NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 3 |
| Chapter Name | Matrices |
| Exercise | Ex 3.1 |
| Number of Questions Solved | 10 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1
Ex 3.1 Class 12 Maths Question 1.
In the matrix \(A=\left[ \begin{matrix} 2 \\ 35 \\ \sqrt { 3 } \end{matrix}\begin{matrix} 5 \\ -2 \\ 1 \end{matrix}\begin{matrix} 19 \\ 5/2 \\ -5 \end{matrix}\begin{matrix} -7 \\ 12 \\ 17 \end{matrix} \right] \)
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a13, a21, a33, a24, a23
Solution:
(i) The matrix A has three rows and 4 columns.
The order of the matrix is 3 x 4.
(ii) There are 3 x 4 = 12 elements in the matrix A
(iii) a13 = 19, a21 = 35, a33 = – 5, a24 = 12, a23 = \(\\ \frac { 5 }{ 2 } \)
Ex 3.1 Class 12 Maths Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
(i) 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6
Thus there are 8 matrices having 24 elements their order are (1 x 24), (24 x 1), (2 x 12), (12 x 2),(3 x 8), (8 x 3), (4 x 6), (6 x 4).
(ii) 13 = 1 x 13,
There are 2 matrices of 13 elements of order (1 x 13) and (13 x 1).
Ex 3.1 Class 12 Maths Question 3.
If a matrix has 18 elements, what are the possible orders it can have ? What, if it has 5 elements.
Solution:
We know that if a matrix is of order m × n, it has mn elements.
=> 18 = 1 x 18 = 2 x 9 = 3 x 6
Thus, all possible ordered pairs of the matrix
having 18 elements are:
(1,18), (18,1), (2,9), (9,2), (3,6), (6,3)
If it has 5 elements, then possible order are: (1,5), (5,1)
Ex 3.1 Class 12 Maths Question 4.
Construct a 2 x 2 matrix, A= [aij] whose elements are given by:
\((i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 } \)
\((ii)\quad { a }_{ ij }=\frac { i }{ j } \)
\((iii)\quad { a }_{ ij }=\frac { { (i+2j) }^{ 2 } }{ 2 } \)
Solution:
\(A={ \left[ { a }_{ ij } \right] }_{ 2\times 2 }=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } \\ { a }_{ 21 } & { a }_{ 22 } \end{bmatrix}\)
\((i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 } \)
Ex 3.1 Class 12 Maths Question 5.
Construct a 3 x 4 matrix , whose elements are given by:
\((i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right| \)
\((ii){ a }_{ ij }=2i-j\)
Solution:
\(A={ \left[ { a }_{ ij } \right] }_{ 3\times 4 }=\left[ \begin{matrix} { a }_{ 11 } \\ { a }_{ 21 } \\ { a }_{ 31 } \end{matrix}\begin{matrix} { a }_{ 12 } \\ { a }_{ 22 } \\ { a }_{ 32 } \end{matrix}\begin{matrix} { a }_{ 13 } \\ { a }_{ 23 } \\ { a }_{ 33 } \end{matrix}\begin{matrix} { a }_{ 14 } \\ { a }_{ 24 } \\ { a }_{ 34 } \end{matrix} \right] \)
\((i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right| \)
Ex 3.1 Class 12 Maths Question 6.
Find the values of x, y, z from the following equations:
\((i)\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}\)
\((ii)\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\)
\((iii)\left[ \begin{matrix} \begin{matrix} x+ & y+ & z \end{matrix} \\ \begin{matrix} x & +y \end{matrix} \\ \begin{matrix} y & +z \end{matrix} \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 5 \\ 7 \end{matrix} \right] \)
Solution:
\((i)\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}\)
Clearly x = 1,y = 4,z = 3
\((ii)\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\)
Now 5 + z = 5 => z = 0
Now x + y = 6 and xy = 8
∴ y = 6 – x and x(6 – x) = 8
6x – x² = 8
x² – 6x + 8 = 0
(x – 4)(x – 2) = 0
=>x = 2,4
When x = 2, y = 6 – 2 = 4
and when x = 4,y = 6 – 4 = 2
Hence x = 2,y = 4,z = 0 or x = 4,y = 2,z = 0.
(iii) Equating the corresponding elements.
=> x+y+z=9 …..(i)
x+z = 5 …(ii)
y+ z = 7 …(iii)
Adding eqs. (ii) & (iii)
x + y + 2z = 12
=> (x+y+z) + z = 12,
9+z = 12 (from equ (i))
z = 3
x + z = 5
=>x + 3 = 5 => x = 2
and y+z = 7
=>y+3 = 7
=> y = 4
=> x = 2, y = 4 and z = 3
Ex 3.1 Class 12 Maths Question 7.
Find the values of a,b,c and d from the equation:
\(\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\)
Solution:
\(\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\)
Ex 3.1 Class 12 Maths Question 8.
A = [aij]m×n is a square matrix, if
(a) m < n (b) n > n
(c) m = n
(d) none of these
Solution:
For a square matrix m=n.
Thus option (c) m = n, is correct.
Ex 3.1 Class 12 Maths Question 9.
Which of the given values of x and y make the following pairs of matrices equal:
\(\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix},\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix} \)
(a) \(x=\frac { -1 }{ 3 } ,y=7\)
(b) Not possible to find
(c) \(y=7,x=\frac { -2 }{ 3 } \)
(d) \(x=\frac { -1 }{ 3 } ,y=\frac { -2 }{ 3 } \)
Solution:
\(\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix},\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix} \)
(a) \(x=\frac { -1 }{ 3 } ,y=7\)
Ex 3.1 Class 12 Maths Question 10.
The number of all possible matrices of order 3×3 with each entry 0 or 1 is
(a) 27
(b) 18
(c) 81
(d) 512
Solution:
There are 3 x 3 matrix or 9 entries in matrix each place can be filled with 0 or 1
∴ 9 Places can be filled in 29 = 512 ways
Number of such matrices = 512
Option (d) is correct.
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