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Examples (Revised) - Chapter 3 Matrices class 12 ncert solutions Maths - SaraNextGen [2024-2025]

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NCERT Solutions Class 12 Maths: Chapter 3 - Matrices | Comprehensive Guide

Example 1

Consider the following information regarding the number of men and women workers in three factories I, II and III

Represent the above information in the form of a $3 \times 2$ matrix. What does the entry in the third row and second column represent?

Solution

The information is represented in the form of a 3 x 2 matrix as follows:

The entry in the third row and second column represents the number of women workers in factory III.

Example 2

If a matrix has 8 elements, what are the possible orders it can have?
Solution

We know that if a matrix is of order $m \times n$, it has $m n$ elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8 .
Thus, all possible ordered pairs are $(1,8),(8,1),(4,2),(2,4)$
Hence, possible orders are $1 \times 8,8 \times 1,4 \times 2,2 \times 4$
Example 3

Construct a $3 \times 2$ matrix whose elements are given by $a_{i j}=\frac{1}{2}|i-3 j|$.

Solution

In general a 3 x 2 matrix is given by

Now $\quad a_{i j}=\frac{1}{2}|i-3 j|, i=1,2,3$ and $j=1,2$.
Therefore $\quad a_{11}=\frac{1}{2}|1-3 \times 1|=1 \quad a_{12}=\frac{1}{2}|1-3 \times 2|=\frac{5}{2}$
$
\begin{array}{ll}
a_{21}=\frac{1}{2}|2-3 \times 1|=\frac{1}{2} & a_{22}=\frac{1}{2}|2-3 \times 2|=2 \\
a_{31}=\frac{1}{2}|3-3 \times 1|=0 & a_{32}=\frac{1}{2}|3-3 \times 2|=\frac{3}{2}
\end{array}
$

Hence the required matrix is given by

Example 4

Find the values of $a, b, c, x, y$ and $z$.
Solution

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

Simplifying, we get
$
a=-2, b=-7, c=-1, x=-3, y=-5, z=2
$

Example 5

Find the values of $a, b, c$, and $d$ from the following equation:

Solution

By equality of two matrices, equating the corresponding elements, we get

Solving these equations, we get
$
a=1, b=2, c=3 \text { and } d=4
$

Example 6

Given and , find $A+B$
Since A, B are of the same order $2 \times 3$. Therefore, addition of A and B is defined and is given by

Example 7

If and , then find $2 A-B$.
Solution

We have

Example 8

If and , then find the matrix $X$, such that $2 \mathrm{~A}+3 \mathrm{X}=5 \mathrm{~B}$.

Solution

We have $2 A+3 X=5 B$

Example 9

Find Xand Y, if

Solution

We have

Example 10

Find the values of x and y from the following equation:

Solution

We have

Example 11

Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices $\mathrm{A}$ and $\mathrm{B}$.

(i) Find the combined sales in September and October for each farmer in each variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive $2 \%$ profit on gross sales, compute the profit for each farmer and for each variety sold in October.

Solution
(i) Combined sales in September and October for each farmer in each variety is given by

(ii) Change in sales from September to October is given by

$\text { (iii) } 2 \% \text { of } B=\frac{2}{100} \times B=0.02 \times B$

Thus, in October Ramkishan receives Rs.100 , Rs.200 and Rs.120 as profit in the sale of each variety of rice, respectively, and Grucharan Singh receives profit of Rs.400, Rs.200 and Rs.200 in the sale of each variety of rice, respectively.

Example 12

Find A B, if

Solution

The matrix A has 2 columns which is equal to the number of rows of $B$. Hence $A B$ is defined. Now

Remark If $\mathrm{AB}$ is defined, then $\mathrm{BA}$ need not be defined. In the above example, $\mathrm{AB}$ is defined but $\mathrm{BA}$ is not defined because $\mathrm{B}$ has 3 column while $\mathrm{A}$ has only 2 (and not 3 ) rows. If $\mathrm{A}, \mathrm{B}$ are, respectively $m \times n, k \times l$ matrices, then both $\mathrm{AB}$ and $\mathrm{BA}$ are defined if and only if $n=k$ and $l=m$. In particular, if both $\mathrm{A}$ and $\mathrm{B}$ are square matrices of the same order, then both $\mathrm{AB}$ and $\mathrm{BA}$ are defined.
Non-commutativity of multiplication of matrices
Now, we shall see by an example that even if $\mathrm{AB}$ and $\mathrm{BA}$ are both defined, it is not necessary that $A B=B A$.

Example13

If then find A B, B A. Show that

$
\mathrm{AB} \neq \mathrm{BA} .
$

Since A is a $2 \times 3$ matrix and B is $3 \times 2$ matrix. Hence AB and BA are both defined and are matrices of order $2 \times 2$ and $3 \times 3$, respectively. Note that.

Clearly $\mathrm{AB} \neq \mathrm{BA}$
In the above example both $\mathrm{AB}$ and $\mathrm{BA}$ are of different order and so $\mathrm{AB} \neq \mathrm{BA}$. But one may think that perhaps $A B$ and $B A$ could be the same if they were of the same order. But it is not so, here we give an example to show that even if $\mathrm{AB}$ and $\mathrm{BA}$ are of same order they may not be same.

Example 14

If then and $\text { Clearly } \mathrm{AB} \neq \mathrm{BA} \text {. }$

Thus matrix multiplication is not commutative.

Example15

Find A B, if

Solution

We have

Thus, if the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix.

Example 16

If find $A(B C),(A B) C \text { and show that }(A B) C=A(B C) \text {. }$

Solution

We have

Example 17

$\text { Calculate } \mathrm{AC}, \mathrm{BC} \text { and }(\mathrm{A}+\mathrm{B}) \mathrm{C} \text {. Also, verify that }(\mathrm{A}+\mathrm{B}) \mathrm{C}=\mathrm{AC}+\mathrm{BC}$

Solution:

Clearly,
$
(A+B) C=A C+B C
$

Example 18

If $\text { then show that } \mathrm{A}^3-23 \mathrm{~A}-40 \mathrm{I}=\mathrm{O}$

Solution

We have

so,

Now,

Example 19

In a legislative assembly election, a political group hired a public relations firm to promote its candidate in three ways: telephone, house calls, and letters. The cost per contact (in paise) is given in matrix $A$ as

$\text { The number of contacts of each type made in two cities } \mathrm{X} \text { and } \mathrm{Y} \text { is given by }$

Find the total amount spent by the group in the two cities X and Y.

Solution

We have

So the total amount spent by the group in the two cities is 340,000 paise and 720,000 paise, i.e., Rs.3400 and Rs.7200, respectively.

Example 20

If and verify that

(i) $\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}$,
(ii) $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$,
(iii) $(k \mathrm{~B})^{\prime}=k \mathrm{~B}^{\prime}$, where $k$ is any constant.

Solution
(i) We have

Thus $\quad\left(A^{\prime}\right)^{\prime}=A$
(ii) We have

Therefore,

Now,

So,

Thus
$
(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}
$
(iii) We have

Then,

Thus,

$(k \mathrm{~B})^{\prime}=k \mathrm{~B}^{\prime}$

Example 21

If $\text { verify that }(A B)^{\prime}=B^{\prime} A^{\prime} \text {. }$

Solution

We have

then,

Now,

Clearly
$
(A B)^{\prime}=B^{\prime} A^{\prime}
$

Example 22

Express the matrix as the sum of a symmetric and a skew symmetric matrix.
Solution

Here

Let,

$\text { Thus } \quad P=\frac{1}{2}\left(B+B^{\prime}\right) \text { is a symmetric matrix. }$

Also, let

Then

$\text { Thus } \quad Q=\frac{1}{2}\left(B-B^{\prime}\right) \text { is a skew symmetric matrix. }$

Now,

Thus, B is represented as the sum of a symmetric and a skew symmetric matrix.

Example 23

If then prove that

Solution

We shall prove the result by using principle of mathematical induction.

We have

Therefore, $\quad$ the result is true for $n=1$.
Let the result be true for $n=k$. So

Now, we prove that the result holds for n=k+1

$\text { Therefore, the result is true for } n=k+1 \text {. Thus by principle of mathematical induction, }$ We have,

holds for all natural numbers.

Example 24

If $A$ and $B$ are symmetric matrices of the same order, then show that $A B$ is symmetric if and only if $\mathrm{A}$ and $\mathrm{B}$ commute, that is $\mathrm{AB}=\mathrm{BA}$.

Solution

Since A and B are both symmetric matrices, therefore $A^{\prime}=A$ and $B^{\prime}=B$.

Conversely, if $\mathrm{AB}=\mathrm{BA}$, then we shall show that $\mathrm{AB}$ is symmetric.
Now
$
\begin{aligned}
(\mathrm{AB})^{\prime} & =\mathrm{B}^{\prime} \mathrm{A}^{\prime} \\
& =\mathrm{BA} \text { (as A and B are symmetric) } \\
& =\mathrm{AB}
\end{aligned}
$

Hence $A B$ is symmetric.

Example 25

Let $\text { Find a matrix } D \text { such that }$