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**Long Answer (L.A.)****Q.49. If AB = BA for any two sqaure matrices, prove by mathematical induction that (AB) ^{n} = A^{n} B^{n}.**

Let P(n) : (AB)^{n} = A^{n}B^{n}

Step 1:

Put n = 1, P(1) : AB = AB (True)

Step 2:

Put n = k, P(k) : (AB)^{k} = A^{k}B^{k}

(Let it be true for any k âˆˆ N)

Step 3:

Put n = k + 1, P(k + 1) : (AB)^{k + 1} = A^{k + 1 }B^{k + 1}

L.H.S. (AB)^{k + 1} = (AB)^{k}.AB

= A^{k}B^{k}.AB [from Step 2]

= A^{k} ^{+ 1}B^{k + 1} R.H.S.

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

Hence, if P(n) is true for P(k) then it is true for P(k + 1).**Q.50. Find x, y, z ifsatisfies Aâ€² = A ^{â€“1}.**

Given that:

Pre-multiplying both sides by A we get,

AAâ€² = AA^{â€“ 1}

â‡’ AAâ€² = I

Equating the corresponding elements, we get,

4y^{2} + z^{2} = 1 ...(i)

2y^{2} â€“ z^{2} = 0 ...(ii)

Adding (i) and (ii) we get,

From eqn. (i), we get,

4y^{2} + z^{2 }= 1

âˆ´

x^{2} + y^{2} + z^{2} = 1 ...(iii)

x^{2} â€“ y^{2} â€“ z^{2} = 0 ...(iv)

Adding (iii) and (iv) we get,

Hence,**Q.51. If possible, using elementary row transformations, find the inverse of the following matrices****(i)****(ii)**** (iii)****Ans.****(i)** Here, A =for elementary row transformation we put

A = IA

R_{2} â†’ R_{2} + R_{1}

R_{3} â†’ R_{3} - R_{2}

R_{1} â†’ R_{1} + R_{2}

R_{2} â†’ R_{2} - 3R_{1}

R_{1} â†’ R_{1} + R_{2} and R_{3} â†’ - 1 . R_{3}

R_{1} â†’ R_{1} + 10R_{3} and R_{2} â†’ R_{2} + 17R_{3}

R_{1} â†’ - 1. R_{1} and R_{2} â†’ - 1. R_{2}

Hence,**(ii) **Here,

Put A = IA

R_{1} â†’ R_{1} - 2R_{3} and R_{2} â†’ R_{2} + R_{1}

R_{1} â†’ R_{1} + R_{2}

First row on L.H.S. contains all zeros, so the inverse of the given matrix A does not exist.

Hence, matrix A has no inverse.**(iii)** Here,

Put A = IA

R_{1 }â†’ 3R_{1} - R_{2}

R_{2} â†’ R_{2} - 5R_{1}

R_{2 }â†’ R_{2 }- 5R_{3}

R_{3} â†’ R_{3} - R_{2}

R_{1} â†’ R_{1} + R_{2}

R_{1} â†’ R_{1} + 3R_{3}

Hence,**Q.52. Express the matrix****as the sum of a symmetric and a skew symmetric ****matrix.****Ans.**

We know that any square matrix can be expressed as the sum of symmetric and skew symmetric matrix i.e.

So

As Pâ€² = P âˆ´ P is a symmetric matrix.

As Q = â€“Q âˆ´ Q is a skew symmetric matrix.

So A = P + Q

Hence, the required relation is

**Objective Type Questions**

**Q.53. The matrix P =****is a****(a) square matrix ****(b) diagonal matrix ****(c) unit matrix ****(d) none****Ans. **(a)**Solution.**

Given that A =

Here number of columns and the number of rows are equal i.e., 3.

So, A is a square matrix.

Hence, the correct option is (a).**Q.54. Total number of possible matrices of order 3 Ã— 3 with each entry 2 or 0 is****(a) 9 ****(b) 27 ****(c) 81 ****(d) 512****Ans. **(d)**Solution.**

Total number of possible matrices of order 3 Ã— 3 with each entry 0 or 2 = 2^{3Ã—3} = 2^{9} = 512.

Hence, the correct option is (d).**Q.55. Ifthen the value of x + y is****(a) x = 3, y = 1 ****(b) x = 2, y = 3 ****(c) x = 2, y = 4 ****(d) x = 3, y = 3****Ans. **(b)

Solution.

Given that:

Equating the corresponding elements, we get,

2x + y = 7 ...(i)

and 4x = x + 6 ...(ii)

from eqn. (ii) 4x â€“ x = 6

3x = 6

âˆ´ x = 2

from eqn. (i) 2 Ã— 2 + y = 7

4 + y = 7 âˆ´ y = 7 â€“ 4 = 3

Hence, the correct option is (b).**Q.56. If A =****then ****A â€“ B is equal to****(a) I ****(b) O ****(c) 2I ****(d)****Ans. **(d)

Solution.

Given that: A =and B =

A â€“ B =

Hence, the correct option is (d).**Q.57. If A and B are two matrices of the order 3 Ã— m and 3 Ã— n, respectively and m = n, then the order of matrix (5A â€“ 2B) is ****(a) m Ã— 3 ****(b) 3 Ã— 3 ****(c) m Ã— n ****(d) 3 Ã— n****Ans. **(d)

Solution.

As we know that the addition and subtraction of two matrices is only possible when they have same order. It is also given that m = n.

âˆ´ Order of (5A â€“ 2B) is 3 Ã— n

Hence, the correct option is (d).**Q.58. If A =then A ^{2} is equal to**

Solution.

Given that A =

A

Hence, the correct option is (d).

a) I

Solution.

Given that: A = [a

Leta_{11} = 0 [âˆµ i = j]

a_{12} = 1 [âˆµ i â‰ j]

a_{21} = 1 [âˆµ i â‰ j]

a_{22} = 0 [âˆµ i = j]

âˆ´

Now, A^{2} = A . A =

Hence, the correct option is (a).**Q.60. The matrix****is ****(a) Identity matrix ****(b) symmetric matrix ****(c) skew symmetric matrix ****(d) none of these****Ans. **(b)

Solution.

Let

Aâ€² = A, so A is a symmetric matrix.

Hence, the correct option is (b).**Q.61. The matrix****is a****(a) diagonal matrix ****(b) symmetric matrix ****(c) skew symmetric matrix ****(d) scalar matrix****Ans. **(c)

Solution.

Let â‡’

Aâ€² = - A, so A is a skew symmetric matrix.

Hence, the correct option is (c).**Q.62. If A is matrix of order m Ã— n and B is a matrix such that ABâ€² and Bâ€²A are both defined, then order of matrix B is****(a) m Ã— m ****(b) n Ã— n ****(c) n Ã— m ****(d) m Ã— n****Ans. **(d)

Solution.

Order of matrix A = m Ã— n

Let order of matrix B be K Ã— P

Order of matrix Bâ€² = P Ã— K

If ABâ€² is defined then the order of ABâ€² is m Ã— K if n = P

If Bâ€²A is defined then order of Bâ€²A is P Ã— n when K = m

Now, order of Bâ€² = P Ã— K

âˆ´ Order of B = K Ã— P

= m Ã— n [âˆµ K = m, P = n]

Hence, the correct option is (d).**Q.63. If A and B are matrices of same order, then (ABâ€²â€“BAâ€²) is a****(a) skew symmetric matrix ****(b) null matrix ****(c) symmetric matrix ****(d) unit matrix****Ans. **(a)

Solution.

Let P = (ABâ€² - BAâ€²) Pâ€² = (ABâ€² - BAâ€²)â€²

= (ABâ€²)â€² -(BAâ€²)â€²

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

= BAâ€² - ABâ€²

= â€“(ABâ€² - BAâ€²) = -P

Pâ€² = â€“P, so it is a skew symmetric matrix.

Hence, the correct option is (a).**Q.64. If A is a square matrix such that A ^{2} = I, then (Aâ€“I)^{3} + (A + I)^{3} â€“7A is equal to**

Solution.

(A â€“ I)

= 2A^{3} + 6AI^{2} â€“ 7A

= 2A.A^{2} + 6AI â€“ 7A

= 2AI + 6AI â€“ 7A [A^{2} = I]

= 8AI â€“ 7A = 8A â€“ 7A

= A

Hence, the correct option is (a).**Q.65. For any two matrices A and B, we have****(a) AB = BA ****(b) AB â‰ BA ****(c) AB = O ****(d) None of the above****Ans. **(d)

Solution.

We know that for any two matrices A and B, we may have

AB = BA, AB â‰ BA and AB = 0, but it is not always true.

Hence, the correct option is (d).**Q.66. On using elementary column operations C _{2} â†’ C_{2} â€“ 2C_{1 }in the following matrix equation**

Solution.

Given that:

Using C

Hence, the correct option is (d).

Solution.

We have,

Using elementary row transformation R

Hence, the correct option is (a).

**Fill in the blanks**

**Q.68. _________ matrix is both symmetric and skew symmetric matrix.****Ans.**

Null matrix i.e.is both symmetric and skew symmetric matrix.**Q.69. Sum of two skew symmetric matrices is always _________ matrix.****Ans.**

Let A and B be any two matrices

âˆ´ For skew symmetric matrices

A = â€“Aâ€² ...(i)

and B = â€“Bâ€² ...(ii)

Adding (i) and (ii) we get

A + B = â€“Aâ€² - Bâ€²

â‡’ A + B = â€“(Aâ€² + Bâ€²), so A + B is skew symmetric matrix.

Hence, the sum of two skew symmetric matrices is always skew symmetric matrix.**Q.70. The negative of a matrix is obtained by multiplying it by _________.****Ans.**

Let A be a matrix

âˆ´ â€“ A = â€“ 1 . A

Hence, negative of a matrix is obtained by multiplying it by â€“ 1.**Q.71. The product of any matrix by the scalar _________ is the null matrix.****Ans.**

Let A be any matrix

âˆ´ 0 . A = A . 0 = 0

Hence, the product of any matrix by the scalar 0 is the null matrix.

Q.72. A matrix which is not a square matrix is called a _________ matrix.**Ans.**

A matrix which is not a square matrix is called a rectangular matrix.**Q.73. Matrix multiplication is _________ over addition.****Ans.**

Matrix multiplication is distributive over addition.

Let A, B and C be any matrices.

So, (i) A(B + C) = AB + AC

(ii) (A + B)C = AC + BC**Q.74. If A is a symmetric matrix, then A ^{3} is a _________ matrix.**

Let A be a symmetric matrix

âˆ´ Aâ€² = A

(A

Hence, if A is a symmetric matrix, then A

Q.75. If A is a skew symmetric matrix, then A

If A is a skew symmetric matrix,

âˆ´ Aâ€² = â€“A

(A

= (â€“A)

= A

Hence, A

If A is a skew symmetric matrix

âˆ´ Aâ€² = â€“ A

(kA)â€² = kAâ€² = k(- A) = - kA

Hence, kA is a skew symmetric matrix.

Pâ€² = (AB â€“ BA)â€²

= (AB)â€² - (BA)â€²

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

= BA â€“ AB [âˆµ Aâ€² = A and Bâ€² = B]

= -(AB - BA)

= â€“P

Hence, (AB â€“ BA) is a skew symmetric matrix.

Qâ€² = (BA â€“ 2AB)â€²

= (BA)â€² - (2AB)â€²

= Aâ€²Bâ€² - 2(AB)â€² [âˆµ (kA)â€² = kAâ€²]

= Aâ€²Bâ€² - 2Bâ€²Aâ€²

= AB â€“ 2BA [âˆµ Aâ€² = A and Bâ€² = B]

= - (2BA - AB)

Hence, (BA â€“ 2AB) is neither a symmetric nor a skew symmetric matrix.

If A is a symmetric matrix

âˆ´ Aâ€² = A

Let P = Bâ€²AB

Pâ€² = (Bâ€²AB)â€²

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

= Bâ€²AB [âˆµ Aâ€² = A and (Bâ€²)â€² = B]

âˆ´ Pâ€² = P

So, P is a symmetric matrix.

Hence, Bâ€²AB is a symmetric matrix.

Given that Aâ€² = A

and Bâ€² = B

Let P = AB

Pâ€² = (AB)â€²

= Bâ€²Aâ€²

Pâ€² = BA [âˆµ Aâ€² = A and Bâ€² = B]

= P

Hence, AB is symmetric if and only if AB = BA.

A

A matrix is an array of elements, numbers or functions having rows and columns.

The matrices having same order can only be added.

The two matrices are said to be equal if their corresponding elements are same.

For addition and subtraction, the order of the two matrices should be same.

If A, B and C are the matrices of addition then

A + (B + C) = (A + B) + C (associative)

A + B = B + A (commutative)

Q.87. Matrix multiplication is commutative.

Since AB â‰ BA if AB and BA are well defined.

False.

Since, in identity matrix all the elements of principal diagonal are unity rest are zero.

e.g.,

If A and B are square matrices then their addition is commutative i.e., A + B = B + A.

Since subtraction of any two matrices of the same order is not commutative i.e.,

A â€“ B â‰ B - A.

Since for any two non-zero matrices A and B, we may get AB = 0.

Transpose of a column matrix is a row matrix.

For two square matrices A and B, AB = BA is not always true.

Let A, B and C be three matrices of the same order.

Given that Aâ€² = A, Bâ€² = B and Câ€² = C

Let P = A + B + C

â‡’ Pâ€² = (A + B + C)â€²

= Aâ€² + Bâ€² + Câ€²

= A + B + C

= P

So, A + B + C is also a symmetric matrix.

Since (AB)â€² = Bâ€²Aâ€².

Let A = [a

AB is defined when n = P

âˆ´ Order of AB = m Ã— q

â‡’ Order of (AB)â€² = q Ã— m

Order of Bâ€² is q Ã— p and order of Aâ€² is n Ã— m

âˆ´ Bâ€²Aâ€² is defined when P = n

and the order of Bâ€²Aâ€² is q Ã— m

Hence, order of (AB)â€² = Order of Bâ€²Aâ€² i.e., q Ã— m.

Let A =

âˆ´

Here AB = AC = 0 but B â‰ C.

Let

P = AAâ€²

Pâ€² = (AAâ€²)â€²

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

= AAâ€²

= P

So, P is symmetric matrix.

Hence, AAâ€² is always a symmetric matrix.

Since AB is definedâˆ´

BA is also defined.

So AB â‰ BA

(A^{2})â€² = (Aâ€²)^{2}

= [â€“ A]^{2} [âˆµ Aâ€² = - A]

= A^{2}

So, A^{2} is a symmetric matrix.**Q.101. (AB) ^{â€“1} = A^{â€“1}. B^{â€“1}, where A and B are invertible matrices satisfying commutative property with respect to multiplication.**

If A and B are invertible matrices of the same order

âˆ´ (AB)

But (AB)^{â€“1} = A^{â€“1}B^{â€“1} [Given]

âˆ´ (BA)^{â€“1} = B^{â€“1}A^{â€“1}

So A^{â€“1}B^{â€“1} = B^{â€“1}A^{â€“1}

âˆ´ A and B satisfy commutative property w.r.t. multiplication.

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