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# 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, a21, a33, 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 amn where 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 = [aij], whose elements are given by

(i)  aij= (i+j)²/2   (ii) aij= i/j   (iii) aij= (i+2j)²/2

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

(i) If elements are designed  aij= (i+j)²/2,then a11,, a12  , a21 , a22 are 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

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

(ii) If elements are designed  aij= i/j,then a11,, a12  , a21 , a22 are 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

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

(iii) If elements are designed  aij= (i+2j)²/2,then a11,, a12  , a21 , a22 are 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

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

Q5. Construct a 3 × 4 matrix ,whose elements are given by

(i) aij= (1/2)⌊-3i + j⌋  (ii)aij  =2i – j

(i) If elements are designed  aij=(1/2)(-3i + j),then a11,, a12  , a13 , a14 , a21 , a22  , a23, a24  ,a31, a32  , a33 , and a34 are 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

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

(ii) If elements are designed  aij  =2i – j ,then a11, a12  , a13 , a14 , a21 , a22  , a23, a24  ,a31, a32  , a33 , and a34 are 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

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

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

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

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

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

Years.

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

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

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

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 -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

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

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

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

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 = [aij]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.

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

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

(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 = 29

= 2×2×2×2×2×2×2×2×2 = 512

Hence, the answer is (D) 512

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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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