NCERT Exemplar - Matrices(Part-1) | Mathematics (Maths) Class 12 - JEE PDF Download

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 × 1, 1 × 28, 2 × 14, 14 × 2, 4 × 7, 7 × 4}. The possible orders of a matrix having 13 elements are {1 × 13, 13 × 1}.

Q.2. In the matrix A = write:
(i) The order of the matrix A 
(ii) The number of elements 
(iii) Write elements  a₂₃, a₃₁, a₁₂
Ans.
(i) The order of the given matrix A is 3 × 3
(ii) The number of elements in matrix A = 3 × 3 = 9
(iii) a_ij = the elements of i^th row and j^th column.
So, a₂₃ = x² – y, a₃₁ = 0, a₁₂ = 1.

Q.3. Construct a_{2 × 2} matrix where

(ii) a_ij = |−2i + 3j|
Ans.
Let

(i) Given that


Hence, the matrix A =

(ii) Given that


Hence, the matrix A =


Q.4. Construct a 3 × 2 matrix whose elements are given by a_ij = e^ixsinjx
Ans. Let

Given that a_ij = e^ix sin jx
a₁₁ = e^x sin x    a₁₂ = e^x sin 2x
a₂₁ = e^2x sin x   a₂₂ = e^2x sin 2x
a₃₁ = e^3x sin x    a₃₂ = e^3x sin 2x
Hence, the matrix


Q.5. Find values of a and b if A = B, where

Ans.
Given that A = B
⇒
Equating the corresponding elements, we get
a + 4 = 2a + 2, 3b = b² + 2 and b² – 5b = – 6
⇒ 2a - a = 2, b² - 3b + 2 = 0, b² - 5b + 6 = 0
∴ a = 2

but here 2 is common.
Hence, the value of a = 2 and b = 2.

Q.6. If possible, find the sum of the matrices A and B, where A = and B =
Ans.
The order of matrix A = 2 × 2 and the order of matrix B = 2 × 3 Addition of matrices is only possible when they have same order. So, A + B is not possible.

Q.7. Iffind
(i) X +Y 
(ii) 2X – 3Y 
(iii) A matrix Z such that X + Y + Z is a zero matrix.
Ans.
Given that

(i) X + Y  =


(ii) 2X – 3Y =


(iii) X + Y + Z = 0


Equating the corresponding elements, we get
5 + a = 0 ⇒ a = – 5, 2 + b = 0 ⇒ b = – 2, – 2 + c = 0 ⇒ c = 2
12 + d = 0 ⇒ d = – 12, e = 0, 1 + f = 0 ⇒ f = – 1
Hence, the matrix,


Q.8. Find non-zero values of x satisfying the matrix equation:

Ans.
The given equation can be written as

Equating the corresponding elements we get
12x = 48, 3x + 8 = 20, x² + 8x = 12x
∴ x = 48/12 = 4, 3x = 20 – 8 = 12, ⇒ x² = 12x – 8x = 4x
∴ x = 4, ⇒ x² – 4x = 0 x = 0, x = 4
Hence, the non-zero values of x is 4.

Q.9. If show that (A + B) (A – B) ≠ A² – B².
Ans.
Given that

A + B =

⇒ A + B =


⇒
∴ (A + B) × (A - B) =

Now, R.H.S. = A² – B² 
= A × A - B × B

Hence,

Hence, (A + B) . (A – B) ≠ A² - B²

Q.10. Find the value of x if

Ans.
Given that,

⇒
⇒

⇒ [2x + 16 + 10x + 12 + x² + 4x] = 0;  ⇒ x² + 16x + 28 = 0
⇒ x² + 14x + 2x + 28 = 0; ⇒ x(x + 14) + 2(x + 14) = 0
⇒ (x + 2) (x + 14) = 0;  x + 2 = 0 or x + 14 = 0
∴ x = – 2 or x = – 14
Hence, the values of x are – 2 and – 14.

Q.11. Show thatsatisfies the equation A² – 3A – 7I = O and hence find A^–1.
Ans.
Given that

A² = A × A

A² – 3A – 7I = O

We are given A² – 3A – 7I = O
⇒ A^–1 [A² – 3A – 7I] = A^–1O
[Pre-multiplying both sides by A^–1]
⇒ A^–1A × A – 3A^–1 × A – 7A^–1 I = O [A^–1O = O]
⇒ I × A – 3I – 7A^–1 I  = O
⇒ A – 3I – 7A^–1  = O
⇒ –7A^–1 = 3I – A


Hence,


Q.12. Find the matrix A satisfying the matrix equation:

Ans.
Let


Equating the corresponding elements, we get,
– 6a – 3c + 10b + 5d = 1 ...(1)
–9a – 6c + 15b + 10d = 0 ...(2)
4a + 2c – 6b – 3d = 0 ...(3)
6a + 4c – 9b – 6d = 1 ...(4)
Multiplying eq. (1) by 2 and subtracting eq. (2), we get,

– 3a + 5b = 2 ...(5)
Now, multiplying eq. (3) by 2 and subtracting eq. (4), we get

2a – 3b = – 1 ...(6)
Solving eq. (5) and (6) i.e.,
– 3a + 5b = 2
2a – 3b = – 1
2 x (- 3a + 5b = 2) ⇒ 6a + 10b = 4
3 x (2a - 3b = -1) ⇒ 6a - 9b = - 3
Adding b = 1
Putting the value of b in eq. (6), we get,
2a – 3 x 1 = - 1
⇒ 2a - 3 = - 1 ⇒ 2a = 3 - 1 ⇒ 2a = 2
∴ a = 1
Now, putting the values of a and b in equations (1) and (3)
–6 x 1 - 3c + 10 x 1 + 5d = 1
⇒ - 6 - 3c + 10 + 5d = 1
⇒ – 3c + 5d + 4 = 1 ⇒ 3c + 5d = - 3 ...(7)
and from eq. (3)
4 x 1 + 2c - 6 x 1 - 3d = 0; ⇒  4 + 2c - 6 - 3d = 0
⇒ - 2 + 2c - 3d = 0 ⇒ 2c - 3d = 2 ...(8)
Solving eq. (7) and (8) we get,

Adding  
Putting the value of d in eq. (8) we get,
2c – 3 x 0 = 2 ⇒ 2c= 2 ⇒ c = 1
Hence,

Q.13. Find A, if

Ans.
Order ofis 3 × 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 = – 4, 4b = 8, 4c = 4
∴ a = – 1 ∴ b = 2 ∴ c = 1
Hence, matrix A = [- 1  2  1]

Q.14. If A =and B =then verify (BA)² ≠ B²A²
Ans.
Here,

∴

Here, number of columns of first i.e., 3 is not equal to the number of rows of second matrix i.e., 2.
So, B² is not possible. Similarly, A² is also not possible.
Hence, (BA)² ≠ B²A²

Q.15. If possible, find BA and AB, where

Ans.


Hence,


Q.16. Show by an example that for  A ≠ O, B ≠ O, AB = O.
Ans.
Let



Hence,


Q.17. Given Is (AB)′ = B′A′?
Ans.
Here,


L.H.S.



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

Q.18. Solve for x and y:

Ans.
Given that:
 

Comparing the corresponding elements of both sides, we get,
2x + 3y – 8 = 0 ⇒ 2x + 3y = 8 ...(1)
x + 5y – 11 = 0 ⇒ x + 5y = 11 ...(2)
Multiplying eq. (1) by 1 and eq. (2) by 2, and then on subtracting, we get,

∴ y = 2
Putting y = 2 in eq. (2) we get,
x + 5 x 2 = 11 ⇒ x + 10 = 11
∴ x = 11 – 10 = 1
Hence, the values of x and y are 1 and 2 respectively.

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

Ans.
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,


Q.20. If A = [3 5],  B = [7 3], then find a non-zero matrix C such that AC = BC.
Ans.
Given that:

Let,


AC = BC (Given)
⇒ [3α + 5β] = [7α + 3β]
⇒ 3α + 5β = 7α + 3β
⇒ 3α – 7α = 3β – 5β
⇒ – 4α = – 2β
∴
So, let α = K and β = 2K, K is some real number.
Hence, matrix C =


Q.21. Give an example of matrices A, B and C such that AB = AC, where A is nonzero matrix, but B ≠ C.
Ans.
Let,


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

Q.22. If

(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.

⇒

Hence, matrix A = [– 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 A² + A  = A (A + I), where I is 3 × 3 unit matrix.
Ans.
Given that:

A² = A × A

L.H.S. A² + A =


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



L.H.S. = R.H.S.
A² + A = A(A + I). Hence verified.

Q.27. If then verify  that :
(i) (A′)′ = A
(ii) (AB)′ = B′A′
(iii) (kA)′ = (kA′).
Ans.
Given that:

(i)


(ii) L.H.S.


R.H.S.




L.H.S. = R.H.S.
Hence,(AB)′ = B′A′is verified.
(iii) L.H.S.


R.H.S.

Hence, L.H.S. = R.H.S.
(kA)′ = (kA′) is verified.

Q.28. If then  verify that:
(i) (2A + B)′ = 2A′ + B′
(ii) (A – B)′ = A′ – B′.
Ans.
Given that:

(i) To verify that:(2A + B)′ = 2A′ + B′
L.H.S. (2A + B)′ =


R.H.S. 2A′ + B′ =


Hence, L.H.S. = R.H.S.
(2A + B)′ = 2A′ + B′is verified.
(ii) To verify that:(A – B)′ = A′ – B′.
L.H.S. (A – B)′ =


R.H.S.A′ – B′=


Hence, L.H.S. = R.H.S.
(A – B)′ = A′ – B′ is verified.

Q.29. Show that A′A and AA′ are both symmetric matrices for any matrix A.
Ans.
Let P = A′A
⇒ P′ = (A′A)′
⇒ P′ = A′(A′)′ [(AB)′ = B′A′]
⇒ P′ = A′A [∵ (A′)′ = A]
⇒ P′ = P
Hence, A¢A is a symmetric matrix.
Now, Let Q = AA′
⇒ Q′ = (AA′)′
⇒ Q′ = (A′)′ A′ [(AB)′ = B′A′]
⇒ Q′ = AA′ [∵ (A′)′ = A]
⇒ Q′ = Q
Hence, AA′ is also a symmetric matrix.

Q.30. Let A and B be square matrices of the order 3 × 3. Is (AB)² = A² B² ? Give reasons.
Ans.
Given that A and B are the matrices of the order 3 x 3.
(AB)² = AB × AB
= AA × BB
= A² × B²
Hence, (AB)² = A²B²

Q.31. how that if A and B are square matrices such that AB = BA, then 
(A + B)² = A² + 2AB + B².
Ans.
To prove that (A + B)² = A² + 2AB + B² 
L.H.S. (A + B)² = (A + B) × (A + B) [∵ A² = A × A]
= A × A + AB + BA + B × B
= A² + AB + AB + B² [AB = BA]
= A² + 2AB + B² R.H.S.
So, L.H.S. = R.H.S.

Q.32. Let and a = 4, b = –2.
Show that: 
(a) A + (B + C) = (A + B) + C 
(b) A (BC) = (AB) C
(c) (a + b)B = aB + bB 
(d) a (C–A) = aC – aA 
(e) (A^T)^T = A 
(f) (bA)^T = b A^T 
(g) (AB)^T = B^T A^T 
(h) (A –B)C = AC – BC 
(i) (A – B)^T = A^T – B^T
Ans.
(a) To prove that: A + (B + C) = (A + B) + C
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.
(b) To prove that: A(BC) = (AB)C
L.H.S. A(BC) =


R.H.S. (AB)C


Hence, L.H.S. = R.H.S.
A(BC) = (AB)C Hence proved.
(c) To prove that: (a + b)B = aB + bB
Here, a  = 4  and b = – 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.
(d) To prove that: a(C – A) =  aC – aA
L.H.S. a(C – A) =


R.H.S. aC – aA =


Hence, L.H.S. = R.H.S.
a(C – A) = aC – aA Hence proved.
(e) To prove that: (A^T)^T = A
L.H.S.

R.H.S.
Hence, (A^T)^T = A
(f) To prove that: (bA)^T = bA^T
L.H.S. (bA)^T =

R.H.S. bA^T =

Hence, L.H.S. = R.H.S.
(bA)^T = bA^T Hence proved.
(g) To prove that: (AB)^T = B^TA^T
L.H.S. (AB)^T


R.H.S. B^TA^T=


Hence, L.H.S. = R.H.S.
(AB)^T = B^TA^T Hence proved.
(h) To prove that: (A – B)C = AC – BC
L.H.S. (A – B)C =


R.H.S. AC – BC =


Hence, L.H.S. = R.H.S.
(A – B)C = AC – BC
(i) To prove that: (A – B)^T = A^T – B^T
L.H.S. (A – B)^T =


R.H.S. A^T – B^T =


Hence, L.H.S. = R.H.S.
(A – B)^T = A^T – B^T Hence proved.

Q.33. If A =then show that A² =
Ans. Given that

A = A . A=

Hence proved.

Q.34. If A = and x² = –1, then show that (A + B)² = A² + B².
Ans. Given that:

L.H.S. (A + B)² = (A + B) × (A + B)

Put x² = –1 (given)

R.H.S.
A² + B² = A ×  A + B × B


Hence, L.H.S. = R.H.S.
(A + B)² = A² + B²

Q.35. Verify that A² = I when A =
Ans. Given that:

L.H.S. A² = A × A =


Hence, A² = I is verified.

Q.36. Prove by Mathematical Induction that (A′)″ = (A″)′, where n ∈ N for any square matrix A.
Ans.
To prove that (A′)″ = (A″)′
Let P (n): (A′)″ = (A″)′
Step 1: Put n = 1, P(1): A′ = A′ which is true for n = 1
Step 2: Put n = K, P(K): (A′)^K = (A^K)′ Let  it be true for n = K
Step 3: Put n = K + 1, P(K + 1): (A′)^K + 1 = (A^K + 1)′
L.H.S. (A′)^K + 1 = (A′)^K × (A′)
= (A^K)′ × (A′) (From step 2)
= (A^K × A)′
= (A^K+1)′ R.H.S.
The given statement is true for P(K+1) whenever it is true for P(K), where K ∈ N.

Q.37. Find inverse, by elementary row operations (if possible), of the following matrices
(i)
(ii)
Ans.
(i) Let

|A| = 1 x 7 - (- 5) x 3 = 7 + 15 = 22 ≠ 0
So, A is invertible.
Let A = IA

R₂ → R₂ + 5R₁
⇒

R₁ → R₁ - 3R₂
⇒=
So
Hence, inverse of
(ii) Let

|A| = 1 x 6 - (- 3) (- 2) = 6 - 6 = 0
|A| = 0 so A is not invertible.
Hence, inverse ofis not possible.

Q.38. Ifthen find values of x, y, z and w.
Ans.
Given that:

Equating the corresponding elements,
xy = 8, w = 4, z + 6 = 0 ⇒ z = – 6, x + y = 6
Now, solving x + y = 6 ...(i)
and xy = 8 ...(ii)
From eqn. (i), y = 6 – x ...(iii)
Putting the value of y in eqn. (ii) we get,
x(6 – x) = 8 ⇒ 6x – x² = 8
⇒ x² – 6x + 8 = 0 ⇒ x² – 4x – 2x + 8 = 0
⇒ x(x – 4) – 2(x – 4) = 0 ⇒ (x - 4) (x - 2) = 0
∴ x = 4, 2
From eqn. (iii), y = 2, 4.
Hence, x = 4 or  2, y = 2 or 4, z = – 6 and w = 4.

Q.39. Iffind a matrix C such that 3A + 5B + 2C is a null matrix.
Ans.
Order of matrices A and B is 2 × 2.
∴ Order of matrix C must be 2 × 2.
Let
∴ 3A + 5B + 2C = 0

Equating the corresponding elements, we get,
48 + 2a = 0 ⇒ 2a = - 48 ⇒ a = - 24
20 + 2b = 0 ⇒ 2b = - 20 ⇒ b = - 10
56 + 2c = 0 ⇒ 2c = - 56 ⇒ c = - 28
76 + 2d = 0 ⇒ 2d = - 76 ⇒ d = - 38
Hence,


Q.40. If  A =then find A² – 5A – 14I. Hence, obtain A³.
Ans.
Given that:

A² = A . A =

∴ A² – 5A – 14I


Hence, A² – 5A – 14I = O
Now, multiplying both sides by A, we get,
A² . A – 5A . A – 14IA = OA
⇒ A³ – 5A² – 14A = 0
⇒ A³ = 5A² + 14A


Hence, A³ =

Q.41. Find the values of a, b, c and d, if

Ans.
Given that:

Equating the corresponding elements, we get,
3a = a + 4 ⇒ 3a – a = 4 ⇒ 2a = 4 ⇒ a = 2
3b = 6 + a + b ⇒ 3b – b – a = 6 ⇒ 2b – a = 6 ⇒ 2b – 2 = 6
⇒ 2b = 8
⇒ b = 4 3c = – 1 + c + d
⇒ 3c – c – d = – 1 ⇒ 2c – d = – 1
and 3d = 2d + 3 ⇒ 3d – 2d = 3 ⇒ d = 3
Now 2c – d = – 1
⇒ 2c – 3 = – 1 ⇒ 2c = 3 - 1 ⇒ 2c = 2
∴ c = 1
∴ a = 2, b = 4, c = 1 and d = 3.

Q.42. Find the matrix A such that

Ans.
Order of matrix is 3 × 2 and the matrix
is 3 × 3
∴ Order of matrix A must be 2 × 3
Let A =

So,


Equating the corresponding elements, we get,
2a – d = – 1 and a = 1 ⇒ 2 × 1 – d = – 1 ⇒ d = 2 + 1 ⇒ d = 3
2b – e = – 8 and b = – 2 ⇒ 2(– 2) – e =  – 8 ⇒ – 4 – e = – 8
⇒ e = 4
2c – f = – 10 and c = – 5 ⇒ 2(– 5)  – f = – 10 ⇒ – 10 – f = – 10
⇒ f = 0
a = 1, b = – 2, c = – 5, d = 3, e = 4 and f = 0
Hence,


Q.43. If A =find A² + 2A + 7I.
Ans.
Given that:


Hence, A² + 2A + 7I =

Q.44. If A =and A ^–1 = A′ , find value of α.
Ans.
Here,
A=
Given that: A^{– 1} = A′
Pre-multiplying both sides by
AA^{– 1} = AA′
⇒ I = AA′ [∵ AA^-1 = I]

Hence, it is true for all values of α.

Q.45. If the matrix is a skew symmetric matrix, find the values of a, b and c.
Ans.
Let,  

For skew symmetric matrix, A′ = - A.

⇒
Equating the corresponding elements, we get
 a = – 2, b = – b ⇒ 2b = 0 ⇒ b = 0 and c = - 3
Hence, a = - 2,  b = 0 and c = - 3.

Q.46. If P (x) =then show that 
P(x) . P(y) = P(x + y) = P(y) . P(x).
Ans.
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).

Q.47. If A is square matrix such that A² = A, show that (I + A)³ = 7A + I.
Ans.
To show that: (I + A)³ = 7A + I
L.H.S. (I + A)³ = I³ + A³ + 3I²A + 3IA²
⇒ I + A².A + 3IA + 3IA2
⇒ I + A.A + 3IA + 3IA [∵ A² = A]
⇒ I + A² + 3IA + 3IA
⇒ I + A + 3IA + 3IA [∵ A² = A]
⇒ I + A + 3A + 3A  ⇒ 7A + I  R.H.S.
L.H.S. = R.H.S. Hence, Proved.

Q.48. If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A′BA is skew symmetric.
Ans.
Given that B is a skew symmetric matrix
∴ B′ = - B
Let P = A′BA
⇒ P′ = (A′BA)′
= A′B′(A′)′ [(AB)′ = B′A′]
= A′(- B) A
= – A′BA = - P
So P′ = – P
Hence, A′BA is a skew symmetric matrix.