2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.1

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Karnataka 2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.1

2nd PUC Maths Matrices NCERT Text Book Questions and Answers
Ex 3.1

Question 1.
In the matrix
\(\mathbf{A}=\left[\begin{array}{rrrr}{2} & {5} & {19} & {-7} \\{35} & {-2} & {\frac{5}{2}} & {12} \\{\sqrt{3}} & {1} & {-5} & {17}\end{array}\right]\)
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements a13, a21, a33, a24 ,a23,
Answer:
(i) The order of the matrix is 3 x 4
(ii) Number of elements = 12
(iii) a13= 19
a21 = 35
a33 = -5
a24 = 12
a23 = \(\frac{5}{2}\)

Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Answer:
1 x 24; 24 x 1; 2 x 12;  12 x 2;
3 x 8; 8 x 3; 4 x 6; 6 x 4
13 elements
1 x 13;   13 x 1

KSEEB Solutions

Question 3.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Answer:
18 elements
1 x 18;      18 x 1;      2 x 9;      9 x 2;       3 x 6;  6 x 3
5 elements
1 x 5;    5 x 1;

Question 4.
Constract a 2 x 2 matrix A =[aij] ,whose element are given by:
\(a_{u}=\frac{(i+j)^{2}}{2}\)
\(a_{i j}=\frac{i}{j}\)
\(a_{i j}=\frac{(i+2 j)^{2}}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.1 1
2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.1 2

KSEEB Solutions

Question 5.
Construct a 3 x 4 matrix, whose elements are given by:
(i) \(a_{i j}=\frac{1}{2}|-3 i+j| \)
(ii) \(a_{i j}=2 i-j\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.1 3

Question 6.
Find the values of x, y and z from the following equations:
\(\text { (i) }\left[\begin{array}{ll}{4} & {3} \\{x} & {5}\end{array}\right]=\left[\begin{array}{ll}{y} & {z} \\{1} & {5}\end{array}\right]\)
Answer:
x = 1, y = 4, z = 3

(ii) \(\left[\begin{array}{cc}{x+y} & {2} \\{5+z} & {xy}\end{array}\right]=\left[\begin{array}{cc}{6} & {2} \\{5} & {8}\end{array}\right]\)
Answer:
5 + z = 5 = \(\succ \)– z = 0
x + y = 6 = \(\succ \) x = 2
xy = 8 = \(\succ \) x = 4
y = 4
4 = 2

KSEEB Solutions

(iii) \(\left[\begin{array}{r}{\mathbf{x}+\mathbf{y}+\mathbf{z}} \\{\mathbf{x}+\mathbf{z}} \\{\mathbf{y}+\mathbf{z}}\end{array}\right]=\left[\begin{array}{l}{\mathbf{9}} \\{\mathbf{5}} \\{\mathbf{7}}\end{array}\right]\)
Answer:
a – b = -1
2a – b = 0
a = 1
∴ 2a – b = 0
⇒ b = 2
2a + c = 5
c = 5 – 2 = 3 ∴ c = 3
d= 13-9 = 4
a = 1, b = 2 ,c = 3, d = 4

Question 8.
A = [aij]m x n\ is a square matrix, if
(A) m < n
(B) m>n
(C) m = n
(D) None of these
Answer:
(C) m = n as in a square matrix, the number of rows = number of colums.

KSEEB Solutions

Question 9.
Which of the given values of x and y make the following pair of matrices equal
\(\left[\begin{array}{cc}{3 x+7} & {5} \\{y+1} & {2-3 x}\end{array}\right],\left[\begin{array}{cc}{0} & {y-2} \\{8} & {4}\end{array}\right]\)
(A) \(x=\frac{-1}{3}, y=7\)
(B) Not possible to find
(C) \(y=7, x=\frac{-2}{3}\)
(D) \(x=\frac{-1}{3}, y=\frac{-2}{3}\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.1 4
since x has 2 values, matrices are not equal. Hence, it is not possible to find the value. a option (B). is correct.

Question 10.
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512
Answer:
There are 9 elements in the matrix. Each elements can be placed in 2 ways is all posible matrices of order 3 x 3 is 29 = 512 a option (D) is correct.

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