# NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.1 (Ex 3.1)

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## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.1 (Ex 3.1)

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There is a basic format in which matrices are solved, they are written in rows and columns in a format like this-

⎡ 1 9 -13 ⎤

⎣ 20 5 -6 ⎦

There are two rows and three columns of matrix. It can also be referred to as a two-by-three matrix. Square matrices are the most essential type of matrices and are also the most applied matrices. Square matrices have the same number of rows and columns. The determinant of a square matrix is a number associated with the matrix, which is the base for the study of a square matrix. Some students can find it difficult to understand the concepts and calculations of matrices. The NCERT Solutions Class 12 Maths Chapter 3 Exercise 3.1 provided by Extramarks makes it easier for students to understand the complicated calculations involved in Chapter 3 Matrices. Extramarks provides students with expert solutions and live problem-solving classes so that students can learn efficiently and score well in their board exams.

### NCERT Solutions Class 12 Maths Chapter 3 Exercise 3.1.

 Chapter 3 Matrix Exercises Exercise 3.2 22 Questions & Solutions (3 Short Answers, 19 Long Answers) Exercise 3.3 12 Questions & Solutions (4 Short Answers, 8 Long Answers) Exercise 3.4 18 Questions & Solutions (18 Short Answers)

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Q.1

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{matrix}\mathrm{A}=\left[\begin{array}{l}2 519-7\\ 35-2 \frac{5}{2}12\\ \sqrt{3} 1-517\end{array}\right], \mathrm{write}:\\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{order}\mathrm{of}\mathrm{the}\mathrm{matrix}\\ \left(\mathrm{ii}\right)\mathrm{Then}\mathrm{umber}\mathrm{of}\mathrm{elements},\\ \left(\mathrm{iii}\right)\mathrm{Write}\mathrm{the}\mathrm{elements}{\mathrm{a}}_{13},{\mathrm{a}}_{21},{\mathrm{a}}_{33},{\mathrm{a}}_{24},{\mathrm{a}}_{23}.\end{array}$

Ans

(i) The order of matrix A is 3×4.

(ii) The number of elements in matrix A is 12.

(iii) The value of a13 = 19, a21 = 35, a33 = – 5, a24 = 12, a23 = 5/2.

Q.2 If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Ans

Possible order of matrix

1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 12 × 2, 24 × 1, 8 × 3.
Possible order for 13 elements: 1 × 13, 13 × 1.

Q.3 If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Ans

Possible order for 18 elements

1 ×18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1

Possible order for 5 elements = 1 × 5, 5 × 1.

Q.4

$\begin{array}{l}\mathrm{Constructa}2×2\mathrm{matrix},\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right],\mathrm{whose}\mathrm{}\mathrm{elementsare}\\ \mathrm{givenby}:\mathrm{ }\left(\mathrm{i}\right)\mathrm{}{\mathrm{a}}_{\mathrm{ij}}=\frac{{\left(\mathrm{i}+\mathrm{j}\right)}^{2}}{2}\left(\mathrm{ii}\right){\mathrm{a}}_{\mathrm{ij}}=\frac{\mathrm{i}}{\mathrm{j}}\left(\mathrm{iii}\right){\mathrm{a}}_{\mathrm{ij}}=\frac{{\left(\mathrm{i}+2\mathrm{j}\right)}^{2}}{2}\end{array}$

Ans

$\begin{array}{l} \mathrm{ }\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\mathrm{\right]}\\ \left(\mathrm{i}\right){\mathrm{a}}_{\mathrm{ij}}=\frac{{\left(\mathrm{i}+\mathrm{j}\right)}^{2}}{2}\\ \mathrm{A}=\left(\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}\end{array}\right)=\left(\begin{array}{l}2\frac{9}{2}\\ \frac{9}{2}8\end{array}\right)\\ \left(\mathrm{ii}\right) {\mathrm{a}}_{\mathrm{ij}}=\frac{\mathrm{i}}{\mathrm{j}}\\ \mathrm{A}=\left(\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}\end{array}\right)=\left(\begin{array}{l}1\frac{1}{2}\\ 21\end{array}\right)\\ \left(\mathrm{iii}\right)\mathrm{ }{\mathrm{a}}_{\mathrm{ij}}=\frac{{\left(\mathrm{i}+2\mathrm{j}\right)}^{2}}{2}\\ \mathrm{A}=\left(\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}\end{array}\right)=\left(\begin{array}{l}\frac{9}{2}\frac{25}{2}\\ 818\end{array}\right)\end{array}$

Q.5

$\begin{array}{l}\mathbf{Construct}\mathrm{}\mathbf{a}\mathrm{}\mathbf{3}\mathrm{}×\mathrm{}\mathbf{4}\mathrm{}\mathbf{matrix},\mathrm{}\mathbf{whose}\mathrm{}\mathbf{elements}\mathrm{}\mathbf{are} \mathbf{given}\mathrm{}\mathbf{by}:\\ \left(\mathrm{i}\right)\mathrm{ }{\mathrm{a}}_{\mathrm{ij}}=\frac{1}{2}|-3\mathrm{i}+\mathrm{j}|\left(\mathrm{ii}\right){\mathrm{a}}_{\mathrm{ij}}=2\mathrm{i}-\mathrm{j}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{aij}=\frac{1}{2}|-3\mathrm{i}+\mathrm{j}|\\ \mathrm{A}=\left[\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}{\mathrm{a}}_{13}{\mathrm{a}}_{14}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}{\mathrm{a}}_{23}{\mathrm{a}}_{24}\\ {\mathrm{a}}_{31}{\mathrm{a}}_{32}{\mathrm{a}}_{33}{\mathrm{a}}_{34}\end{array}\right]\\ \mathrm{A}=\left[\begin{array}{l}1\frac{1}{2}0\frac{1}{2}\\ \frac{5}{2}2\frac{3}{2}1\\ 4\frac{7}{2}3\frac{5}{2}\end{array}\right]\\ \left(\mathrm{ii}\right)\mathrm{ }{\mathrm{a}}_{\mathrm{ij}}=2\mathrm{i}-\mathrm{j}\\ \mathrm{A}=\left[\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}{\mathrm{a}}_{13}{\mathrm{a}}_{14}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}{\mathrm{a}}_{23}{\mathrm{a}}_{24}\\ {\mathrm{a}}_{31}{\mathrm{a}}_{32}{\mathrm{a}}_{33}{\mathrm{a}}_{34}\end{array}\right]\\ =\left[\begin{array}{l}10-1-2\\ 32 \mathrm{ }1 \mathrm{ }0\\ 54 \mathrm{ }3 2\end{array}\right]\end{array}$

Q.6

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}\mathrm{from}\mathrm{the}\mathrm{following} \mathrm{equations}:\\ \left(\mathrm{i}\right) \left[\begin{array}{l}43\\ \mathrm{x}5\end{array}\right]=\left[\begin{array}{l}\mathrm{y}\mathrm{z}\\ 15\end{array}\right] \\ \left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{x}+\mathrm{y}3\\ 5+\mathrm{z}\mathrm{xy}\end{array}\right]=\left[\begin{array}{l}62\\ 58\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}\\ \mathrm{x}+\mathrm{z}\\ \mathrm{y}+\mathrm{z}\end{array}\right]=\left[\begin{array}{l}9\\ 5\\ 7\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right) \left[\begin{array}{l}\mathrm{4}\mathrm{3}\\ \mathrm{x}\mathrm{5}\end{array}\right]=\left[\begin{array}{l}\mathrm{y}\mathrm{z}\\ \mathrm{1}\mathrm{5}\end{array}\right]\\ ⇒\mathrm{x}=1,\mathrm{y}=4\mathrm{and}\mathrm{z}=\mathrm{3}\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{x}+\mathrm{y}\mathrm{3}\\ 5+\mathrm{z}\mathrm{xy}\end{array}\right]=\left[\begin{array}{l}6\mathrm{2}\\ \mathrm{5}\mathrm{8}\end{array}\right]\\ ⇒5+\mathrm{z}=\mathrm{5}⇒\mathrm{z}=5-5=0\\ \mathrm{ }\mathrm{x}+\mathrm{y}=6,\mathrm{and}\mathrm{xy}=8⇒\mathrm{y}=\frac{8}{\mathrm{x}}\\ ⇒\mathrm{ }\mathrm{x}+\mathrm{y}=6\\ \mathrm{x}+\frac{8}{\mathrm{x}}=6\\ \frac{{\mathrm{x}}^{2}+8}{\mathrm{x}}=6\\ ⇒{\mathrm{x}}^{2}+8=6\mathrm{x}\\ {\mathrm{x}}^{2}-6\mathrm{x}+8=0⇒{\mathrm{x}}^{2}-4\mathrm{x}-2\mathrm{x}+8=0\\ ⇒ \mathrm{ }\left(\mathrm{x}-4\right)\left(\mathrm{x}-2\right)=0\\ ⇒\mathrm{ }\mathrm{x}=2,4\\ \mathrm{If}\mathrm{x}=2\mathrm{then}\mathrm{y}=\frac{8}{2}=4\\ \mathrm{If} \mathrm{x}=4 \mathrm{then} \mathrm{y}=\frac{8}{4}=2\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}\\ \mathrm{x}+\mathrm{z}\\ \mathrm{y}+\mathrm{z}\end{array}\right]\mathrm{=}\left[\begin{array}{l}\mathrm{9}\\ \mathrm{5}\\ \mathrm{7}\end{array}\right]\\ \mathrm{x}+\mathrm{y}+\mathrm{z}=9...\left(\mathrm{i}\right)\\ \mathrm{x}+\mathrm{z}=5...\left(\mathrm{ii}\right)\\ \mathrm{y}+\mathrm{z}=7...\left(\mathrm{iii}\right)\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{have}\\ \mathrm{y}+5=9⇒\mathrm{y}=4\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{have}\\ \mathrm{x}+7=9⇒\mathrm{x}=2\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{have}\\ 2+4+\mathrm{z}=9⇒\mathrm{z}=9-6=3\\ \mathrm{Thus},\mathrm{ }\mathrm{x}=2,\mathrm{ }\mathrm{y}=4\mathrm{ }\mathrm{and} \mathrm{z}=3.\end{array}$

Q.7

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{ }\mathrm{a},\mathrm{b},\mathrm{c} \mathrm{and}\mathrm{ }\mathrm{d}\mathrm{from}\mathrm{thee}\mathrm{quation}:\\ \left[\begin{array}{l}\mathrm{a}–\mathrm{b}2\mathrm{a}+\mathrm{c}\\ 2\mathrm{a}–\mathrm{b}3\mathrm{c}+\mathrm{d}\end{array}\right]=\left[\begin{array}{l}-1 5\\ 013\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\left[\begin{array}{l}\mathrm{a}-\mathrm{b}2\mathrm{a}+\mathrm{c}\\ 2\mathrm{a}-\mathrm{b}3\mathrm{c}+\mathrm{d}\end{array}\right]=\left[\begin{array}{l}-1 5\\ 013\end{array}\right]\\ ⇒\mathrm{ }\mathrm{a}-\mathrm{b}=-1...\left(\mathrm{i}\right)\mathrm{ }\\ \mathrm{ }2\mathrm{a}+\mathrm{c}=5\mathrm{ }...\left(\mathrm{ii}\right)\\ \mathrm{ }2\mathrm{a}-\mathrm{b}=0\mathrm{ }...\left(\mathrm{iii}\right)\mathrm{ }\\ 3\mathrm{c}+\mathrm{d}=\mathrm{13}...\left(\mathrm{iv}\right)\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right) \mathrm{and}\mathrm{ }\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{have}\\ \mathrm{a}=1\mathrm{and}\mathrm{b}=\mathrm{2}\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \mathrm{2}\left(1\right)+\mathrm{c}=5⇒\mathrm{c}=5-2=3\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{c}\mathrm{in}\mathrm{equation}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\\ \mathrm{3}\left(3\right)+\mathrm{d}=13⇒\mathrm{d}=13-9=4\end{array}$

Q.8

$\begin{array}{l}\mathrm{A}={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{\mathrm{m}×\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{matrix},\mathrm{if}\\ \left(\mathrm{A}\right)\mathrm{m}<\mathrm{n}\\ \left(\mathrm{B}\right)\mathrm{m}>\mathrm{n}\\ \left(\mathrm{C}\right)\mathrm{m}=\mathrm{n}\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)\mathrm{None}\mathrm{of}\mathrm{these}\end{array}$

Ans

A is a square matrix if m=n.

So, option (C) is correction answer.

Q.9

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{make}\mathrm{the}\mathrm{following}\\ \mathrm{pair}\mathrm{of}\mathrm{matrice}\mathrm{sequal}\\ \left[\begin{array}{l}3\mathrm{x}+75\\ \mathrm{y}+12–3\mathrm{x}\end{array}\right],\mathrm{ }\left[\begin{array}{l}0\mathrm{y}–2\\ 84\end{array}\right]\\ \left(\mathrm{A}\right)\mathrm{x}=-\frac{1}{3}, \mathrm{y}=7\left(\mathrm{B}\right)\mathrm{Not}\mathrm{possiblet}\mathrm{of}\mathrm{ind}\\ \left(\mathrm{C}\right)\mathrm{y}=7,\mathrm{x}=\frac{-2}{3} \left(\mathrm{D}\right)\mathrm{x}=-\frac{1}{3},\mathrm{ }\mathrm{y}=\frac{-2}{3}\end{array}$

Ans

$\begin{array}{l}\mathrm{If}\mathrm{both}\mathrm{matrices}\mathrm{are}\mathrm{equal},\mathrm{then}\\ \left[\begin{array}{l}3\mathrm{x}+7\mathrm{5}\\ \mathrm{y}+12-3\mathrm{x}\end{array}\right]= \left[\begin{array}{l}0\mathrm{y}-2\\ 8\mathrm{4}\end{array}\right]\\ ⇒ \mathrm{ }3\mathrm{x}+7=0, \mathrm{ }\mathrm{y}-\mathrm{2}=5\\ \mathrm{and} \mathrm{y}+1=8, \mathrm{ }2-3\mathrm{x}=4\\ \mathrm{Then},\mathrm{ }\mathrm{x}=-\frac{7}{3},\mathrm{ }\mathrm{y}=7\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{x}=-\frac{2}{3}\\ \mathrm{Since},\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{different}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{so}\mathrm{it}\mathrm{is}\mathrm{difficult}\\ \mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{to}\mathrm{make}\mathrm{matrices}\mathrm{equal}\mathrm{.}\end{array}$

Q.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

Ans

Number of elements is 9 and each element can be filled by 0 or 1.

The number of all possible matrices of order 3 x 3
with each entry 0 or 1 = 29 = 512

So, the option ‘D’ is correct.

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