## NCERT Solutions Class 12th Maths Chapter 3: Matrices## Exercise 3.1
Number of columns in the given matric = 4 Therefore, order of the given matrix A = 3 × 4
Therefore, number of elements in the matrix A = 12
a a a a a
We know that the number of elements in a matrix of order n Pairs of natural numbers whose product is 24 are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4) Therefore, possible orders a matrix of 24 elements can have are:
Similarly, if a matrix has 13 elements, then its possible orders will be the pair of natural numbers whose product is 13, i.e., (1, 13), and (13, 1). Therefore, possible orders a matrix of 13 elements can have are:
We know that the number of elements in a matrix of order n Pairs of natural numbers whose product is 18 are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3) Therefore, possible orders a matrix of 18 elements can have are:
Similarly, if a matrix has 5 elements, then its possible orders will be the pair of natural numbers whose product is 5, i.e., (1, 5), and (5, 1). Therefore, possible orders a matrix of 5 elements can have are:
**a**_{ij}= (i + j)^{2}/2**a**_{ij}= i/j**a**_{ij}= (i + 2 j)^{2}/2
a a a a Therefore, required 2 × 2 matrix =
a a a a Therefore, required 2 × 2 matrix =
a a a a Therefore, required 2 × 2 matrix =
**a**_{ij}= |-3i + j|/2**a**_{ij}= 2i - j
a a a a a a a a a a a a Therefore, required 3 × 4 matrix =
a a a a a a a a a a a a Therefore, required 3 × 4 matrix =
(i) = (ii) = (iii) =
5 + z = 5
x + y = 6 y = 6 - x xy = 8 x(6 - x) = 8 6x - x x x x(x - 2) - 4(x - 2) = 0 (x - 2)(x - 4) = 0 So, x - 2 = 0 ⇒ y = 6 - 2 ⇒
x - 4 = 0 ⇒ y = 6 - 4 ⇒
x + y + z = 9 x + z = 5 y + z = 7 So, x + (y + z) = 9 x + 7 = 9
x + z = 5 z = 5 - x z = 5 - 2
y + z = 7 y = 7 - z y = 7 - 3
=
The given matrices are equal. Therefore, their corresponding elements are also equal. a - b = -1 2a + c = 5 2a - b = 0 3c + d = 13 Now, 2a - b = 0 a + a - b = 0 a + (-1) = 0
a - b = -1 1 - b = -1
2a + c = 5 2(1) + c = 5
3c + d = 13 3(3) + d = 13 9 + d = 13
Hence, a = 1, b = 2, c = 3, and d = 4.
The number of rows is same as the number of columns in a square matrix, i.e., m = n. Hence, (C) is the correct answer.
,
Let us assume that = If two matrices are equal then their corresponding elements are also equal. Therefore, 3x + 7 = 0 3x = -7 x = -7/3 2 - 3x = 4 3x = -2 x = -2/3 But x = -7/3 and x = -2/3 cannot be true at the same time. Thus, the given matrices are not equal. Hence, the correct answer is (B).
Number of elements in a matrix of order 3 × 3 = 9 Each entry in the matrix has to be 0 or 1. So, number of ways to fill each entry = 2 Number of entries to be filled = 9 Total permutations for matrix of order 3 × 3 = 2 Hence, the correct answer is (D). ## Exercise 3.2
**A + B****A - B****3A - C****AB****BA**
Now, to verify that A + (B - C) = (A + B) - C LHS = A + (B - C) and RHS = (A + B) - C
When two matrices are equal then their corresponding elements are also equal. Therefore, 2 + y = 5 and 2x + 2 = 8 Solve for y: 2 + y = 5
Solve for x: 2x + 2 = 8 2x = 6
When two matrices are equal then their corresponding elements are also equal. Therefore, 2x + 3 = 9, 2z - 3 = 15, 2y = 12 and 2t + 6 = 18 Solve for x: 2x + 3 = 9 2x = 6
Solve for y: 2y = 12
Solve for z: 2z - 3 = 15 2z = 18
Solve for t: 2t + 6 = 18 2t = 12
Hence, x = 3, y = 6, z = 9, and t = 6.
When two matrices are equal then their corresponding elements are also equal. Therefore, 2x - y = 10 and 3x + y = 5 Add both equations 2x - y + 3x + y = 10 + 5 5x = 15
Solve for y: 2x - y = 10 2(3) - y = 10 6 - y = 10
Hence, x = 3 and y = -4.
When two matrices are equal then their corresponding elements are also equal. Therefore, 3x = 4 + x, 3y = x + y + 6, 3z = -1 + z + w, and 3w = 2w + 3 Solve for x: 3x = 4 + x 2x = 4
Solve for y: 3y = x + y + 6 2y = x + 6 2y = 2 + 6 2y = 8
Solve for w: 3w = 2w + 3
Solve for z: 3z = -1 + z + w 2z = w - 1 2z = 3 - 1 2z = 2
Hence, x = 2, y = 4, z = 1 and w = 3.
LHS = F (x) F (y) = F (x + y) = RHS Hence, proved.
LHS ≠ RHS Hence, proved.
A
So, LHS = A = 0 = RHS Hence, proved.
When two matrices are equal, their corresponding elements are also equal. Therefore, 3k = 3, -2 = -2k, 4 = 4k, -2 = -2k
LHS = I + A
Let the amount invested in the first bond (in rupees) be x. Then, the amount invested in the second bond (in rupees) = 30,000 - x
So, Investment in bonds (in rupees) Annual Interest rate Interest (in rupees) According to matrix multiplication x × 5% + (30000 - x) × 7% = 1800 x × 5/100 + (30000 - x) × 7/100 = 1800 5x/100 + (210000 - 7x)/100 = 1800 (5x + 210000 - 7x) = 1800 × 100 210000 - 2x = 180000 2x = 30000 x = 15,000 30,000 - x = 30,000 - 15,000 = 15,000 Therefore, the amount invested in the first bond is Rs 15,000 and the amount invested in the second bond is Rs 15,000.
So, Investment in bonds (in rupees) Annual Interest rate Interest (in rupees) According to matrix multiplication x × 5% + (30000 - x) × 7% = 2000 x × 5/100 + (30000 - x) × 7/100 = 2000 5x/100 + (210000 - 7x)/100 = 2000 (5x + 210000 - 7x) = 2000 × 100 210000 - 2x = 200000 2x = 10000 x = 5,000 30,000 - x = 30,000 - 5,000 = 25,000 Therefore, the amount invested in the first bond is Rs 5,000 and the amount invested in the second bond is Rs 25,000.
Let the total amount bookshop will receive from selling all books (in rupees) = x. Total amount (in rupees) According to matrix multiplication 120 × 80 + 96 × 60 + 120 × 40 = x 9600 + 5760 + 4800 = x 20160 = x Hence, the bookshop will receive a total of Rs 20160 from selling all the books.
Order of the matrix P = p × k Order of the matrix Y = 3 × k PY will be defined if k = 3 Order of the matrix PY = p × k Order of the matrix W = n × 3 Order of the matrix Y = 3 × k Order of the matrix WY = n × k PY + WY is defined only if the order of PY is equal to the order of WY. p × k = n × k p = n Hence, (A) is the correct answer.
Addition or subtraction operations on two matrices doesn't change their order. Therefore, order of 7X - 5Z is equal of the order of matrix X and order of matrix Z, i.e., 2 × n. Hence. (B) is the correct answer. ## Exercise 3.3
**(A + B)' = A' + B'****(A - B)' = A' - B'**
RHS = A' + B' Now, = LHS Hence, verified that (A - B)' = A' - B'. **(A + B)' = A' + B'****(A - B)' = A' - B'**
= LHS Hence, verified that (A + B)' = A' + B'.
= LHS Hence, verified that (AB)' = B' A'.
**(A + A') is a symmetric matrix****(A - A') is a skew symmetric matrix**
A = 1/2 × {(A + A') + (A - A')} A = ½ (A + A') + ½ (A - A') Let ½ (A + A') = P and ½ (A - A') = Q. Therefore, P is a symmetric matrix. Q' = -Q Therefore, Q is a skew symmetric matrix. Hence, given matrix can be expressed as the sum of a symmetric and a skew symmetric matrix as A = P + Q. Q' = -Q Therefore, Q is a skew symmetric matrix. Hence, given matrix can be expressed as the sum of a symmetric and a skew symmetric matrix as A = P + Q. Q' = -Q Therefore, Q is a skew symmetric matrix. Hence, given matrix can be expressed as the sum of a symmetric and a skew symmetric matrix as A = P + Q.
(AB - BA)' = (AB)' - (BA)' (AB - BA)' = B'A' - A'B' (AB - BA)' = BA - AB as A = A' and B = B' since both matrices are symmetric (AB - BA)' = -(AB - BA) Therefore, (AB - BA) is a skew symmetric matrix. Hence, (A) is the correct answer. When two matrices are equal then their corresponding elements are also equal. Therefore, 2 cos α = 1 cos α = 1/2 cos α = cos π/3 α = π/3 Hence, (B) is the correct answer. ## Exercise 3.4
We know that A A AB = I must be true and BA = I must be true as well. Thus, AB = BA = I. Hence, (D) is the correct answer. ## Miscellaneous Exercise
(AB - BA)' = (AB)' - (BA)' (AB - BA)' = B'A' - A'B' (AB - BA)' = BA - AB
Therefore, (AB - BA) is a skew symmetric matrix.
CASE I: Matrix A is symmetric. A = A' Then, (B' AB)' = (AB)'(B')' (B' AB)' = (AB)' B (B' AB)' = B' A' B (B' AB)' = B' A B
Therefore, (B' AB)' is a symmetric matrix when A is a symmetric matrix. CASE II: Matrix A is skew symmetric. A' = -A Then, (B' AB)' = (AB)'(B')' (B' AB)' = (AB)' B (B' AB)' = B' A' B (B' AB)' = -B' A B
Therefore, (B' AB)' is a skew symmetric matrix when A is a skew symmetric matrix. When two matrices are equal then their corresponding elements are also equal. Therefore, 2x 6y z Solve for x: 2x x
Solve for y: 6y y
Solve for z: 3z z
Hence, x = ± 1/√2, y = ± 1/√6, and z = ± 1/√3. [6 × 0 + 2 × 2 + 4 × x] = O [0 + 4 + 4x] = [0] When two matrices are equal then their corresponding elements are also equal. Therefore, 4 + 4x = 0 4x = -4
A matrix can be formed representing the annual sales of all three products A matrix can also be formed representing the selling price of the three products x, y and z Then total revenue of the markets I and II can be calculated with the product of above two matrices. Hence, the revenue of market I is Rs 46000 and that of market II is Rs 53000.
A matrix can be formed representing the annual sales of all three products A matrix can also be formed representing the cost prices of the three products x, y and z Then total cost of the markets I and II can be calculated with the product of above two matrices. Therefore, cost of market I is Rs 31000 and that of market II is Rs 36000. Gross profit from the markets = Total revenue - Total cost Hence, the gross profit of market I is Rs 15000 and that of market II is Rs 17000.
When two matrices are equal then their corresponding elements are also equal. Therefore, a + 4b = -7, 2a + 5b = -8, 3a + 6b = -9, c + 4d = 2, 2c + 5d = 4, and 3c + 6d = 6 Solve for b: (3a + 6b) - (2a + 5b) - (a + 4b) = -9 - (-8) - (-7) 3a + 6b - 2a - 5b - a - 4b = -9 + 8 + 7 3a - 3a + 6b - 9b = -1 + 7 -3b = 6
Solve for a: a + 4b = -7 a + 4(-2) = -7 a - 8 = -7
Solve for d: (3c + 6d) - (2c + 5d) - (c + 4d) = 6 - 4 - 2 3c + 6d - 2c - 5d - c - 4d = 6 - 6 3c - 3c + 6d - 9d = 0 -3d = 0
Solve for c: c + 4d = 2 c + 4(0) = 2
A When two matrices are equal then their corresponding elements are also equal. Therefore, α 1 - α² - βγ = 0 Hence, (C) is the correct answer.
A zero matrix is always symmetric as well as skew symmetric. Hence, (B) is the correct answer.
(I + A)³ - 7 A = I = I + A = I + AA + 3A + 3IA - 7A = I + A + 3A + 3A - 7A = I + 7A - 7A = I Hence, (C) is the correct answer. |