NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices
NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices are solved for helping the students of 12 class in boosting their preparation of the exams.NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices will provide you information about matrix which is one of the technics of solving all kinds of algebraic problems. In this chapter 3-matrix, you will learn the properties of matrix, like addition,multiplication, subtraction,division and solutions of equations by using the properties of matrix.
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Download PDF of NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices
PDF of NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices
NCERT solutions of class 12 maths
Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
Study notes of Maths and Science NCERT and CBSE from class 9 to 12
Q1. In the following matrix
(i) The order of matrix
(ii) The number of elements
(iii) Write the elements a_{13}, a_{21, }a_{33,} a_{24}, a_{23}
Ans.
(i) The order of matrix = Number of rows × Number of column = 3 × 4
(ii) The number of elements in matrix are = 3× 4= 12
(iii) The number of elements in a matrix are represented as a_{mn }where m is number of row and n is number of column,so a_{13},=19, a_{21 }=35 , a_{33 }= – 5, a_{24}=12, a_{23 }= 5/2
Q2. If a matrix has 24 elements ,what are the possible orders it can have ?What ,if it has 13 elements.
Ans. The number of elements in the matrix are = 24
The possible orders of the matrix are , 1 × 24, 2× 12, 3 × 8, 4 × 6, 24 ×1,12 × 2,8 ×3 and 6 ×4, if the matrix has 13 elements then the possible orders of the matrix are,1 × 13 and 13 × 1
Q3.If the matrix has 18 elements, what are the possible orders it can have ? What, if it has 5 elements.
Ans. The number of elements in the matrix are = 18
The possible orders of the matrix are, 1 × 18, 2× 9, 3 × 6, 18 ×1,9 × 2 and 6 ×3 if the matrix has 5 elements then the possible orders of the matrix are,1 × 5 and 5 × 1
Q4. Construct a 2 × 2 matrix A = [a_{ij}], whose elements are given by
(i) a_{ij}= (i+j)²/2 (ii) a_{ij}= i/j (iii) a_{ij}= (i+2j)²/2
Ans. The matrix to be constructed of the order 2 × 2, so the elements of matrix A = [a_{ij}] are determined as follows.
(i) If elements are designed a_{ij}= (i+j)²/2,then a_{11},, a_{12 } , a_{21 }, a_{22 }are determined as follows
a_{11}= (1+1)²/2 = 2²/2 = 2
a_{12}=(1+2)²/2 = 3²/2 =9/2
a_{21 }= (2+1)²/2 = 3²/2 =9/2
a_{22 }= (2+2)²/2 = 8
Hence, matrix A will be as follows
(ii) If elements are designed a_{ij}= i/j,then a_{11},, a_{12 } , a_{21 }, a_{22 }are determined as follows
a_{11}= 1/1 =1
a_{12}=1/2
a_{21 }= 2/1 = 2
a_{22 }= 2/2 = 1
Hence, matrix A will be as follows
(iii) If elements are designed a_{ij}= (i+2j)²/2,then a_{11},, a_{12 } , a_{21 }, a_{22 }are determined as follows
a_{11}= (1+2×1)²/2 = 3²/2 = 9/2
a_{12}=(1+2×2)²/2 = 5²/2 =25/2
a_{21 }= (2+2×1)²/2 = 4²/2 =16/2 = 8
a_{22 }= (2+2×2)²/2 = 36/2 = 18
Hence, matrix A will be as follows
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Q5. Construct a 3 × 4 matrix ,whose elements are given by
(i) a_{ij}= (1/2)⌊-3i + j⌋ (ii)a_{ij} =2i – j
(i) If elements are designed a_{ij}=(1/2)(-3i + j),then a_{11},, a_{12 } , a_{13} , a_{14 }_{, }a_{21 }, a_{22 } , a_{23}, a_{24} ,a_{31}, a_{32 } , a_{33} , and a_{34 }are determined as follows
a_{11}= (1/2)⌊-3×1+ 1⌋ = 1, a_{12}= (1/2)⌊-3×1 + 2⌋ = 1/2, a_{13 }= (1/2)⌊-3×1 + 3⌋=0
a_{14 }= (1/2)⌊-3×1 + 4⌋=1/2, a_{21 }= (1/2)⌊-3×2 + 1⌋ = 5/2 , a_{22 }= (1/2)⌊-3×2 + 2⌋ =2
a_{23 }= (1/2)⌊-3×2+ 3⌋= 3/2, a_{24 }= (1/2)⌊-3×2 + 4⌋ = 1, a_{31 }= (1/2)⌊-3×3 + 1⌋= 4
a_{32 }= (1/2)⌊-3×3 + 2⌋= 7/2, a_{33 }= (1/2)⌊-3×3 + 3⌋=3, a_{34 }= (1/2)⌊-3×3 + 4⌋= 5/2
Hence, matrix will be as follows
(ii) If elements are designed a_{ij} =2i – j ,then a_{11}, a_{12 } , a_{13} , a_{14 }_{, }a_{21 }, a_{22 } , a_{23}, a_{24} ,a_{31}, a_{32 } , a_{33} , and a_{34 }are determined as follows
a_{11}=2×1 – 1 = 1, a_{12}= 2×1 – 2 = 0, a_{13 }= 2×1- 3=-1
a_{14 }=2×1 – 4=-2, a_{21 }= 2×2 – 1 = 3 , a_{22 }= 2×2 – 2 =2
a_{23 }=2×2 – 3= 1, a_{24 }=2×2 – 4 = 0, a_{31 }= 2×3 – 1= 5
a _{32} = 2 x 3 – 2 = 4, a _{33} = 2 x 3 – 3 = 3, a _{34} = 2 x 3 – 4 = 2
Hence, matrix will be as follows
Q6. Find the values of x, y and z from the following equations
Years.
Two matrices are equal ,so their corresponding element should be equal to each other
On comparing LHS and RHS matrices,we get
y = 4 and x = 1
Two matrices are equal ,so their corresponding element should be equal to each other
On comparing LHS and RHS matrices,we get
x +y = 6….(i)
5 + z = 5 ⇒ z = 0 …. (ii)
xy = 8…….(iii)
Substituting the value of y = 6- x from equation (i) in equation (iii)
( 6- x)x = 8
-x² +6x -8 = 0
-x² +6x -8 = 0
x² -6x +8 = 0
x² -4x -2x+8 = 0
x(x – 4) – 2(x – 4) = 0
(x – 4)(x – 2) = 0
x = 4,2
putting the value x = 4 in equation (i), we get the value of y= 2
So,required values are x = 4, y= 2 and z = 0
putting the value x = 2 in equation (i), we get the value of y= 4
So,required values are x = 2, y= 4 and z = 0
Two matrices are equal ,so their corresponding element should be equal to each other
On comparing LHS and RHS matrices,we get
x+ y + z = 9….(i)
x + z = 5….(ii)
y + z = 7….(iii)
On solving equation (i) and equation (ii),we get y= 4
On solving equation (i) and equation (iii),we get x= 2
Putting the value of y = 4 in equation (iii),we get z = 3
Hence required values are, x = 2, y = 4 and z = 3
Q7. Find the value of a,b,c and d from the equation
Two matrices are equal ,so their corresponding element should be equal to each other
On comparing LHS and RHS matrices,we get
a – b = -1….(i), 2a – b = 0….(ii), 2a + c = 5…..(iii), 3c + d = 13…..(iv)
From equation (i) and equation (ii), we get
b = 2 and a = 1
Putting the value of a = 1 in equation in equation (iii),we get c = 3
Putting the value of c = 3 in equation (iv), we get d = 4
Therefore the required values are a = 1, b = 2, c = 3 and d = 4
Q8. A = [a_{ij}]_{m×n }is a square matrix,if
(A) m < n (B) m > n (C) m = n (D) None of these
Ans. The square matrix has same number of rows and columns,therefore answer is (C) m = n
Q9. Which of the given values of x and y make the following pair of matrices equal.
(A) x = -1/3, y = 7 (B) Not possible to find
(C) y = 7, x = -2/3 (D) x = -1/3, y = -2/3
Ans. Two matrices are equal,so their corresponding element should be equal to each other
On comparing LHS and RHS matrices, we get
3x + 7 =0⇒ x = -7/3, 2 – 3x = 4⇒ x = -2/3
Since,the value of x is not unique, therefore the answer is (B) Not possible to find
Q10.The number of all possible matrices of the order 3×3 with each entry of 0 or 1 is
(A) 27 (B) 18 (C) 81 (D) 512
Ans. Permutation of filling one element of the matrix = 2
Total number of elements = 3 × 3 = 9
Therefore the total number of permutations filling all the elements are = 2^{9}
= 2×2×2×2×2×2×2×2×2 = 512
Hence, the answer is (D) 512
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Chapter 1-Sets | Chapter 9-Sequences and Series |
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Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
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NCERT solutions for class 12 maths
Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
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