NCERT Class 12 Mathematics — Matrices, Exercise 3.1. All 10 questions solved.

Exercise 3.1 sets up the vocabulary the rest of the chapter runs on. Three ideas cover everything in it.

The order of a matrix is written rows × columns, in that order, and a matrix of order $m \times n$ has $mn$ elements. The element $a_{ij}$ sits in row $i$, column $j$ — again row first. And two matrices are equal only if they have the same order and every corresponding pair of entries is equal.

Key insight. Matrix equality is not one statement but many: equating two $2\times2$ matrices gives four simultaneous equations at once. That is what makes questions 6, 7 and 9 solvable — and it is also why question 9’s answer is that no solution exists, because the four equations it produces contradict each other.

Question 1

In the matrix $\mathrm{A} = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$, 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) Counting: there are 3 horizontal rows and 4 vertical columns, so the order is $3 \times 4$. Rows always come first.

(ii) The number of elements is the product of the order: $3 \times 4 = 12$.

(iii) Read each subscript as (row, column):

$$a_{13} = 19 \ \text{(row 1, column 3)}, \qquad a_{21} = 35 \ \text{(row 2, column 1)}$$

$$a_{33} = -5, \qquad a_{24} = 12, \qquad a_{23} = \frac{5}{2}$$

Question 2

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

Solution. An order $m \times n$ is possible exactly when $mn$ equals the number of elements, so this is a question about factor pairs — and the pairs are ordered, since $2\times12$ and $12\times2$ are different matrices.

The factors of 24 are $1, 2, 3, 4, 6, 8, 12, 24$, giving eight orders:

$$1\times24, \quad 2\times12, \quad 3\times8, \quad 4\times6, \quad 6\times4, \quad 8\times3, \quad 12\times2, \quad 24\times1$$

13 is prime, so its only factors are 1 and 13:

$$1\times13, \quad 13\times1$$

Question 3

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

Solution. The factors of 18 are $1, 2, 3, 6, 9, 18$, giving six ordered pairs:

$$1\times18, \quad 2\times9, \quad 3\times6, \quad 6\times3, \quad 9\times2, \quad 18\times1$$

5 is prime, so:

$$1\times5, \quad 5\times1$$

A prime number of elements always gives exactly two orders, both of which are a single row or a single column.

Question 4

Construct a $2 \times 2$ matrix, $\mathrm{A} = [a_{ij}]$, whose elements are given by:

Question 4 (i)

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

Solution. Substitute each of the four $(i, j)$ pairs in turn.

$$a_{11} = \frac{(1+1)^2}{2} = 2, \qquad a_{12} = \frac{(1+2)^2}{2} = \frac{9}{2}$$

$$a_{21} = \frac{(2+1)^2}{2} = \frac{9}{2}, \qquad a_{22} = \frac{(2+2)^2}{2} = 8$$

Question 4 (ii)

$a_{ij} = \dfrac{i}{j}$

Solution.

$$a_{11} = 1, \qquad a_{12} = \frac{1}{2}, \qquad a_{21} = 2, \qquad a_{22} = 1$$

Note $a_{12} \neq a_{21}$ here, whereas in part (i) they were equal — the formula $(i+j)^2$ is symmetric in $i$ and $j$, and $i/j$ is not.

Question 4 (iii)

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

Solution.

$$a_{11} = \frac{(1+2)^2}{2} = \frac{9}{2}, \qquad a_{12} = \frac{(1+4)^2}{2} = \frac{25}{2}$$

$$a_{21} = \frac{(2+2)^2}{2} = 8, \qquad a_{22} = \frac{(2+4)^2}{2} = 18$$

Question 5

Construct a $3 \times 4$ matrix, whose elements are given by:

Question 5 (i)

$a_{ij} = \dfrac{1}{2}\,|-3i + j|$

Solution. Twelve entries, with $i$ running from 1 to 3 and $j$ from 1 to 4. The modulus means every entry is non-negative, whatever the sign inside.

Row 1 ($i = 1$): $\tfrac12|-3+1| = 1$, $\tfrac12|-3+2| = \tfrac12$, $\tfrac12|-3+3| = 0$, $\tfrac12|-3+4| = \tfrac12$

Row 2 ($i = 2$): $\tfrac12|-6+1| = \tfrac52$, $\tfrac12|-6+2| = 2$, $\tfrac12|-6+3| = \tfrac32$, $\tfrac12|-6+4| = 1$

Row 3 ($i = 3$): $\tfrac12|-9+1| = 4$, $\tfrac12|-9+2| = \tfrac72$, $\tfrac12|-9+3| = 3$, $\tfrac12|-9+4| = \tfrac52$

Question 5 (ii)

$a_{ij} = 2i – j$

Solution. No modulus this time, so negative entries are allowed and do occur.

Row 1: $1, 0, -1, -2$. Row 2: $3, 2, 1, 0$. Row 3: $5, 4, 3, 2$.

Question 6

Find the values of $x$, $y$ and $z$ from the following equations:

Question 6 (i)

$\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$

Solution. Equate corresponding entries. Each position gives one equation, and here each is immediate:

$$y = 4, \qquad z = 3, \qquad x = 1$$

Question 6 (ii)

$\begin{bmatrix} x + y & 2 \\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$

Solution. Equating entries gives

$$x + y = 6, \qquad xy = 8, \qquad 5 + z = 5 \Rightarrow z = 0$$

The first two say $x$ and $y$ are numbers with sum 6 and product 8, so they are the roots of

$$t^2 – 6t + 8 = 0 \quad\Rightarrow\quad (t-2)(t-4) = 0 \quad\Rightarrow\quad t = 2, 4$$

Nothing distinguishes which is $x$ and which is $y$, so both assignments are valid.

Question 6 (iii)

$\begin{bmatrix} x + y + z \\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$

Solution. Three equations in three unknowns, and they are easier than they look — subtracting the second from the first isolates $y$ straight away.

$$(x + y + z) – (x + z) = 9 – 5 \quad\Rightarrow\quad y = 4$$

Then from $y + z = 7$ we get $z = 3$, and from $x + z = 5$ we get $x = 2$.

Question 7

Find the value of $a$, $b$, $c$ and $d$ from the equation $\begin{bmatrix} a – b & 2a + c \\ 2a – b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$.

Solution. Four positions, four equations:

$$a – b = -1, \qquad 2a + c = 5, \qquad 2a – b = 0, \qquad 3c + d = 13$$

The first and third share the term $b$, so subtracting eliminates it:

$$(2a – b) – (a – b) = 0 – (-1) \quad\Rightarrow\quad a = 1$$

Then $b = 2a = 2$, and from $2a + c = 5$ we get $c = 3$, and from $3c + d = 13$ we get $d = 13 – 9 = 4$.

Question 8

$\mathrm{A} = [a_{ij}]_{m \times n}$ is a square matrix, if

Solution. A square matrix is one with as many rows as columns — that is the definition, and it makes the array literally square in shape.

Question 9

Which of the given values of $x$ and $y$ make the following pair of matrices equal: $\begin{bmatrix} 3x + 7 & 5 \\ y + 1 & 2 – 3x \end{bmatrix}$, $\begin{bmatrix} 0 & y – 2 \\ 8 & 4 \end{bmatrix}$

Solution. Equality requires all four positions to match, so write down all four equations before choosing.

$$3x + 7 = 0 \quad\Rightarrow\quad x = -\frac{7}{3}$$

$$2 – 3x = 4 \quad\Rightarrow\quad -3x = 2 \quad\Rightarrow\quad x = -\frac{2}{3}$$

These two demand different values of $x$, and one variable cannot be two things at once. The matrices therefore cannot be made equal for any $x$ and $y$ — the $y$ equations ($5 = y – 2$ and $y + 1 = 8$, both giving $y = 7$) are consistent, but that is not enough.

This is why options (A), (C) and (D) are all wrong: each satisfies some of the four conditions and fails at least one.

Question 10

The number of all possible matrices of order $3 \times 3$ with each entry 0 or 1 is:

Solution. A $3 \times 3$ matrix has $9$ entries, and each may be filled independently in $2$ ways.

$$\underbrace{2 \times 2 \times \cdots \times 2}_{9 \text{ times}} = 2^9 = 512$$

Option (A) is $3^3$ and (C) is $3^4$ — both come from multiplying the wrong two numbers. The rule is (number of choices) raised to (number of positions), not the other way round.

Common mistakes

• Reading $a_{ij}$ as (column, row). It is row first: $a_{24}$ is row 2, column 4. Question 1(iii) is entirely this.
• Treating $m \times n$ and $n \times m$ as the same order. They are different matrices, so both count. Question 2 has eight answers, not four.
• Stopping at one solution in question 6(ii). $x + y = 6$ and $xy = 8$ are symmetric in $x$ and $y$, so both assignments are valid and both should be given.
• Checking only some entries for equality. Question 9 is designed so the $y$ equations agree and the $x$ equations do not. Every position must match.
• Computing $9^2$ or $3^3$ in question 10. There are 9 positions with 2 choices each, so it is $2^9$.
• Forgetting the modulus in question 5(i). Every entry there is non-negative; part (ii), with no modulus, does produce negatives.

Practise next

• Exercise 3.2 — addition, scalar multiplication and matrix multiplication, the longest exercise in the chapter at 22 questions.
• Exercise 3.3 — transpose, symmetric and skew-symmetric matrices.