ICSE Class 10

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
Find x and y, if:

Solution:

Question 2.
Find x and y, if :

Solution:

Question 3.

Solution:

Question 4.

(i) the order of the matrix X
(ii) the matrix X.
Solution:

Question 5.
Evaluate

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

(i) A (BA)
(ii) (AB) A
Solution:

Question 9.
Find x and y, if

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:
(i) Order of matrix A is 2 x 2
Order of matrix B is 2 x 1
Order of matrix X is 2 x 1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
Evaluate if possible :

If not possible, give a reason.
Solution:

(iv) It is not possible, because number of columns of the first matrix is not equal to number of rows of the second matrix.

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Find x and y, if:

Solution:

Question 5.

Solution:

Question 6.

(i) AB
(ii) BA
(iii) A²
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:


Comparing the elements, we get:
-2b =-2
b = 1
a = 2

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Simplify : A² + BC.
Solution:

Question 16.
Solve for x and y :

Solution:

Question 17.
In each case given below, find :
(a) the order of matrix M.
(b) the matrix M.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.
If A and B are any two 2 x 2 matrices such that AB = BA = B and B is not a zero matrix, What can you say about the matrix A?
Solution:
AB = BA = B
But it is possible, when A = 0 or B = 0
But B is not a zero matrix (given)
A is a zero matrix or A is an identity matrix

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.
State, with reason, whether the following are true or false. A, B and C are matrices of order 2 x 2.
(i) A + B = B + A
(ii) A – B = B – A
(iii) (B . C). A = B . (C . A)
(iv) (A + B) . C = A . C + B . C
(v) A . (B – C) = A . B – A . C
(vi) (A – B) . C = A . C – B . C
(vii) A² – B² = (A + B) (A – B)
(viii) (A – B)² = A² – 2 A . B + B²
Solution:
(i) True : Because addition of matrices is commutative.
(ii) False : Subtraction of matrices is not commutative.
(iii) True : Multiplication of matrices is associative.
(iv) True: Multiplication of matrices is distributive over addition.
(v) True : As given above in (iv)
(vi) True : As given above in (iv)
(vii) False : Laws of algebra for factorization and expansion are not applicable to matrices.
(viii) False, As given above in (vii)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
Evaluate

Solution:

Question 2.
Find x and y if :

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If I is the unit matrix of order 2 x 2; find the matrix M, such that:

Solution:

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 x 2 and 2 x 3 respectively; then their sum A + B is possible,
(ii) The matrices A 2 x3, and B 2 x 3, are conformable for subtraction,
(iii) Transpose of a 2 x 1 matrix is a 2 x 1 matrix,
(iv) Transpose of a square matrix is a square matrix,
(v) A column matrix has many columns and only one row.
Solution:
(i) False, because the orders of both the matrices are not same,
(ii) True,
(iii) False, because transpose of a 2 x 1 matrix is a 1 x 2 matrix,
(iv) True,
(v) False, because it has only one column and may have many rows.

Question 2.

Solution:
Comparing the elements in order, we get that :
x = 3
y + 2 = 1 ⇒ y = 1 – 2 = -1
z – 1 = 2 ⇒ z = 2 + 1 = 3
Hence x = 3, y = -1 , z = 3

Question 3.
Solve for a, b and c; if :

Solution:
(i) Comparing the elements in order, we get
-4 = b + 4 ⇒ – b = 4 + 4 = 8 ⇒ b = – 8
a + 5 = 2 ⇒ a = 2 – 5= -3
2 = c – 1 ⇒ c = 2 + 1 = 3
Hence a = -3, b = -8, c = 3
(ii) Comparing the elements in order, we get
a = 3
a – b = – 1 ⇒ 3 – b = -1 ⇒ -b = -1 – 3 ⇒ -b = -4 ⇒ b = 4
b + c = 2 ⇒ 4 + c = 2 ⇒ c = 2 – 4 = -2
Hence a = 3, b = 4, c = – 2

Question 4.

Solution:

Question 5.

Solution:

Question 6.
Wherever possible, write each of the following as a single matrix.

Solution:

This is not possible because both the matrices are not of the same order.

Question 7.
Find, x and y from the following equations:

(ii) [-8 x] + [y -2] = [-3 2]
Solution:

Comparing, the elements of two equal martrices:
3 – x = 7 ⇒ x = -7 + 3 = -4
and y + 2 = 2 ⇒ y = 2 – 2 = 0
Hence x = – 4, y = 0
(ii) [-8 x] + [y -2] = [-3 2]
⇒ [-8 + y x – 2] = [-3 2]
Comparing the elements, we get :
-8 + y = -3 ⇒ y = -3 + 8 = 5
x – 2 = 2 ⇒ x = 2 + 2 = 4
Hence x = 4, y = 5

Question 8.

Solution:

Question 9.
Write the additive inverse of matrices A, B and C

Solution:

Question 10.
Given A = [ 2 -3], B = [0 2] and C = [-1 4]; find the matrix X in each of the following :
(i) X + B = C – A
(ii) A – X = B + C
Solution:
A = [2 -3], B = [0 2], C = [-1 4]
(i) X + B = C – A
X + [0 2] = [-1 4] – [2 -3]
⇒ X + [0 2] = [- 1 – 2 4 – (-3)]
X + [0 2] = [-3 4 + 3] = [-3 7]
X = [-3 7] – [0 2] = [-3 – 0 7 – 2] = [-3 5]
(ii) A – X = B + C
⇒ -X = B + C – A
⇒ X = A – B – C = [2 -3] – [0 2] – [-1 4] = [2 – 0 – (-1) -3 – 2 – 4] = [ 2 + 1 – 9] = [3 – 9]

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

Question 1.
Show that (x – 1) is a factor of x³ – 7x² + 14x – 8. Hence, completely factorise the given expression.
Solution:

Question 2.
Using Remainder Theorem, factorise : x³ + 10x² – 37x + 26 completely. (2014)
Solution:
f(x) = x³ + 10x² – 37x + 26
f(1) = (1)³ + 10(1)² – 37(1) + 26 = 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f(x)

Question 3.
When x³ + 3x² – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution:
Let f(x) = x³ + 3x² – mx + 4
and x – 2 = 0 then x = 2
f(2) = (2)³ + 3(2)² – m(2) + 4 = 8 + 12 – 2m + 4 = 24 – 2m
Remainder = 24 – 2m
But, remainder is given m + 3
m + 3 = 24 – 2m
⇒ m + 2m = 24 – 3
⇒ 3m = 21
⇒ m = 7
Hence m = 7

Question 4.
What should be subtracted from 3x³ – 8x² + 4x – 3, so that the resulting expression has x + 2 as a factor ?
Solution:
The number to be subtracted = Remainder obtained by dividing 3x³ – 8x² + 4x – 3 by x + 2
Let f(x) = 3x³ – 8x² + 4x – 3
and x + 2 = 0, then x = – 2
Remainder = f(-2) = 3 (-2)³ – 8 (-2)² + 4 (-2) – 3 = -24 – 32 – 8 – 3 = -67
Hence the number to be subtracted = – 67

Question 5.
If (x + 1) and (x – 2) are factors of x³ + (a + 1) x² – (b – 2) x – 6, find the values of a and 6. And then, factorise the given expression completely.
Solution:

Question 6.
If x – 2 is a factor of x² – ax + b and a + b = 1, find the values of a and b.
Solution:
(x – 2) is a factor of x² + ax + b
Let x – 2 = 0 ⇒ x = 2
Now x² + ax + b = (2)² + a x 2 + b = 4 + 2a + b = 2a + b + 4
x – 2 is the factor Remainder = 0 or 2a + b + 4 = 0
⇒ 2a + b = -4 …(i)
But a + b = 1 (given) …(ii)
Subtracting, we get : a = -5
Substituting the value of a in (ii)
-5 + b = 1 ⇒ b = 1 + 5 ⇒ b = 6
Hence a = -5, b = 6

Question 7.
Factorise x³ + 6x² + 11x + 6 completely using factor theorem.
Solution:

Question 8.
Find the value of ‘m’ if mx³ + 2x² – 3 and x² – mx + 4 leave the same remainder when each is divided by x – 2.
Solution:
Let f(x) = mx³ + 2x² – 3
g (x) = x² – mx + 4
Let x – 2 = 0, then x = 2
f(2) = m (2)³ + 2 (2)² – 3 = 8m + 8 – 3 = 8m + 5
g(2) = (2)² – mx² + 4 = 4 – 2m + 4 = 8 – 2m
In both cases the remainder is same
8m + 5 = 8 – 2m
⇒ 8m + 2m = 8 – 5
⇒ 10m = 3
⇒ m = \(\frac { 3 }{ 10 }\)

Question 9.
The polynomial px³ + 4x² – 3x + q is completely divisible by x² – 1; find the values of p and q. Also, for these values of p and q, factorize the given polynomial completely.
Solution:

Question 10.
Find the number which should be added to x² + x + 3 so that the resulting polynomial is completely divisible by (x + 3).
Solution:
Let k be added to f(x), then f(x) = x² + x + 3 + k
Let x + 3 = 0, then x = -3
f(-3) = (-3)2 + (-3) + 3 + k = 9 – 3 + 3 + k = 9 + k
f(x) is divisible by x + 3, then remainder will be 0.
9 + k = 0 ⇒ k = -9
-9 should be added

Question 11.
When the polynomial x³ + 2x² – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x³ + ax² – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
Solution:
Let f(x) = x³ + 2x² – 5ax – 1
and let x – 1 = 0, then x = 1
f(1) = (1)³ + 2(1)² – 5a x 1 – 7 = 1 + 2 – 5a – 7 = -5a – 4
-5a – 4 = A ….(i)
Let g (x) = x³ + ax² +12x + 16
and let x + 2 = 0, then x = -2
g (-2) = (-2)³ + a (-2)² – 12 (-2) + 16 = -8 + 4a + 24 + 16 = 32 + 4a
32 + 4a = B ….(ii)
2A + B = 0
2 (-5a – 4) + 32 + 4a = 0
⇒ -10a – 8 + 32 + 4a = 0
⇒ -6a + 24 = 0
⇒ 6a = 24
⇒ a = 4
a = 4

Question 12.
(3x + 5) is a factor of the polynomial (a – 1) x³ + (a + 1) x² – (2a + 1) x – 15. Find the value of ‘a’. For this value of ‘a’, factorise the given polynomial completely.
Solution:
Let f(x) = (a – 1) x³ + (a + 1) x² – (2a + 1) x – 15

Question 13.
When divided by x – 3 the polynomials x³ – px² + x + 6 and 2x³ – x² – (p + 3) x – 6 leave the same remainder. Find the value of ‘p.’
Solution:
When (x – 3) divides x³ – px² + x + 6,
then Remainder = p(3) = (3)³ – p(3)² + (3) + 6 = 27 – 9p + 9 = 36 – 9p
When (x – 3) divides 2x³ – x² – (p + 3) x – 6,
then Remainder = p(3) = 2(3)³ – (3)² – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p
⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
⇒ p = 1

Question 14.
Use the Remainder Theorem to factorise the following expression : 2x³ + x² – 13x + 6
Solution:
(a) By hit and trial, putting x = 2, we have
2 (8) + 4 – 26 + 6 = 0
⇒ (x – 2) is the factor of 2x³ + x² – 13x + 6

Question 15.
Using remainder theorem, find the value of k if on dividing 2x³ + 3x² – kx + 5 by x – 2, leaves a remainder 7. (2016)
Solution:
Let f(x) = 2x³ + 3x² – kx + 5 By the remainder theorem,
f(2) = 7
⇒ 2(2)³ + 3(2)² – k(2) + 5 = 7
⇒ 2(8) + 3(4) – k(2) + 5 = 7
⇒ 16 + 12 – 2k + 5 = 7
⇒ 2k = 16 + 12 + 5 – 7
⇒ 2k = 26
⇒ k = 13
The value of k is 13.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.