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NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE PDF Download

Long Answer (L.A.)
Q.49. If AB = BA for any two sqaure matrices, prove by mathematical induction that (AB)n = An Bn.
Ans.

Let P(n) : (AB)n = AnBn
Step 1:
Put n = 1, P(1) : AB = AB (True)
Step 2:
Put n = k, P(k) : (AB)k = AkBk
(Let it be true for any k ∈ N)
Step 3:
Put n = k + 1, P(k + 1) : (AB)k + 1 = Ak + 1 Bk + 1
L.H.S. (AB)k + 1 = (AB)k.AB
= AkBk.AB [from Step 2]
= Ak + 1Bk + 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 ifNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEsatisfies A′ = A–1.
Ans.
Given that:NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE

Pre-multiplying both sides by A we get,
 AA′ = AA– 1

⇒ AA′ = I
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Equating the corresponding elements, we get,
4y2 + z2 = 1 ...(i)
2y2 – z2 = 0 ...(ii)
Adding (i) and (ii) we get,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
From eqn. (i), we get,
4y2 + z= 1
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
x2 + y2 + z2 = 1 ...(iii)
x2 – y2 – z2 = 0 ...(iv)
Adding (iii) and (iv) we get,
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE

Q.51. If possible, using elementary row transformations, find the inverse of the following matrices
(i)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(ii)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
 (iii)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Ans.
(i) Here, A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEfor elementary row transformation we put
A = IA
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R2 → R2 + R1
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R3 → R3 - R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R2 → R2 - 3R1
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + R2 and R3 → - 1 . R3
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + 10R3 and R2 → R2 + 17R3
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → - 1. R1 and R2 → - 1. R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(ii) Here,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Put A = IA
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - 2R3 and R2 → R2 + R1
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
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,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Put A = IA
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R→ 3R1 - R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R2 → R2 - 5R1
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R→ R- 5R3
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R3 → R3 - R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + R2
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + 3R3
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE

Q.52. Express the matrixNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEas 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.
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
SoNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
As P′ = P ∴ P is a symmetric matrix.
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE

As Q = –Q ∴ Q is a skew symmetric matrix.

So A = P + Q
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence, the required relation isNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE


Objective Type Questions

Q.53. The matrix P =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEis a
(a) square matrix 
(b) diagonal matrix 
(c) unit matrix 
(d) none
Ans. (a)
Solution.
Given that A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
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 × 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 × 3 with each entry 0 or 2 = 23×3 = 29 = 512.
Hence, the correct option is (d).

Q.55. IfNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEthen 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: NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Equating the corresponding elements, we get,
2x + y = 7 ...(i)
and 4x = x + 6 ...(ii)
from eqn. (ii) 4x – x = 6
3x = 6
∴ x = 2
from eqn. (i) 2 × 2 + y = 7
4 + y = 7 ∴ y = 7 – 4 = 3
Hence, the correct option is (b).

Q.56. If A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEthen A – B is equal to
(a) I 
(b) O 
(c) 2I 
(d)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Ans. (d)
Solution.

Given that: A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEand B =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
A – B = NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence, the correct option is (d).

Q.57. If A and B are two matrices of the order 3 × m and 3 × n, respectively and m = n, then the order of matrix (5A – 2B) is 
(a) m × 3 
(b) 3 × 3 
(c) m × n 
(d) 3 × 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.
∴ Order of (5A – 2B) is 3 × n
Hence, the correct option is (d).

Q.58. If A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEthen A2 is equal to
(a)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(b)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(c)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(d)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Ans. (d)
Solution.

Given that A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
A2 = A . A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence, the correct option is (d).

Q.59. If matrix A = [aij]2 × 2 , where aij = 1 if i ≠ j  = 0 if i  = j then A2 is equal to
a) I 

(b) A 
(c) 0 
(d) None of these
Ans. (a)
Solution.

Given that: A = [aij]2 × 2

LetNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEa11 = 0 [∵ i = j]
a12 = 1 [∵ i ≠ j]
a21 = 1 [∵ i ≠ j]
a22 = 0 [∵ i = j]
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Now, A2 = A . A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence, the correct option is (a).

Q.60. The matrixNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEis 
(a) Identity matrix 
(b) symmetric matrix 
(c) skew symmetric matrix 
(d) none of these
Ans. (b)
Solution.

LetNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
A′ = A, so A is a symmetric matrix.
Hence, the correct option is (b).

Q.61. The matrixNCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEis a
(a) diagonal matrix 
(b) symmetric matrix 
(c) skew symmetric matrix 
(d) scalar matrix
Ans. (c)
Solution.

Let NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
A′ = - A, so A is a skew symmetric matrix.
Hence, the correct option is (c).

Q.62. If A is matrix of order m × n and B is a matrix such that AB′ and B′A are both defined, then order of matrix B is
(a) m × m 
(b) n × n 
(c) n × m 
(d) m × n
Ans. (d)
Solution.

Order of matrix A = m × n
Let order of matrix B be K × P
Order of matrix B′ = P × K
If AB′ is defined then the order of AB′ is m × K if n = P
If B′A is defined then order of B′A is P × n when K = m
Now, order of B′ = P × K
∴ Order of B = K × P
= m × n [∵ K = m, P = n]
Hence, the correct option is (d).

Q.63. If A and B are matrices of same order, then (AB′–BA′) is a
(a) skew symmetric matrix 
(b) null matrix 
(c) symmetric matrix 
(d) unit matrix
Ans. (a)
Solution.

Let P = (AB′ - BA′) P′ = (AB′ - BA′)′
= (AB′)′ -(BA′)′
= (B′)A′ - (A′)′B′ [∵ (AB)′ = B′A′]
= BA′ - AB′
= –(AB′ - BA′) = -P
P′ = –P, so it is a skew symmetric matrix.
Hence, the correct option is (a).

Q.64. If A is a square matrix such that A2 = I, then (A–I)3 + (A + I)3 –7A is equal to
(a) A 
(b) I – A 
(c) I + A 
(d) 3A
Ans. (a)
Solution.

(A – I)3 + (A + I)3 – 7A = A3 – I3 – 3A2I + 3AI2 + A3 + I3 + 3A2I + 3 AI2 – 7A

= 2A3 + 6AI2 – 7A

= 2A.A2 + 6AI – 7A

= 2AI + 6AI – 7A [A2 = I]
= 8AI – 7A = 8A – 7A
= A
Hence, the correct option is (a).

Q.65. For any two matrices A and B, we have
(a) AB = BA 
(b) AB ≠ 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 ≠ BA and AB = 0, but it is not always true.
Hence, the correct option is (d).

Q.66. On using elementary column operations C2 → C2 – 2Cin the following matrix equation
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEwe have :
(a)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(b)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(c)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(d)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Ans. (d)
Solution.

Given that:NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Using C→ C2 - 2C1, we get
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Hence, the correct option is (d).

Q.67. On using elementary row operation R1 → R1 – 3R2 in the following matrix equation:
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEwe have :
(a)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(b)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(c)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
(d)NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Ans. (a)
Solution.

We have,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Using elementary row transformation R1 → R1 - 3R2,
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
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.NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEEis 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
∴ For skew symmetric matrices
A = –A′ ...(i)
and B = –B′ ...(ii)
Adding (i) and (ii) we get
A + B = –A′ - B′
⇒ A + B = –(A′ + B′), 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
∴ – A = – 1 . A
Hence, negative of a matrix is obtained by multiplying it by – 1.

Q.71. The product of any matrix by the scalar _________ is the null matrix.
Ans.
Let A be any matrix
∴ 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 A3 is a _________ matrix.
Ans.
Let A be a symmetric matrix
∴ A′ = A
(A3)′ = (A′)3 = A3 [∵ (A′)k = (Ak)′]
Hence, if A is a symmetric matrix, then A3 is a symmetric matrix.

Q.75. If A is a skew symmetric matrix, then A2 is a _________.

Ans.
If A is a skew symmetric matrix,
∴ A′ = –A
(A2)′ = (A′)2
= (–A)2
= A2
Hence, A2 is a symmetric matrix.

Q.76. If A and B are square matrices of the same order, then 
(i) (AB)′ = _________. 
(ii) (kA)′ = _________. (k is any scalar) 
(iii) [k (A – B)]′ = _________.
Ans.
(i) (AB)′ = B′A′
(ii) (kA)′ = k . A′
(iii) [k(A - B)]′ = k(A - B)′ = k(A′ - B′)

Q.77. If A is skew symmetric, then kA is a _________. (k is any scalar)
Ans.
If A is a skew symmetric matrix
∴ A  = – A
(kA)  = kA  = k(- A) = - kA
Hence, kA is a  skew symmetric matrix.

Q.78. If A and B are symmetric matrices, then 
(i) AB – BA is a _________. 
(ii) BA – 2AB is a _________.
Ans.
(i) Let P = (AB – BA)
P′ = (AB – BA)
= (AB) - (BA)
= BA - AB [∵ (AB) = BA]
= BA – AB [∵ A = A and B = B]
= -(AB - BA)
= –P
Hence, (AB – BA) is a skew symmetric matrix.
(ii) Let Q = (BA – 2AB)
Q = (BA – 2AB)
= (BA) - (2AB)
= AB - 2(AB) [∵ (kA) = kA]
= AB - 2BA
= AB – 2BA [∵ A = A and B = B]
= - (2BA - AB)
Hence, (BA – 2AB) is neither a symmetric nor a skew symmetric matrix.

Q.79. If A is symmetric matrix, then B′AB is _________.
Ans.
If A is a symmetric matrix
∴ A = A
Let P = BAB
P = (BAB)
= BA(B) [∵ (AB) = BA]
= BAB [∵ A = A and (B) = B]
∴ P = P
So, P is a symmetric matrix.
Hence, BAB is a symmetric matrix.

Q.80. If A and B are symmetric matrices of same order, then AB is symmetric if and only if _________.
Ans.
Given that A = A
and B = B
Let P = AB
P = (AB)
= BA
P = BA [∵ A = A and B = B]
= P
Hence, AB is symmetric if and only if AB = BA.

Q.81. In applying one or more row operations while finding A–1 by elementary row operations, we obtain all zeros in one or more, then A–1 _________.
Ans.
A–1 does not exist if we apply one or more row operations while finding A–1 by elementary row operations, obtain all zeroes in one or more rows.

True or False
Q.82. A matrix denotes a number.
Ans. False.
A matrix is an array of elements, numbers or functions having rows and columns.

Q.83. Matrices of any order can be added.
Ans. False.
The matrices having same order can only be added.

Q.84. Two matrices are equal if they have same number of rows and same number of columns.
Ans. False.
The two matrices are said to be equal if their corresponding elements are same.

Q.85. Matrices of different order can not be subtracted.
Ans. True.
For addition and subtraction, the order of the two matrices should be same.

Q.86. Matrix addition is associative as well as commutative.
Ans. True.
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.

Ans. False.
Since AB ≠ BA if AB and BA are well defined.

Q.88. A square matrix where every element is unity is called an identity matrix.
Ans.
False.
Since, in identity matrix all the elements of principal diagonal are unity rest are zero.
e.g.,NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE

Q.89. If A and B are two square matrices of the same order, then A + B = B + A.
Ans. True.
If A and B are square matrices then their addition is commutative i.e., A + B = B + A.

Q.90. If A and B are two matrices of the same order, then A – B = B – A.
Ans. False.
Since subtraction of any two matrices of the same order is not commutative i.e.,
A – B ≠ B - A.

Q.91. If matrix AB = O, then A = O or B = O or both A and B are null matrices.
Ans. False.
Since for any two non-zero matrices A and B, we may get AB = 0.

Q.92. Transpose of a column matrix is a column matrix.
Ans. False.
Transpose of a column matrix is a row matrix.
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE

Q.93. If A and B are two square matrices of the same order, then AB = BA.
Ans. False.
For two square matrices A and B, AB = BA is not always true.

Q.94. If each of the three matrices of the same order are symmetric, then their sum is a symmetric matrix.
Ans. True.
Let A, B and C be three matrices of the same order.
Given that A′ = A, B′ = B and C′ = C
Let P = A + B + C
⇒ P′ = (A + B + C)′
= A′ + B′ + C′
= A + B + C
= P
So, A + B + C is also a symmetric matrix.

Q.95. If A and B are any two matrices of the same order, then (AB)′ = A′B′.
Ans. False.
Since (AB)′ = B′A′.

Q.96. If (AB)′ = B′ A′, where A and B are not square matrices, then number of rows in A is equal to number of columns in B and number of columns in A is equal to number of rows in B.
Ans. True.
Let A = [aij]m × n and B = [bij]p × q 
AB is defined when n = P
∴ Order of AB = m × q
⇒ Order of (AB)′ = q × m
Order of B′ is q × p and order of A′ is n × m
∴ B′A′ is defined when P = n
and the order of B′A′ is q × m
Hence, order of (AB)′ = Order of B′A′ i.e., q × m.

Q.97. If A, B and C are square matrices of same order, then AB = AC always implies that B = C.
Ans. False.
Let A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Here AB = AC  = 0 but B ≠ C.

Q.98. AA′ is always a symmetric matrix for any matrix A.
Ans. True.
Let
P = AA′
P′ = (AA′)′
= (A′)′ . A′ [(AB)′ = B′A′]
= AA′
= P
So, P is symmetric matrix.
Hence, AA′ is always a symmetric matrix.

Q.99. If A =NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE then AB and BA are defined and equal.
Ans. False.
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
Since AB is defined∴NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
BA is also defined.
NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE
So AB ≠ BA

Q.100. If A is skew symmetric matrix, then A2 is a symmetric matrix.
Ans. True.

(A2)′ = (A′)2

= [– A]2 [∵ A′ = - A]

= A2
So, A2 is a symmetric matrix.

Q.101. (AB)–1 = A–1. B–1, where A and B are invertible matrices satisfying commutative property with respect to multiplication.
Ans. True.
If A and B are invertible matrices of the same order
∴ (AB)–1 = (BA)–1 [∵ AB = BA]

But (AB)–1 = A–1B–1 [Given]
∴ (BA)–1 = B–1A–1

So A–1B–1 = B–1A–1
∴ A and B satisfy commutative property w.r.t. multiplication.

The document NCERT Exemplar - Matrices(Part-2) | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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FAQs on NCERT Exemplar - Matrices(Part-2) - Mathematics (Maths) Class 12 - JEE

1. What are matrices and how are they used in mathematics?
Ans. Matrices are rectangular arrays of numbers or symbols arranged in rows and columns. They are used in various fields of mathematics, such as linear algebra, statistics, and computer science. Matrices allow us to perform operations like addition, subtraction, multiplication, and finding determinants, which help in solving complex mathematical problems.
2. How can matrices be represented and manipulated in computer programming?
Ans. In computer programming, matrices can be represented as two-dimensional arrays. Each element of the matrix is stored at a specific index in the array. Various programming languages provide built-in functions and libraries to perform matrix operations, such as matrix multiplication, transpose, and finding inverses. These operations can be easily implemented by accessing the elements of the array and applying the necessary mathematical calculations.
3. What is the significance of the determinant of a matrix?
Ans. The determinant of a square matrix is a scalar value that provides important information about the matrix. It helps in determining whether a matrix has an inverse or not. If the determinant is non-zero, the matrix is invertible and has a unique inverse. A determinant of zero indicates that the matrix is singular and does not have an inverse. The determinant is also used to find the area or volume of geometric shapes represented by matrices.
4. How can matrices be used to solve systems of linear equations?
Ans. Matrices can be used to solve systems of linear equations by representing the coefficients and constants of the equations in matrix form. The augmented matrix is formed by combining the coefficient matrix and the constant matrix. Using various matrix operations, such as row operations, the augmented matrix can be transformed into row-echelon form or reduced row-echelon form. The solutions to the system of equations can be obtained by analyzing the transformed matrix.
5. Can matrices be used to solve real-world problems?
Ans. Yes, matrices are widely used to solve real-world problems in various fields. For example, in physics, matrices are used to represent and solve problems related to forces, motion, and energy. In computer science, matrices are used in computer graphics, image processing, and data analysis. Matrices are also used in economics, engineering, and social sciences to model and solve complex systems of equations and optimize various parameters.
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