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NCERT Solutions · Class 12 · Chapter 3: Matrices

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.1

Exercise 3.1: Order, elements and equality of matrices. An \(m \times n\) matrix has m rows, n columns and \(mn\) elements; \(a_{ij}\) sits in row i, column j. Two matrices are equal only when they have the same order and every pair of corresponding elements is equal, which turns one matrix equation into several ordinary equations.

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Exercise 3.1 questions and solutions

Exercise 3.1, Question 1

\(A = \begin{bmatrix}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{bmatrix}\)
(i) The order of A
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  1. 3 rows and 4 columns.
Answer: \(3 \times 4\)
(ii) The number of elements
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  1. \(3 \times 4\).
Answer: 12
(iii) \(a_{13}, a_{21}, a_{33}, a_{24}, a_{23}\)
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  1. Read row, then column: \(a_{13}\) is row 1, column 3, and so on.
Answer: \[\begin{gathered}a_{13} = 19, \\ a_{21} = 35, \\ a_{33} = -5, \\ a_{24} = 12, \\ a_{23} = \tfrac52\end{gathered}\]

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Exercise 3.1, Question 2

A matrix has 24 elements: list its possible orders. What if it has 13 elements?
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  1. An \(m \times n\) matrix has \(mn\) elements, so list the factor pairs.
  2. \[\begin{aligned}24 &= 1 \cdot 24 \\ &= 2 \cdot 12 \\ &= 3 \cdot 8 \\ &= 4 \cdot 6\end{aligned}\], each in either order.
  3. 13 is prime: only \(1 \times 13\) and \(13 \times 1\).
Answer: 24: \[\begin{gathered}1 \times 24, \\ 24 \times 1, \\ 2 \times 12, \\ 12 \times 2, \\ 3 \times 8, \\ 8 \times 3, \\ 4 \times 6, \\ 6 \times 4\end{gathered}\] 13: \(1 \times 13,\ 13 \times 1\)

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Exercise 3.1, Question 3

A matrix has 18 elements: list its possible orders. What if it has 5 elements?
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  1. \(18 = 1 \cdot 18 = 2 \cdot 9 = 3 \cdot 6\), each in either order.
  2. 5 is prime: \(1 \times 5\) and \(5 \times 1\).
Answer: 18: \[\begin{gathered}1 \times 18, \\ 18 \times 1, \\ 2 \times 9, \\ 9 \times 2, \\ 3 \times 6, \\ 6 \times 3\end{gathered}\] 5: \(1 \times 5,\ 5 \times 1\)

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Exercise 3.1, Question 4

Construct the \(2 \times 2\) matrix \(A = [a_{ij}]\) with:
(i) \(a_{ij} = \dfrac{(i + j)^2}{2}\)
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  1. \[\begin{gathered}a_{11} = \tfrac42, \\ a_{12} = a_{21} = \tfrac92, \\ a_{22} = \tfrac{16}{2}\end{gathered}\]
Answer: \[\begin{bmatrix}2 & \frac{9}{2} \\ \frac{9}{2} & 8\end{bmatrix}\]
(ii) \(a_{ij} = \dfrac{i}{j}\)
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  1. \[\begin{gathered}a_{11} = 1, \\ a_{12} = \tfrac12, \\ a_{21} = 2, \\ a_{22} = 1\end{gathered}\]
Answer: \[\begin{bmatrix}1 & \frac{1}{2} \\ 2 & 1\end{bmatrix}\]
(iii) \(a_{ij} = \dfrac{(i + 2j)^2}{2}\)
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  1. \[\begin{gathered}a_{11} = \tfrac92, \\ a_{12} = \tfrac{25}{2}, \\ a_{21} = \tfrac{16}{2}, \\ a_{22} = \tfrac{36}{2}\end{gathered}\]
Answer: \[\begin{bmatrix}\frac{9}{2} & \frac{25}{2} \\ 8 & 18\end{bmatrix}\]

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Exercise 3.1, Question 5

Construct the \(3 \times 4\) matrix with:
(i) \(a_{ij} = \tfrac12\left|-3i + j\right|\)
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  1. Row i: \(\tfrac12|j - 3i|\) for \(j = 1, 2, 3, 4\); e.g. \(a_{11} = \tfrac12|-2| = 1\), \(a_{24} = \tfrac12|-2| = 1\).
Answer: \[\begin{bmatrix}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{bmatrix}\]
(ii) \(a_{ij} = 2i - j\)
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  1. Row i: \(2i - 1, 2i - 2, 2i - 3, 2i - 4\).
Answer: \[\begin{bmatrix}1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2\end{bmatrix}\]

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Exercise 3.1, Question 6

Find x, y and z.
(i) \(\begin{bmatrix}4 & 3 \\ x & 5\end{bmatrix} = \begin{bmatrix}y & z \\ 1 & 5\end{bmatrix}\)
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  1. Equate corresponding elements: \(y = 4,\ z = 3,\ x = 1\).
Answer: \(x = 1,\ y = 4,\ z = 3\)
(ii) \(\begin{bmatrix}x + y & 2 \\ z + 5 & x y\end{bmatrix} = \begin{bmatrix}6 & 2 \\ 5 & 8\end{bmatrix}\)
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  1. \(x + y = 6,\ xy = 8,\ 5 + z = 5\).
  2. \(z = 0\); x and y are the roots of \(t^2 - 6t + 8 = 0\), i.e. 2 and 4.
Answer: \(z = 0\) and \(x = 2, y = 4\) or \(x = 4, y = 2\)
(iii) \(\begin{bmatrix}x + y + z \\ x + z \\ y + z\end{bmatrix} = \begin{bmatrix}9 \\ 5 \\ 7\end{bmatrix}\)
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  1. \(x + y + z = 9\) and \(x + z = 5\) give \(y = 4\); \(y + z = 7\) gives \(z = 3\); then \(x = 2\).
Answer: \(x = 2,\ y = 4,\ z = 3\)

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Exercise 3.1, Question 7

Find a, b, c, d: \(\begin{bmatrix}a - b & 2 a + c \\ 2 a - b & 3 c + d\end{bmatrix} = \begin{bmatrix}-1 & 5 \\ 0 & 13\end{bmatrix}\)
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  1. \(a - b = -1\) and \(2a - b = 0\): subtracting, \(a = 1\), so \(b = 2\).
  2. \(2a + c = 5 \Rightarrow c = 3\); \(3c + d = 13 \Rightarrow d = 4\).
Answer: \(a = 1,\ b = 2,\ c = 3,\ d = 4\)

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Exercise 3.1, Question 8

\(A = [a_{ij}]_{m \times n}\) is a square matrix if: (A) \(m < n\) (B) \(m > n\) (C) \(m = n\) (D) none of these
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  1. Square means as many rows as columns.
Answer: (C)

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Exercise 3.1, Question 9

Which values make \(\begin{bmatrix}3 x + 7 & 5 \\ y + 1 & 2 - 3 x\end{bmatrix}\) and \(\begin{bmatrix}0 & y - 2 \\ 8 & 4\end{bmatrix}\) equal? (A) \(x = -\tfrac13, y = 7\) (B) not possible (C) \(y = 7, x = -\tfrac23\) (D) \(x = -\tfrac13, y = -\tfrac23\)
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  1. Equality needs \(3x + 7 = 0\), so \(x = -\tfrac73\), and also \(2 - 3x = 4\), so \(x = -\tfrac23\).
  2. The two conditions on x contradict each other (and \(5 = y - 2\), \(y + 1 = 8\) both give \(y = 7\), but that cannot rescue x).
Answer: (B) Not possible to find

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Exercise 3.1, Question 10

The number of \(3 \times 3\) matrices with every entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512
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  1. 9 entries, 2 choices each: \(2^9\).
Answer: (D) 512

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Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.