The Matrices chapter in 12th standard NCERT Maths introduces the concept of matrices, their types, operations, and applications. Here’s a summary:
1. Definition of a Matrix:
- A matrix is a rectangular array of numbers (real or complex) enclosed in square brackets or parentheses.
- It has rows (horizontal lines) and columns (vertical lines).
- A matrix with m rows and n columns is said to be of order m × n (or m by n). The number of elements is mn.
2. Types of Matrices:
- Row Matrix: A matrix with only one row (order 1 × n).
- Column Matrix: A matrix with only one column (order m × 1).
- Square Matrix: A matrix with an equal number of rows and columns (order n × n).
- Identity Matrix: A scalar matrix where all diagonal elements are 1 (denoted by Iₙ, where n is the order).
3. Operations on Matrices:
- Equality of Matrices: Two matrices are equal if they have the same order and corresponding elements are equal.
- Addition of Matrices: Two matrices can be added only if they have the same order. Addition is performed element-wise.
- Subtraction of Matrices: Similar to addition, matrices can be subtracted only if they have the same order. Subtraction is also element-wise.
- Scalar Multiplication: Multiplying a matrix by a scalar (a number) involves multiplying each element of the matrix by that scalar.
- Matrix Multiplication: Two matrices A (m × n) and B (n × p) can be multiplied. The resulting matrix AB will have the order m × p. Matrix multiplication is not commutative in general (AB ≠ BA). It is performed by taking the dot product of rows of the first matrix with columns of the second matrix.
4. Transpose of a Matrix:
- The transpose of a matrix A (denoted by Aᵀ or A’) is obtained by interchanging its rows and columns.
- Properties:
- (Aᵀ)ᵀ = A
- (A + B)ᵀ = Aᵀ + Bᵀ
- (kA)ᵀ = kAᵀ (where k is a scalar)
- (AB)ᵀ = BᵀAᵀ
5. Elementary Operations (Transformations):
These are operations performed on the rows or columns of a matrix:
- Interchanging two rows or columns.
6. Invertible Matrices:
- A square matrix A is invertible (or non-singular) if there exists another square matrix B such that AB = BA = I (the identity matrix). B is called the inverse of A and is denoted by A⁻¹.
- The inverse of a matrix can be found using elementary operations or adjoints.
7. Applications:
Matrices are used in various fields, including:
- Solving systems of linear equations.
- Computer graphics.
- Physics.
- Economics.
Exercise 3.1
1.
write: (i) The order of the matrix. (ii) The number of elements. (iii) Write the elements
Ans :
(i) The order of the matrix:
The order of a matrix is defined as (number of rows) × (number of columns).
- The matrix A has 3 rows.
- The matrix A has 4 columns.
Therefore, the order of matrix A is 3 × 4.
(ii) The number of elements:
The number of elements in a matrix is the product of the number of rows and the number of columns.
- Number of rows = 3
- Number of columns = 4
Number of elements = 3 × 4 = 12
(iii)
a₁₃ = 19
a₂₁ = 35
a₃₃ = -5
a₂₄ = 12
a₂₃ = 5/2
2. If a matrix has 24 elements, what are possible orders it can order? What, if it has 13 elements?
Ans :
1. Matrix with 24 elements:
If a matrix has 24 elements, its order (number of rows × number of columns) must result in a product of 24. We need to find all pairs of positive integers whose product is 24.
The pairs of factors of 24 are:
- 1 × 24
- 24 × 1
- 2 × 12
- 12 × 2
- 3 × 8
- 8 × 3
- 4 × 6
- 6 × 4
2. Matrix with 13 elements:
If a matrix has 13 elements, its order must result in a product of 13. We need to find all pairs of positive integers whose product is 13.
The pairs of factors of 13 are:
- 1 × 13
- 13 × 1
4. Construct a 2 x 2 matrix A = [a ij ] whose elements are given by:
(i)
(ii) a ij = i/j
(iii)
Ans :
(ii)
(iii)
5. Construct a 3 x 4 matrix, whose elements are given by:
(i)
(ii) a ij = 2i – j
Ans :
6. Find the values of x,y and z from the following equations:
(i)
(ii)
(iii)
Ans :
Comparing the corresponding elements, we get: x = 1, y = 4 = z = 3
(ii)
Comparing the corresponding elements, we get: x + y = 6, xy = 8 and 5 + z = 5 Now, 5 + z = 5 ⇒ z = 0 We know that: (x – y) 2 = (x + y) 2 – 4xy ⇒ (x – y) 2 = 36 – 32 = 4 ⇒x – y = ± 2 Now, when x – y = 2 and x + y = 6 , we get x = 4, y = 2 When x – y = -2 and x + y = 6, we get x = 2, y = 4 ∴ x = 4, y = 2, and z = 0, or x = 2, y = 4 and z = 0
(iii)
7. Find the values of a,b, c and d from the equation
Ans :
- Equate Corresponding Elements: Since the matrices are equal, their corresponding elements must be equal. This gives us the following equations:
- a – b = -1 …(1)
- 2a + c = 5 …(2)
- 2a – b = 0 …(3)
- 3x + d = 13 …(4) (Note: There seems to be a typo in the original question. It should likely be 3d + d = 13)
- Solve for a and b: Subtract equation (1) from equation (3): (2a – b) – (a – b) = 0 – (-1) 2a – b – a + b = 1 a = 1
Substitute a = 1 into equation (1): 1 – b = -1 -b = -2 b = 2 - Solve for c: Substitute a = 1 into equation (2): 2(1) + c = 5 2 + c = 5 c = 3
- Solve for d (Assuming the typo is corrected): If the equation was 3d + d = 13 (which is likely the intended equation): 4d = 13 d = 13/4
If the equation was 3x + d = 13, we cannot solve for d without knowing x.
Therefore, the values are:
- a = 1
- b = 2
- c = 3
- d = 13/4 (assuming the typo is corrected to 3d + d = 13)
8. A = [a ij ] m x n is a square matrix if:
(A) m < n (B) m > n (C) m = n (D) None of these
Ans :
The correct answer is (C) m = n.
Here’s why:
- A square matrix is defined as a matrix with an equal number of rows and columns.
- ‘m’ represents the number of rows.
- ‘n’ represents the number of columns.
Therefore, for a matrix to be a square matrix, m must be equal to n.
9. Which of the given values of x and y make the following pairs of matrices equal:
(B) Not possible to find
Ans :
Option (B) is correct
10. The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512
Ans :
- Number of Elements: A 3 × 3 matrix has 3 rows and 3 columns, so it has 3 × 3 = 9 elements.
- Choices for Each Element: Each of the 9 elements can be either 0 or 1, which means there are 2 choices for each element.
- Total Possible Matrices: Since each of the 9 elements has 2 independent choices, the total number of possible matrices is 2 multiplied by itself 9 times, which is 2⁹.
- Calculate 2⁹: 2⁹ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512
Therefore, the correct answer is (D) 512.
Exercise 3.2
1. Let
Find each of the following: (i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA
Ans :
(i)
(ii)
(ii)
(iv)
(v)
2. Compute the following
(i)
(ii)
(iii)
(iv)
Ans :
(i)
(ii)
(iii)
(iv)
3. Compute the indicated products:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ans :
(i)
(ii)
(iv)
(v)
(vi)
4. If
then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Ans :
Hence, L.H.S. = R.H.S. Proved.
5. If
then compute 3A – 5B.
Ans :
6. Simplify:
Ans :
7. Find X and Y, if:
(i)
(ii)
Ans :
(i)
(ii)
8. Find X if
Ans :
9. Find x and y if
Ans :
10. Solve the equation for x,y,z and t if
Ans :
11. If
find the values of x and y
Ans :
Comparing the corresponding elements of these two matrices, we get:
2x − y = 10
and 3x + y = 5
Adding these two equations, we have:
5x = 15 ⇒
x = 3
Now,
3x + y = 5
⇒ y = 5 − 3x
⇒ y = 5 − 9
= −4
∴x = 3 and y = −4
12. Given:
find the values of x ,y,z and w
Ans :
13. If
show that F(x) F(y) = F(x + y)
Ans :
14. Show that
(i)
(ii)
Ans :
(i)
(ii)
(15) Find A 2 – 5A + 6I if A
Ans :
16. If
prove that A 3 – 6A 2 + 7A + 2I = 0.
Ans :
Hence Prove.
17. If
find k so that A 2 = kA – 2I
Ans :
18. If
and I is the identity matrix of order 2, show that I + A = (I-A)
Ans :
19. A trust fund has ` 30,000 that must be invested in two different types of bond. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 in two types of bonds, if the trust fund must obtain an annual interest of (a) ₹ 1800, (b)₹ 2000.
Ans :
Let ₹ x be invested in the first bond and ₹ y be invested in the second bond. Then, we have the following system of equations:
- x + y = 30000
- 0.05x + 0.07y = I
where I is the annual interest earned.
[1 1] [x] = [30000]
[0.05 0.07] [y] = [I]
We can solve this system of equations using matrix multiplication. First, we need to find the inverse of the matrix on the left-hand side:
[1 1]^-1 = [0.07 -1] / (0.07 – 0.05) = [3.5 -50]
[0.05 0.07] [-0.05 1]
Then, we can multiply both sides of the equation by the inverse matrix to get:
[x] = [3.5 -50] [30000]
[y] = [-0.05 1] [I]
This gives us the following equations:
- x = 3.5 * 30000 – 50 * I
- y = -0.05 * 30000 + I
Now, we can substitute the values of I to find the amounts to be invested in each bond.
(a) If the trust fund must obtain an annual interest of ₹ 1800, then I = 1800. Substituting this into the equations above, we get:
- x = 3.5 * 30000 – 50 * 1800 = ₹ 15000
- y = -0.05 * 30000 + 1800 = ₹ 3000
Therefore, ₹ 15000 should be invested in the first bond and ₹ 3000 should be invested in the second bond.
(b) If the trust fund must obtain an annual interest of ₹ 2000, then I = 2000. Substituting this into the equations above, we get:
- x = 3.5 * 30000 – 50 * 2000 = ₹ 5000
- y = -0.05 * 30000 + 2000 = ₹ 5000
Therefore, ₹ 5000 should be invested in the first bond and ₹ 5000 should be invested in the second bond.
20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Ans :
1. Calculate the number of each book:
- Chemistry books: 10 dozen × 12 books/dozen = 120 books
- Physics books: 8 dozen × 12 books/dozen = 96 books
- Economics books: 10 dozen × 12 books/dozen = 120 books
2. Create matrices:
Matrix A (representing the number of each book):
A = [120 96 120]
Matrix B (representing the selling price of each book):
B = [80]
[60]
[40]
3. Perform matrix multiplication:
To find the total amount, we need to multiply matrix A (number of books) by matrix B (selling price per book).
Total Amount = A × B = [120 96 120] × [80]
[60]
[40]
Calculating the matrix product:
Total Amount = [120 × 80 + 96 × 60 + 120 × 40] = [9600 + 5760 + 4800] = 20160
Result:
The bookshop will receive a total of ₹ 20160 from selling all the books.
21. Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p, n x 3 and p x k repectively. The restriction on n,k and p so that PY + WY will be define are:
A. k = 3, p = n
B. k is arbitrary, p = 2
C. p is arbitrary, k = 3
D. k = 2, p = 3
Ans :
Analyzing the Expression PY + WY:
- PY:
- P has order p × k
- Y has order 3 × k
- For PY to be defined, k (the number of columns in P) must be equal to 3 (the number of rows in Y).
- WY:
- W has order n × 3
- Y has order 3 × k
- For WY to be defined, 3 (the number of columns in W) must be equal to 3 (the number of rows in Y), which is already satisfied.
- PY + WY:
- For PY + WY to be defined, PY and WY must have the same order.
- PY will have order p × k (after satisfying the condition k = 3, it becomes p × 3)
- WY will have order n × k (after satisfying the condition k = 3, it becomes n × 3)
- Therefore, for PY and WY to have the same order, we must have p = n.
Conclusion:
The restrictions on n, k, and p so that PY + WY is defined are:
- k = 3
- p = n
Therefore, the correct answer is (A) k = 3, p = n.
22. Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p, n x 3 and p x k respectively. If n = p then order of matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D)p × n
Ans :
Analyzing 7X – 5Z:
- 7X:
- This is scalar multiplication. The order of 7X will be the same as the order of X, which is 2 × n.
- 5Z:
- This is also scalar multiplication. The order of 5Z will be the same as the order of Z, which is 2 × p.
- 7X – 5Z:
- For this subtraction to be defined, 7X and 5Z must have the same order.
- Since n = p, both 7X and 5Z have the order 2 × n (or equivalently, 2 × p).
Conclusion:If n = p, the order of the matrix 7X – 5Z is (B) 2 × n.
Exercise 3.2
1. Find the transpose of each of the following matrices:
(i)
(ii)
(iii)
Ans :
(i) Let A=
Transpose of A = A’ or A T = [ 5 1/2 -1]
(ii)
Transpose of A = A’ or A T =
(iii)
Transpose of A = A’ or A T
=
2. If
then verify that: (i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’
Ans :
3.If
then verify that: (i) (A + B)’ = A’ + B’ (ii) (A – B)’ = A’ – B’
Ans :
4. If
then find (A + 2B)’.
Ans :
5. For the matrices A and B, verify that (AB)’ = B’A’, where:
(i)
(ii)
Ans :
6. (i) If A=
then verify that A’A = I.
(ii) If A =
then verify that A’A = I.
Ans :
7. (i) (i) Show that the matrix A
=is a symmetric matrix.
(ii) Show that the matrix A
=is a skew symmetric matrix.
Ans :
i) Given: A
=
Changing rows of matrix A as the columns of new matrix A’ =
= A ∴ A’ = A Therefore, by definitions of symmetric matrix, A is a symmetric matrix.
(ii) Given: A =
A’ =
= –
= – A ∴ A’ = – A Therefore, by definition matrix A is a skew-symmetric matrix
8. For a matrix A =
verify that: (i) (A + A’) is a symmetric matrix. (ii) (A – A’) is a skew symmetric matrix.
Ans :
9. Find 1/2 (A + A’) and 1/2(A – A’) when A =
Ans :
10. Express the following matrices as the sum of a symmetric and skew symmetric matrix: (i)
(ii)
(iii)
(iv)
Ans :
(iii)
(iv)
Choose the correct answer in Exercises 11 and 12.
11. If A and B are symmetric matrices of same order, AB – BA is a: (A) Skew-symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix
Ans :
Properties of Symmetric Matrices:
- A matrix A is symmetric if its transpose is equal to itself (Aᵀ = A).
Analyzing AB – BA:
Let C = AB – BA
Now, let’s find the transpose of C:
Cᵀ = (AB – BA)ᵀ
= (AB)ᵀ – (BA)ᵀ (Using the property (A – B)ᵀ = Aᵀ – Bᵀ)
= BᵀAᵀ – AᵀBᵀ (Using the property (AB)ᵀ = BᵀAᵀ)
= BA – AB (Since A and B are symmetric, Aᵀ = A and Bᵀ = B)
= -(AB – BA)
= -C
We found that Cᵀ = -C. This is the property that defines a skew-symmetric matrix.
Conclusion:
Correct Answer (A) Skew-symmetric matrix.
12. If A
, then A + A’ = I, if the value of α is: (A) π/6 (B) π/3 (C) π (D) 3 π/2
Ans :
Therefore , B is correct
Exercise 3.4
1. Matrices A and B will be inverse of each other only if
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0, BA = I
(D) AB = BA = I
Ans :
The correct answer is (D) AB = BA = I.
Here’s why:
- Inverse Matrices: Two square matrices, A and B, are inverses of each other if and only if their product (in both orders) results in the identity matrix (I).
- Identity Matrix: The identity matrix is a square matrix with 1s on the main diagonal and 0s everywhere else. It acts like the number 1 in matrix multiplication.
