Chapter 4 Matrices Exercise 4-a Solutions Plus Two CHSE Odisha Board
Question 1.
State the order of the following matrices.
(i) [abc]
(ii)
(iii)
(iv)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)
Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a square matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p2
Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i)
(ii)
(iii)
(iv) (1, 2, 3)
Question 4.
Let A =
(i) What is the order of A?
(ii) Write down the entries a31, a25, a23
(iii) Write down AT.
(iv) What is the order of AT?
Solution:
A =
(i) Order of A is (3 x 5)
(ii) a31 = 3, a25= 2, a23 = 6
(iii) AT =
(iv) Order of AT is (5 x 3).
Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A =
Solution:
(ii) A =
Solution:
(iii) A =
Solution:
(iv) A =
Solution:
(v)
Solution:
Question 6.
(i) Find the 2×2 matrix X
if X +
Solution:
(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] determine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1
(iii) If
Solution:
(iv) Find a matrix which when added to
Solution:
Question 7.
Calculate whenever possible, the following products.
(i)
Solution:
(ii)
Solution:
(iii)
Solution:
(iv)
Solution:
Question 8.
If A =
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:
Question 9.
Find the following products.
(i)
Solution:
(ii)
Solution:
(iii)
Solution:
(iv)
Solution:
(v)
Solution:
(vi)
Solution:
(vii)
Solution:
(viii)
Solution:
(ix)
Solution:
(x)
Solution:
Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False
(ii) k[0] = 0, k ∈ R
Solution:
False
(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False
(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True
(v)
Solution:
True
(vi)
Solution:
False
(vii) With five elements a matrix can not be constructed.
Solution:
False
(viii)The unit matrix is its own transpose.
Solution:
True
Question 11.
If A =
Solution:
Question 12.
Find x and y in the following.
(i)
Solution:
(ii)
Solution:
(iii)
Solution:
(iv)
Solution:
(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:
Question 13.
The element of ith row and jth column of the following matrix is i +j. Complete the matrix.
Question 14.
Write down the matrix
Question 15.
Construct a 2 x 3 matrix having elements given by
(i) aij = i + j
(ii) aij = i – j
(iii) aij = i × j
(iv) aij = i / j
Solution:
Question 16.
If
Solution:
Question 17.
Find A such that
Solution:
Question 18.
If
Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)
Question 20.
Find A if
Solution:
Question 21.
Find B if B2 =
Solution:
.jpg)
∴ a2 + bc = 17, ab + bd= 8
ca + cd = 8, bc + d2 = 17
∴ a2 + bc = bc + d2
or, a2 + d2 or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a2 + bc = 17
or, a2 + b . b = 17 (∵ b = c)
or, a2 + b2 = 17
Also (a + b)2 = a2 + b2 + 2ab
∴ (a + b)2 = 17 + 8 = 25
or, a + b = 5
And (a – b)2 = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B =
Question 22.
Find x and y when
Question 23.
Find AB and BA given that:
Question 24.
Evaluate
Question 25.
If
.jpg)
Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:
Question 26.
Find A and B where
Question 27.
If A =
Solution:
Question 28.
Verify that [AB]T = BTAT where
Question 29.
Verify that A =
Solution:
Question 30.
If A =
Solution:
Question 31.
Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A2 – B2 = (A – B) (A + B)
(ii) A2 + 2AB + B2 = (A + B)2
(iii) A2 – 2AB + B2 = (A – B)2
Solution:
(i) (A – B) (A + B)
= A2 + AB – BA – B2
= A2 + AB – AB- B2(∵ AB = BA)
= A2 – B2
(ii) (A + B)2 = (A + B) (A + B)
= A2 + AB + BA + B2
= A2 + AB + AB + B2 (∵ AB = BA)
= A2 + 2AB + B2
(iii) (A – B)2 = (A – B) (A – B)
= A2 – AB – BA + B2
= A2 – AB – AB + B2 (∵ AB = BA)
= A2 – 2AB + B2
Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A2 – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A2 – AβI – αIA + αβI2
= A2 – βAI – αA + αβI
(∵ IA = A, I2 = I)
= A2 – βA – αA + αβI) (∵ AI = A)
= A2 – (α + β) A + αβI
Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar multiplication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)
Question 39.
Question 43.
| Men | Women | Children | |
| Family A → | 4 | 6 | 2 |
| Family B → | 2 | 2 | 4 |
| Family B | ||
| Calory | Proteins | |
| Men | 2400 | 45 |
| Women | 1900 | 55 |
| Children | 1800 | 33 |
Solution:
The given information can be written in matrix form as
∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.
Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x 50000-x].jpg)
⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

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