given that:

A

L.H.S. A

R.H.S. A(A + I) =

L.H.S. = R.H.S.

A

(ii) (AB)â€² = Bâ€²Aâ€²

(iii) (kA)â€² = (kAâ€²).

Given that:

R.H.S.

L.H.S. = R.H.S.

Hence,(AB)â€² = Bâ€²Aâ€²is verified.

R.H.S.

Hence, L.H.S. = R.H.S.

(kA)â€² = (kAâ€²) is verified.

(ii) (A â€“ B)â€² = Aâ€² â€“ Bâ€².

Given that:

L.H.S. (2A + B)â€² =

R.H.S. 2Aâ€² + Bâ€² =

Hence, L.H.S. = R.H.S.

(2A + B)â€² = 2Aâ€² + Bâ€²is verified.

L.H.S. (A â€“ B)â€² =

R.H.S.Aâ€² â€“ Bâ€²=

Hence, L.H.S. = R.H.S.

(A â€“ B)â€² = Aâ€² â€“ Bâ€² is verified.

Let P = Aâ€²A

â‡’ Pâ€² = (Aâ€²A)â€²

â‡’ Pâ€² = Aâ€²(Aâ€²)â€² [(AB)â€² = Bâ€²Aâ€²]

â‡’ Pâ€² = Aâ€²A [âˆµ (Aâ€²)â€² = A]

â‡’ Pâ€² = P

Hence, AÂ¢A is a symmetric matrix.

Now, Let Q = AAâ€²

â‡’ Qâ€² = (AAâ€²)â€²

â‡’ Qâ€² = (Aâ€²)â€² Aâ€² [(AB)â€² = Bâ€²Aâ€²]

â‡’ Qâ€² = AAâ€² [âˆµ (Aâ€²)â€² = A]

â‡’ Qâ€² = Q

Hence, AAâ€² is also a symmetric matrix.

Given that A and B are the matrices of the order 3 x 3.

(AB)

= AA Ã— BB

= A

Hence, (AB)

To prove that (A + B)

L.H.S. (A + B)

= A Ã— A + AB + BA + B Ã— B

= A

= A

So, L.H.S. = R.H.S.

L.H.S. A + (B + C) =

R.H.S. (A + B) + C

Hence, L.H.S. = R.H.S.

A + (B + C) = (A + B) + C Hence proved.

L.H.S. A(BC) =

R.H.S. (AB)C

Hence, L.H.S. = R.H.S.

A(BC) = (AB)C Hence proved.

Here, a = 4 and b = â€“ 2

L.H.S. (a + b)B =

R.H.S. aB + bB

Hence, L.H.S. = R.H.S.

(a + b)B = aB + bB Hence proved.

L.H.S. a(C â€“ A) =

R.H.S. aC â€“ aA =

Hence, L.H.S. = R.H.S.

a(C â€“ A) = aC â€“ aA Hence proved.

L.H.S.

R.H.S.

Hence, (A

L.H.S. (bA)

R.H.S. bA

Hence, L.H.S. = R.H.S.

(bA)

L.H.S. (AB)

R.H.S. B

Hence, L.H.S. = R.H.S.

(AB)

L.H.S. (A â€“ B)C =

R.H.S. AC â€“ BC =

Hence, L.H.S. = R.H.S.

(A â€“ B)C = AC â€“ BC

L.H.S. (A â€“ B)

R.H.S. A

Hence, L.H.S. = R.H.S.

(A â€“ B)

A = A . A=

Hence proved.

L.H.S. (A + B)

Put x

R.H.S.

A

Hence, L.H.S. = R.H.S.

(A + B)

L.H.S. A

Hence, A

Q.36. Prove by Mathematical Induction that (Aâ€²)â€³ = (Aâ€³)â€², where n âˆˆ N for any square matrix A.

To prove that (Aâ€²)â€³ = (Aâ€³)â€²

Let P (n): (Aâ€²)â€³ = (Aâ€³)â€²

Step 1: Put n = 1, P(1): Aâ€² = Aâ€² which is true for n = 1

Step 2: Put n = K, P(K): (Aâ€²)

Step 3: Put n = K + 1, P(K + 1): (Aâ€²)

L.H.S. (Aâ€²)

= (A

= (A

= (A

The given statement is true for P(K+1) whenever it is true for P(K), where K âˆˆ N.

|A| = 1 x 7 - (- 5) x 3 = 7 + 15 = 22 â‰ 0

So, A is invertible.

Let A = IA

R

â‡’

R

â‡’=

So

Hence, inverse of

|A| = 1 x 6 - (- 3) (- 2) = 6 - 6 = 0

|A| = 0 so A is not invertible.

Hence, inverse ofis not possible.

Given that:

Equating the corresponding elements,

xy = 8, w = 4, z + 6 = 0 â‡’ z = â€“ 6, x + y = 6

Now, solving x + y = 6 ...(i)

and xy = 8 ...(ii)

From eqn. (i), y = 6 â€“ x ...(iii)

Putting the value of y in eqn. (ii) we get,

x(6 â€“ x) = 8 â‡’ 6x â€“ x

â‡’ x

â‡’ x(x â€“ 4) â€“ 2(x â€“ 4) = 0 â‡’ (x - 4) (x - 2) = 0

âˆ´ x = 4, 2

From eqn. (iii), y = 2, 4.

Hence, x = 4 or 2, y = 2 or 4, z = â€“ 6 and w = 4.

Order of matrices A and B is 2 Ã— 2.

âˆ´ Order of matrix C must be 2 Ã— 2.

Let

âˆ´ 3A + 5B + 2C = 0

Equating the corresponding elements, we get,

48 + 2a = 0 â‡’ 2a = - 48 â‡’ a = - 24

20 + 2b = 0 â‡’ 2b = - 20 â‡’ b = - 10

56 + 2c = 0 â‡’ 2c = - 56 â‡’ c = - 28

76 + 2d = 0 â‡’ 2d = - 76 â‡’ d = - 38

Hence,

Given that:

A

âˆ´ A

Hence, A

Now, multiplying both sides by A, we get,

A

â‡’ A

â‡’ A

Hence, A

Given that:

Equating the corresponding elements, we get,

3a = a + 4 â‡’ 3a â€“ a = 4 â‡’ 2a = 4 â‡’ a = 2

3b = 6 + a + b â‡’ 3b â€“ b â€“ a = 6 â‡’ 2b â€“ a = 6 â‡’ 2b â€“ 2 = 6

â‡’ 2b = 8

â‡’ b = 4 3c = â€“ 1 + c + d

â‡’ 3c â€“ c â€“ d = â€“ 1 â‡’ 2c â€“ d = â€“ 1

and 3d = 2d + 3 â‡’ 3d â€“ 2d = 3 â‡’ d = 3

Now 2c â€“ d = â€“ 1

â‡’ 2c â€“ 3 = â€“ 1 â‡’ 2c = 3 - 1 â‡’ 2c = 2

âˆ´ c = 1

âˆ´ a = 2, b = 4, c = 1 and d = 3.

Order of matrix is 3 Ã— 2 and the matrix

is 3 Ã— 3

âˆ´ Order of matrix A must be 2 Ã— 3

Let A =

So,

Equating the corresponding elements, we get,

2a â€“ d = â€“ 1 and a = 1 â‡’ 2 Ã— 1 â€“ d = â€“ 1 â‡’ d = 2 + 1 â‡’ d = 3

2b â€“ e = â€“ 8 and b = â€“ 2 â‡’ 2(â€“ 2) â€“ e = â€“ 8 â‡’ â€“ 4 â€“ e = â€“ 8

â‡’ e = 4

2c â€“ f = â€“ 10 and c = â€“ 5 â‡’ 2(â€“ 5) â€“ f = â€“ 10 â‡’ â€“ 10 â€“ f = â€“ 10

â‡’ f = 0

a = 1, b = â€“ 2, c = â€“ 5, d = 3, e = 4 and f = 0

Hence,

Q.43. If A =find A

Given that:

Hence, A

Here,

A=

Given that: A

Pre-multiplying both sides by

AA

â‡’ I = AAâ€² [âˆµ AA

Hence, it is true for all values of Î±.

Let,

For skew symmetric matrix, Aâ€² = - A.

â‡’

Equating the corresponding elements, we get

a = â€“ 2, b = â€“ b â‡’ 2b = 0 â‡’ b = 0 and c = - 3

Hence, a = - 2, b = 0 and c = - 3.

Given that:

[Replacing x by y]

P(x).P(y)

P(x + y)

Now

P(y).P(x)

= P(x + y)

Hence, P(x).P(y) = P(x + y) = P(y).P(x).

To show that: (I + A)

L.H.S. (I + A)

â‡’ I + A

â‡’ I + A.A + 3IA + 3IA [âˆµ A

â‡’ I + A

â‡’ I + A + 3IA + 3IA [âˆµ A

â‡’ I + A + 3A + 3A â‡’ 7A + I R.H.S.

L.H.S. = R.H.S. Hence, Proved.

Given that B is a skew symmetric matrix

âˆ´ Bâ€² = - B

Let P = Aâ€²BA

â‡’ Pâ€² = (Aâ€²BA)â€²

= Aâ€²Bâ€²(Aâ€²)â€² [(AB)â€² = Bâ€²Aâ€²]

= Aâ€²(- B) A

= â€“ Aâ€²BA = - P

So Pâ€² = â€“ P

Hence, Aâ€²BA is a skew symmetric matrix.

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