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

$A=\begin{bmatrix}2 &5 & 19 & -7\\ 35 & -2 &5/2 & 12\\ \sqrt{3} & 1& -5 & 17 \end{bmatrix} ,Write$

(i) The order of matrix

(ii) The number of elements

(iii) Write the elements a₁₃, a_21,a_33, a₂₄, a₂₃

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_mnwhere m is number of row and n is number of column,so a₁₃,=19, a₂₁=35 , a₃₃= – 5, a₂₄=12, a₂₃= 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 a₁₁,, a₁₂ , a₂₁, a₂₂are determined as follows

a₁₁= (1+1)²/2 = 2²/2 = 2

a₁₂=(1+2)²/2 = 3²/2 =9/2

a₂₁= (2+1)²/2 = 3²/2 =9/2

a₂₂= (2+2)²/2 = 8

Hence, matrix A will be as follows

$A=\begin{bmatrix}2 & 9/2\\ 9/2 & 8 \end{bmatrix}$

(ii) If elements are designed a_ij= i/j,then a₁₁,, a₁₂ , a₂₁, a₂₂are determined as follows

a₁₁= 1/1 =1

a₁₂=1/2

a₂₁= 2/1 = 2

a₂₂= 2/2 = 1

Hence, matrix A will be as follows

$A=\begin{bmatrix}1 & 1/2\\ 2 & 1 \end{bmatrix}$

(iii) If elements are designed a_ij= (i+2j)²/2,then a₁₁,, a₁₂ , a₂₁, a₂₂are determined as follows

a₁₁= (1+2×1)²/2 = 3²/2 = 9/2

a₁₂=(1+2×2)²/2 = 5²/2 =25/2

a₂₁= (2+2×1)²/2 = 4²/2 =16/2 = 8

a₂₂= (2+2×2)²/2 = 36/2 = 18

Hence, matrix A will be as follows

$A=\begin{bmatrix}9/2 & 25/2\\ 8 & 18 \end{bmatrix}$

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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 a₁₁,, a₁₂ , a₁₃ , a₁₄_,a₂₁, a₂₂ , a₂₃, a₂₄ ,a₃₁, a₃₂ , a₃₃ , and a₃₄are determined as follows

a₁₁= (1/2)⌊-3×1+ 1⌋ = 1, a₁₂= (1/2)⌊-3×1 + 2⌋ = 1/2, a₁₃= (1/2)⌊-3×1 + 3⌋=0

a₁₄= (1/2)⌊-3×1 + 4⌋=1/2, a₂₁= (1/2)⌊-3×2 + 1⌋ = 5/2 , a₂₂= (1/2)⌊-3×2 + 2⌋ =2

a₂₃= (1/2)⌊-3×2+ 3⌋= 3/2, a₂₄= (1/2)⌊-3×2 + 4⌋ = 1, a₃₁= (1/2)⌊-3×3 + 1⌋= 4

a₃₂= (1/2)⌊-3×3 + 2⌋= 7/2, a₃₃= (1/2)⌊-3×3 + 3⌋=3, a₃₄= (1/2)⌊-3×3 + 4⌋= 5/2

Hence, matrix will be as follows

$\begin{bmatrix}1 &1/2 &0 & 1/2\\ 5/2 &2 & 3/2 & 1\\ 4 &7/2 & 3 & 5/2 \end{bmatrix}$

(ii) If elements are designed a_ij =2i – j ,then a₁₁, a₁₂ , a₁₃ , a₁₄_,a₂₁, a₂₂ , a₂₃, a₂₄ ,a₃₁, a₃₂ , a₃₃ , and a₃₄are determined as follows

a₁₁=2×1 – 1 = 1, a₁₂= 2×1 – 2 = 0, a₁₃= 2×1- 3=-1

a₁₄=2×1 – 4=-2, a₂₁= 2×2 – 1 = 3 , a₂₂= 2×2 – 2 =2

a₂₃=2×2 – 3= 1, a₂₄=2×2 – 4 = 0, a₃₁= 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

$\begin{bmatrix}1 &0 &-1 & -2\\ 3 &2 & 1 & 0\\ 5 &4 & 3 & 2 \end{bmatrix}$

Q6. Find the values of x, y and z from the following equations

$\left ( i \right ).\begin{bmatrix}4 & 3\\ x& 5 \end{bmatrix}=\begin{bmatrix}y & 2\\ 1& 5 \end{bmatrix}$

$\left ( ii \right ).\begin{bmatrix}x+y & 2\\ 5+z& xy \end{bmatrix}=\begin{bmatrix}6 & 2\\ 5& 8 \end{bmatrix}$

$\left ( iii \right ).\begin{bmatrix}x+y+z \\ x+z \\ y+z \end{bmatrix}=\begin{bmatrix}9 \\ 5 \\7 \end{bmatrix}$

Years.

$\left ( i \right ).\begin{bmatrix}4 & 3\\ x& 5 \end{bmatrix}=\begin{bmatrix}y & 2\\ 1& 5 \end{bmatrix}$

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

$\left ( ii \right ).\begin{bmatrix}x+y & 2\\ 5+z& xy \end{bmatrix}=\begin{bmatrix}6 & 2\\ 5& 8 \end{bmatrix}$

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 -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

$\left ( iii \right ).\begin{bmatrix}x+y+z \\ x+z \\ y+z \end{bmatrix}=\begin{bmatrix}9 \\ 5 \\7 \end{bmatrix}$

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

$\begin{bmatrix}a-b & 2a+c\\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix}-1 & 5\\ 0& 13 \end{bmatrix}$

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×nis 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.

$\begin{bmatrix}3x+7 & 5\\ y+1& 2-3x \end{bmatrix},\begin{bmatrix}0 &y-2 \\ 8& 4 \end{bmatrix}$

(A) x = -1/3, y = 7 (B) Not possible to find

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 = 2⁹

= 2×2×2×2×2×2×2×2×2 = 512

Hence, the answer is (D) 512

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

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