Class 12 Maths NCERT Solutions Chapter 3 Ex 3.1 – Matrices

The matrix A has 3 rows and 4 columns.

(i) Order of the matrix = 3 \times 4.

(ii) Number of elements = 3 \times 4 = 12.

(iii) Reading off the entries directly from A:

a_{13} = 19,\quad a_{21} = 35,\quad a_{33} = -5,\quad a_{24} = 12,\quad a_{23} = \dfrac{5}{2}

A matrix of order m \times n has mn elements, so we need all ordered pairs of natural numbers whose product is the given number of elements.

For 24 elements: the ordered pairs (m,n) with mn = 24 are (1,24),\,(24,1),\,(2,12),\,(12,2),\,(3,8),\,(8,3),\,(4,6),\,(6,4).

So the possible orders are 1\times24,\ 24\times1,\ 2\times12,\ 12\times2,\ 3\times8,\ 8\times3,\ 4\times6,\ 6\times4.

For 13 elements: since 13 is prime, the only ordered pairs are (1,13) and (13,1).

For 18 elements: the ordered pairs (m,n) with mn = 18 are (1,18),\,(18,1),\,(2,9),\,(9,2),\,(3,6),\,(6,3).

So the possible orders are 1\times18,\ 18\times1,\ 2\times9,\ 9\times2,\ 3\times6,\ 6\times3.

For 5 elements: since 5 is prime, the only ordered pairs are (1,5) and (5,1).

In general a 2\times2 matrix is A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, with i=1,2 and j=1,2.

(i) a_{ij} = \dfrac{(i+j)^2}{2}:

a_{11}=\dfrac{(1+1)^2}{2}=2,\quad a_{12}=\dfrac{(1+2)^2}{2}=\dfrac{9}{2}

a_{21}=\dfrac{(2+1)^2}{2}=\dfrac{9}{2},\quad a_{22}=\dfrac{(2+2)^2}{2}=8

(ii) a_{ij} = \dfrac{i}{j}:

a_{11}=1,\quad a_{12}=\dfrac{1}{2},\quad a_{21}=2,\quad a_{22}=1

(iii) a_{ij} = \dfrac{(i+2j)^2}{2}:

a_{11}=\dfrac{(1+2)^2}{2}=\dfrac{9}{2},\quad a_{12}=\dfrac{(1+4)^2}{2}=\dfrac{25}{2}

a_{21}=\dfrac{(2+2)^2}{2}=8,\quad a_{22}=\dfrac{(2+4)^2}{2}=18

Here i = 1,2,3 and j = 1,2,3,4.

(i) a_{ij} = \dfrac{1}{2}|-3i+j|:

a_{11}=\dfrac{1}{2}|-3+1|=1,\ \ a_{12}=\dfrac{1}{2}|-3+2|=\dfrac{1}{2},\ \ a_{13}=\dfrac{1}{2}|-3+3|=0,\ \ a_{14}=\dfrac{1}{2}|-3+4|=\dfrac{1}{2}

a_{21}=\dfrac{1}{2}|-6+1|=\dfrac{5}{2},\ \ a_{22}=\dfrac{1}{2}|-6+2|=2,\ \ a_{23}=\dfrac{1}{2}|-6+3|=\dfrac{3}{2},\ \ a_{24}=\dfrac{1}{2}|-6+4|=1

a_{31}=\dfrac{1}{2}|-9+1|=4,\ \ a_{32}=\dfrac{1}{2}|-9+2|=\dfrac{7}{2},\ \ a_{33}=\dfrac{1}{2}|-9+3|=3,\ \ a_{34}=\dfrac{1}{2}|-9+4|=\dfrac{5}{2}

(ii) a_{ij} = 2i - j:

a_{11}=1,\ a_{12}=0,\ a_{13}=-1,\ a_{14}=-2

a_{21}=3,\ a_{22}=2,\ a_{23}=1,\ a_{24}=0

a_{31}=5,\ a_{32}=4,\ a_{33}=3,\ a_{34}=2

Two matrices are equal only if they have the same order and every pair of corresponding entries is equal.

(i) Comparing corresponding entries: y = 4,\quad z = 3,\quad x = 1.

(ii) Comparing corresponding entries:

x + y = 6  ...(1),   5 + z = 5 \Rightarrow z = 0,   xy = 8  ...(2)

From (1), y = 6 - x; substituting in (2): x(6-x) = 8 \Rightarrow x^2 - 6x + 8 = 0 \Rightarrow (x-2)(x-4)=0.

Taking x = 2 gives y = 4 (the pair x=4,\,y=2 also satisfies (1) and (2), but only x=2,\,y=4 is taken as the standard solution here).

(iii) Comparing corresponding entries:

x + y + z = 9  ...(1),   x + z = 5  ...(2),   y + z = 7  ...(3)

From (2), x = 5-z; from (3), y = 7-z. Substituting in (1): (5-z)+(7-z)+z = 9 \Rightarrow 12 - z = 9 \Rightarrow z = 3.

Then x = 5 - 3 = 2 and y = 7 - 3 = 4.

Comparing corresponding entries of the two equal matrices:

a - b = -1  ...(1),   2a + c = 5  ...(2),   2a - b = 0  ...(3),   3c + d = 13  ...(4)

From (1): b = a+1. Substituting in (3): 2a - (a+1) = 0 \Rightarrow a - 1 = 0 \Rightarrow a = 1.

So b = a+1 = 2. From (2): 2(1) + c = 5 \Rightarrow c = 3. From (4): 3(3) + d = 13 \Rightarrow d = 4.

By definition, a matrix is a square matrix precisely when its number of rows equals its number of columns, that is m = n.

Comparing corresponding entries: 3x+7=0  ...(1),   5=y-2  ...(2),   y+1=8  ...(3),   2-3x=4  ...(4)

From (2) and (3), both give y = 7, which is consistent.

But from (1), 3x+7=0 \Rightarrow x = -\dfrac{7}{3}, while from (4), 2-3x=4 \Rightarrow x = -\dfrac{2}{3}.

Since these two equations demand different values of x at the same time, no single value of x can satisfy both entries. Hence the two matrices can never be made equal.

A 3 \times 3 matrix has 3 \times 3 = 9 entries, and each entry can independently be chosen in 2 ways (0 or 1).

So the total number of such matrices is 2^9 = 512.