NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices - Future Study Point

# 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 a13, a21, a33, a24, a23

Ans.

(i) The order of matrix = Number of rows ร Number of column = 3 ร 4

(ii) The number of elements in matrix are = 3ร 4= 12

(iii) The number of elements in a matrix are represented as amn where m is number of row and n is number of column,so a13,=19, a21ย =35 , a33ย = – 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 ร 24, 2ร 12, 3 ร 8, 4 ร 6, 24 ร1,12 ร 2,8 ร3 and 6 ร4, if the matrix hasย  13 elements then the possible orders of the matrix are,1 ร 13 and 13 ร 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 ร 18, 2ร 9, 3 ร 6,ย  18 ร1,9 ร 2 and 6 ร3ย  if the matrix hasย  5 elements then the possible orders of the matrix are,1 ร 5 and 5 ร 1

Q4. Construct a 2 ร 2 matrix A = [aij], whose elements are given by

(i)ย  aij= (i+j)ยฒ/2ย  ย (ii) aij= i/jย  ย (iii) aij= (i+2j)ยฒ/2ย

Ans. The matrix to be constructed of the order 2 ร 2, so theย  elements ofย  matrix A = [aij] are determined as follows.

(i) If elements are designedย  aij= (i+j)ยฒ/2,then a11,, a12 ย , a21 , a22 are determined as follows

a11= (1+1)ยฒ/2 = 2ยฒ/2 = 2

a12=(1+2)ยฒ/2 = 3ยฒ/2 =9/2ย

a21 = (2+1)ยฒ/2 = 3ยฒ/2 =9/2

a22ย = (2+2)ยฒ/2 = 8

Hence, matrix Aย  will be as follows

(ii) If elements are designedย  aij= i/j,then a11,, a12 ย , a21 , a22 are determined as follows

a11= 1/1 =1

a12=1/2

a21 = 2/1 = 2

a22ย = 2/2 = 1

Hence, matrix Aย  will be as follows

(iii) If elements are designedย  aij= (i+2j)ยฒ/2,then a11,, a12 ย , a21 , a22 are determined as follows

a11= (1+2ร1)ยฒ/2 = 3ยฒ/2 = 9/2

a12=(1+2ร2)ยฒ/2 = 5ยฒ/2 =25/2ย

a21 = (2+2ร1)ยฒ/2 = 4ยฒ/2 =16/2 = 8

a22ย = (2+2ร2)ยฒ/2 = 36/2 = 18

Hence, matrix Aย  will be as follows

Q5. Construct a 3 ร 4 matrix ,whose elements are given byย

(i) aij= (1/2)โ-3i + jโย  (ii)aijย  =2i – j

(i) If elements are designedย  aij=(1/2)(-3i + j),then a11,, a12 ย , a13 , a14ย , a21 , a22 ย , a23, a24ย  ,a31, a32 ย , a33 , and a34ย are determined as follows

a11= (1/2)โ-3ร1+ 1โ = 1, a12=ย  (1/2)โ-3ร1 + 2โ = 1/2,ย a13 = (1/2)โ-3ร1 + 3โ=0

a14 = (1/2)โ-3ร1 + 4โ=1/2,ย a21 = (1/2)โ-3ร2 + 1โ = 5/2 ,ย a22ย = (1/2)โ-3ร2 + 2โ =2

a23 = (1/2)โ-3ร2+ 3โ= 3/2,ย a24 = (1/2)โ-3ร2 + 4โ = 1,ย a31 = (1/2)โ-3ร3 + 1โ= 4

a32 = (1/2)โ-3ร3 + 2โ= 7/2,ย a33 = (1/2)โ-3ร3 + 3โ=3,ย a34 = (1/2)โ-3ร3 + 4โ= 5/2

Hence, matrixย  ย will be as follows

(ii) If elements are designedย  aijย  =2i – j ,then a11, a12 ย , a13 , a14ย , a21 , a22 ย , a23, a24ย  ,a31, a32 ย , a33 , and a34ย are determined as follows

a11=2ร1 – 1ย = 1, a12= 2ร1 – 2 = 0, a13 = 2ร1- 3=-1

a14 =2ร1 – 4=-2, a21 = 2ร2 – 1 = 3 , a22ย = 2ร2 – 2ย =2

a23 =2ร2 – 3= 1, a24 =2ร2 – 4 = 0, a31 = 2ร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, matrixย  ย will be as follows

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

Years.ย

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 โ z = 0 …. (ii)

xy = 8…….(iii)

Substituting the value of y = 6- xย  fromย  equation (i) in equation (iii)

( 6- x)x = 8

-xยฒ +6x -8 = 0

-xยฒ +6x -8 = 0

xยฒ -6x +8 = 0

xยฒ -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

Puttingย  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 = [aij]mรnย  is a square matrix,ifย

(A)ย  m < nย  ย  ย  ย (B) m > nย  ย  (C) m = nย  ย (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 = 7ย  ย  ย  (B) Not possible to find

(C) y = 7, x = -2/3ย  ย  (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 =0โ x = -7/3, 2 – 3x = 4โ 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 3ร3 with each entry of 0 or 1 is

(A)ย  27ย  ย  ย  ย (B) 18ย  ย  ย  ย (C) 81ย  ย  ย  ย (D) 512

Ans. Permutation of filling one element of the matrix = 2

Total number ofย  elements = 3 ร 3 = 9

Therefore the total number of permutations filling all the elements are = 29

= 2ร2ร2ร2ร2ร2ร2ร2ร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

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 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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