Anand Classes brings you the most reliable and exam-oriented Matrices NCERT Solutions for Exercise 3.1 of Class 12 Chapter 3, designed to help students strengthen their conceptual understanding and score higher in board examinations. These solutions follow the latest NCERT guidelines and offer step-by-step explanations to simplify matrix operations, properties, and fundamental concepts. Perfect for CBSE students, these notes ensure clarity, accuracy, and quick revision before exams. Click the print button to download study material and notes.
Math NCERT Solutions for Exercise 3.1 of Class 12 Chapter 3 Matrices
NCERT Question.1 : In the matrix
$$
A = \begin{bmatrix}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{bmatrix}
$$
Write
(i) The order of the matrix $\;$ (ii) The number of elements$\;$ (iii) Write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$.
Solution:
(i) The order of the matrix
We can see that matrix contains $3$ rows and $4$ columns.
So, the order of this matrix is
$$3 \times 4$$
(ii) The number of elements
We know that the number of elements in a matrix is equal to:
$$\text{Number of rows} \times \text{Number of columns}$$
Therefore, number of elements in a matrix is
$$3 \times 4 = 12$$
(iii) Write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$
- $a_{13}$ = Element in first row and third column = $19$
- $a_{21}$ = Element in second row and first column = $35$
- $a_{33}$ = Element in third row and third column = $-5$
- $a_{24}$ = Element in second row and fourth column = $12$
- $a_{23}$ = Element in second row and third column = $\frac{5}{2}$
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NCERT Question.2 : If a matrix has $24$ elements, what are the possible orders it can have? What, if it has $13$ elements?
Solution:
We know that number of elements in a matrix is the product of number of rows and number of columns.
If a matrix has order $m \times n$, then the number of elements is:
$$mn$$
For 24 elements:
We need to find ordered pairs of natural numbers whose product is $24$:
$$(1, 24),(24, 1),(2, 12),(12, 2),(3, 8),(8, 3),(4, 6),(6, 4)$$
Hence, the possible orders are:
$$1 \times 24, 24 \times 1, 2 \times 12, 12 \times 2, 3 \times 8, 8 \times 3, 4 \times 6, 6 \times 4$$
For 13 elements:
$13$ is a prime number, so the only ordered pairs are:
$$(1, 13),(13, 1)$$
Hence, the possible orders are:
$$1 \times 13, 13 \times 1$$
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NCERT 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 the number of elements in a matrix is the product of number of rows and number of columns.
If a matrix has order $m \times n$, then:
$$mn = \text{number of elements}$$
For 18 elements
We find ordered pairs of natural numbers whose product is $18$:
$$(1, 18),(18, 1),(2, 9),(9, 2),(3, 6),(6, 3)$$
Hence, possible orders are:
$$1 \times 18, 18 \times 1, 2 \times 9, 9 \times 2, 3 \times 6, 6 \times 3$$
For 5 elements
$5$ is a prime number, so ordered pairs are:
$$(1, 5),(5, 1)$$
Hence, possible orders are:
$$1 \times 5, 5 \times 1$$
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NCERT Question.4 : Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by:
$\\[0,5em](i) a_{ij} = \dfrac{(i+j)^2}{2}\\[0,5em]$
(ii) $a_{ij} = \dfrac{i}{j}\\[0,5em]$
(iii) $a_{ij} = \dfrac{(i + 2j)^2}{2}$
Solution :
(i) $a_{ij} = \dfrac{(i+j)^2}{2}$
Elements in this $2 \times 2$ matrix
- $a_{11} = \dfrac{(1+1)^2}{2} = \dfrac{4}{2} = 2\\[1em]$
- $a_{12} = \dfrac{(1+2)^2}{2} = \dfrac{9}{2}\\[1em]$
- $a_{21} = \dfrac{(2+1)^2}{2} = \dfrac{9}{2}\\[1em]$
- $a_{22} = \dfrac{(2+2)^2}{2} = \dfrac{16}{2} = 8$
Resultant matrix:
$$
A = \begin{bmatrix}
2 & \dfrac{9}{2} \\
\dfrac{9}{2} & 8
\end{bmatrix}
$$
(ii) $a_{ij} = \dfrac{i}{j}$
Elements:
- $a_{11} = \dfrac{1}{1} = 1\\[1em]$
- $a_{12} = \dfrac{1}{2}\\[1em]$
- $a_{21} = \dfrac{2}{1} = 2\\[1em]$
- $a_{22} = \dfrac{2}{2} = 1\\[1em]$
Resultant matrix:
$$
A = \begin{bmatrix}
1 & \dfrac{1}{2} \\
2 & 1
\end{bmatrix}
$$
(iii) $a_{ij} = \dfrac{(i + 2j)^2}{2}$
Elements:
- $a_{11} = \dfrac{(1+2\cdot1)^2}{2} = \dfrac{9}{2}\\[1em]$
- $a_{12} = \dfrac{(1+2\cdot2)^2}{2} = \dfrac{25}{2}\\[1em]$
- $a_{21} = \dfrac{(2+2\cdot1)^2}{2} = \dfrac{16}{2} = 8\\[1em]$
- $a_{22} = \dfrac{(2+2\cdot2)^2}{2} = \dfrac{36}{2} = 18\\[1em]$
Resultant matrix:
$$
A = \begin{bmatrix}
\dfrac{9}{2} & \dfrac{25}{2} \\
8 & 18
\end{bmatrix}
$$
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NCERT Question.5 : Construct a $3 \times 4$ matrix whose elements are given by:
$\\[1em](i) a_{ij} = \dfrac{1}{2}\bigl| -3i + j \bigr|\\[1em]$
(ii) $a_{ij} = 2i – j$
Solution :
We compute each entry $\{a_{ij}\}$ for $(i=1,2,3)$ and $(j=1,2,3,4)$.
Row $(i=1)$:
\begin{aligned}
a_{11} &= \dfrac{1}{2}\bigl| -3(1) + 1 \bigr| = \dfrac{1}{2}\lvert -2\rvert = \dfrac{1}{2}\times 2 = 1,\\[1em]
a_{12} &= \dfrac{1}{2}\bigl| -3(1) + 2 \bigr| = \dfrac{1}{2}\lvert -1\rvert = \dfrac{1}{2}\times 1 = \dfrac{1}{2},\\[1em]
a_{13} &= \dfrac{1}{2}\bigl| -3(1) + 3 \bigr| = \dfrac{1}{2}\lvert 0\rvert = 0,\\[1em]
a_{14} &= \dfrac{1}{2}\bigl| -3(1) + 4 \bigr| = \dfrac{1}{2}\lvert 1\rvert = \dfrac{1}{2}.
\end{aligned}
Row $(i=2)$:
\begin{aligned}
a_{21} &= \dfrac{1}{2}\bigl| -3(2) + 1 \bigr| = \dfrac{1}{2}\lvert -5\rvert = \dfrac{5}{2},\\[1em]
a_{22} &= \dfrac{1}{2}\bigl| -3(2) + 2 \bigr| = \dfrac{1}{2}\lvert -4\rvert = 2,\\[1em]
a_{23} &= \dfrac{1}{2}\bigl| -3(2) + 3 \bigr| = \dfrac{1}{2}\lvert -3\rvert = \dfrac{3}{2},\\[1em]
a_{24} &= \dfrac{1}{2}\bigl| -3(2) + 4 \bigr| = \dfrac{1}{2}\lvert -2\rvert = 1.
\end{aligned}
Row $(i=3)$:
\begin{aligned}
a_{31} &= \dfrac{1}{2}\bigl| -3(3) + 1 \bigr| = \dfrac{1}{2}\lvert -8\rvert = 4,\\[1em]
a_{32} &= \dfrac{1}{2}\bigl| -3(3) + 2 \bigr| = \dfrac{1}{2}\lvert -7\rvert = \dfrac{7}{2},\\[1em]
a_{33} &= \dfrac{1}{2}\bigl| -3(3) + 3 \bigr| = \dfrac{1}{2}\lvert -6\rvert = 3,\\[1em]
a_{34} &= \dfrac{1}{2}\bigl| -3(3) + 4 \bigr| = \dfrac{1}{2}\lvert -5\rvert = \dfrac{5}{2}.
\end{aligned}
Resultant matrix
$$
A = \begin{bmatrix}
1 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\
\dfrac{5}{2} & 2 & \dfrac{3}{2} & 1 \\
4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}
\end{bmatrix}
$$
(ii) $a_{ij} = 2i – j$
Compute entries of $a_{ij} = 2i – j$ for $(i=1,2,3)$ and $(j=1,2,3,4)$.
Row $(i=1)$:
\begin{aligned}
a_{11} &= 2(1) – 1 = 1,\\[1em]
a_{12} &= 2(1) – 2 = 0,\\[1em]
a_{13} &= 2(1) – 3 = -1,\\[1em]
a_{14} &= 2(1) – 4 = -2.
\end{aligned}
Row $(i=2)$:
\begin{aligned}
a_{21} &= 2(2) – 1 = 3,\\[1em]
a_{22} &= 2(2) – 2 = 2,\\[1em]
a_{23} &= 2(2) – 3 = 1,\\[1em]
a_{24} &= 2(2) – 4 = 0.
\end{aligned}
Row $(i=3)$:
\begin{aligned}
a_{31} &= 2(3) – 1 = 5,\\[1em]
a_{32} &= 2(3) – 2 = 4,\\[1em]
a_{33} &= 2(3) – 3 = 3,\\[1em]
a_{34} &= 2(3) – 4 = 2.
\end{aligned}
Resultant matrix
$$
A = \begin{bmatrix}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{bmatrix}
$$
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✅ FAQ Section
Q1. What is covered in Matrices Exercise 3.1 of Class 12?
Exercise 3.1 introduces the basic concepts of matrices, including definition, notation, types of matrices, and simple matrix operations such as equality and addition.
Q2. Are these NCERT solutions suitable for board exam preparation?
Yes, Anand Classes provides fully exam-oriented and accurate solutions that follow the NCERT syllabus, making them ideal for CBSE Class 12 board preparation.
Q3. Can I download the Matrices Exercise 3.1 solutions in PDF format?
Absolutely. The complete NCERT Solutions PDF for Class 12 Matrices Exercise 3.1 is available for free download.
Q4. Do these solutions include step-by-step explanations?
Yes. Each question is solved with detailed steps to strengthen understanding and help students learn matrix operations easily.
Q5. Are these notes helpful for competitive exams as well?
Concepts explained in Matrices are useful for JEE, NDA, and other entrance exams, and Anand Classes notes provide strong conceptual clarity.
