NCERT Exemplar Solutions Class 12 Maths Matrices — Short Answer Q1–24

NCERT Exemplar Problems, Class 12 Mathematics — Matrices, Short Answer questions 1 to 24. All 24 questions solved.

Twenty-four questions on the basic skills of matrix algebra: order and number of elements (Questions 1, 2), building a matrix from a rule for $a_{ij}$ (3, 4), equality of matrices (5, 8, 18, 19), addition and scalar multiples (6, 7, 19), and multiplication, where the order of the factors matters (9 to 17, 20 to 24). Several questions ask for an example (16, 21), and those are answered with a specific pair of matrices that is checked by multiplying out.

Key insight. Two matrices are equal only if they have the same order and every corresponding entry is equal, and a product $AB$ exists only if the columns of $A$ equal the rows of $B$. Check orders first, then compare entries one position at a time. Most of the questions below reduce to exactly that.

Question 1

If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

Solution. A matrix of order $m \times n$ has $mn$ elements, so we need every ordered pair of positive integers with $mn = 28$. Since $28 = 2^2 \cdot 7$, its factor pairs are $1 \times 28$, $2 \times 14$, $4 \times 7$ and their reverses.

For 13 elements, $13$ is prime, so the only factors are $1$ and $13$.

Question 2

In the matrix $A = \begin{bmatrix} a & 1 & x \\ 2 & \sqrt{3} & x^2 – y \\ 0 & 5 & \dfrac{-2}{5} \end{bmatrix}$, write: (i) the order of the matrix $A$, (ii) the number of elements, (iii) the elements $a_{23}$, $a_{31}$, $a_{12}$.

Solution. (i) There are three rows and three columns, so the order is $3 \times 3$. (ii) A $3 \times 3$ matrix has $3 \times 3 = 9$ elements. (iii) $a_{ij}$ sits in row $i$, column $j$. So $a_{23}$ is row 2, column 3: $x^2 – y$. $a_{31}$ is row 3, column 1: $0$. $a_{12}$ is row 1, column 2: $1$.

Question 3

Construct a $2 \times 2$ matrix where (i) $a_{ij} = \dfrac{(i – 2j)^2}{2}$ (ii) $a_{ij} = |-2i + 3j|$.

Solution. Substitute each pair $(i, j)$ into the rule.

(i) $a_{11} = \dfrac{(1-2)^2}{2} = \dfrac12$, $a_{12} = \dfrac{(1-4)^2}{2} = \dfrac92$, $a_{21} = \dfrac{(2-2)^2}{2} = 0$, $a_{22} = \dfrac{(2-4)^2}{2} = 2$.

(ii) $a_{11} = |-2+3| = 1$, $a_{12} = |-2+6| = 4$, $a_{21} = |-4+3| = |-1| = 1$, $a_{22} = |-4+6| = 2$.

The answer key prints $a_{21} = -1$ in part (ii); that cannot be right because $a_{21}$ is an absolute value, and an absolute value is never negative.

Question 4

Construct a $3 \times 2$ matrix whose elements are given by $a_{ij} = e^{i \cdot x} \sin jx$.

Solution. Three rows ($i = 1, 2, 3$) and two columns ($j = 1, 2$). Row $i$ carries the factor $e^{ix}$, and column $j$ carries $\sin jx$:

Question 5

Find values of $a$ and $b$ if $A = B$, where $A = \begin{bmatrix} a+4 & 3b \\ 8 & -6 \end{bmatrix}$, $B = \begin{bmatrix} 2a+2 & b^2+2 \\ 8 & b^2-5b \end{bmatrix}$.

Solution. Equal matrices have equal corresponding entries. From the first row, first column: $a + 4 = 2a + 2$, so $a = 2$.

The remaining entries involve $b$ alone. From $3b = b^2 + 2$: $b^2 – 3b + 2 = 0$, so $b = 1$ or $b = 2$. From $-6 = b^2 – 5b$: $b^2 – 5b + 6 = 0$, so $b = 2$ or $b = 3$. Both equations must hold at once, so the common value $b = 2$ is the only one that works.

Question 6

If possible, find the sum of the matrices $A$ and $B$, where $A = \begin{bmatrix} \sqrt{3} & 1 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} x & y & z \\ a & b & 6 \end{bmatrix}$.

Solution. Matrices can be added only when they have the same order. $A$ is $2 \times 2$ and $B$ is $2 \times 3$, so there is no entry in $B$ to pair with some entries of $A$ and vice versa.

Question 7

If $X = \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4 \end{bmatrix}$, find (i) $X + Y$ (ii) $2X – 3Y$ (iii) a matrix $Z$ such that $X + Y + Z$ is a zero matrix.

Solution. Both are $2 \times 3$, so everything is defined; work entry by entry.

(i) $X + Y = \begin{bmatrix} 5 & 2 & -2 \\ 12 & 0 & 1 \end{bmatrix}$.

(ii) $2X = \begin{bmatrix} 6 & 2 & -2 \\ 10 & -4 & -6 \end{bmatrix}$ and $3Y = \begin{bmatrix} 6 & 3 & -3 \\ 21 & 6 & 12 \end{bmatrix}$, so $2X – 3Y = \begin{bmatrix} 0 & -1 & 1 \\ -11 & -10 & -18 \end{bmatrix}$.

(iii) $X + Y + Z = O$ means $Z = -(X + Y)$, the negative of the answer to (i).

Question 8

Find non-zero values of $x$ satisfying the matrix equation $x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix}$.

Solution. Multiply through by the scalars. The left side is $\begin{bmatrix} 2x^2 + 16 & 2x + 10x \\ 3x + 8 & x^2 + 8x \end{bmatrix}$ and the right side is $\begin{bmatrix} 2x^2 + 16 & 48 \\ 20 & 12x \end{bmatrix}$.

Compare entries. The top-left entries agree for every $x$. Top-right: $12x = 48$, so $x = 4$. Bottom-left: $3x + 8 = 20$, so $x = 4$. Bottom-right: $x^2 + 8x = 12x$, so $x^2 – 4x = 0$, giving $x = 0$ or $x = 4$. The question asks for non-zero values, which removes $x = 0$, and $x = 4$ satisfies all four entries.

Question 9

If $A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, show that $(A + B)(A – B) \neq A^2 – B^2$.

Solution. $A + B = \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix}$ and $A – B = \begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix}$, so $$(A+B)(A-B) = \begin{bmatrix} 0 & 0 \\ 0 & 4 + 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}.$$

Next, $A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ and $B^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$, so $A^2 – B^2 = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$.

These are different. The reason is that $(A+B)(A-B) = A^2 – AB + BA – B^2$, and the cross terms cancel only when $AB = BA$, which fails here.

Question 10

Find the value of $x$ if $\begin{bmatrix} 1 & x & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O$.

Solution. Multiply from the right first: the $3 \times 3$ matrix times the column gives $\begin{bmatrix} 1 + 6 + 2x \\ 2 + 10 + x \\ 15 + 6 + 2x \end{bmatrix} = \begin{bmatrix} 7 + 2x \\ 12 + x \\ 21 + 2x \end{bmatrix}$.

The row $[1\ x\ 1]$ times this column is a $1 \times 1$ matrix: $(7 + 2x) + x(12 + x) + (21 + 2x) = x^2 + 16x + 28$.

Setting it to zero: $(x + 2)(x + 14) = 0$.

Question 11

Show that $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$ satisfies the equation $A^2 – 3A – 7I = O$ and hence find $A^{-1}$.

Solution. $A^2 = \begin{bmatrix} 25 – 3 & 15 – 6 \\ -5 + 2 & -3 + 4 \end{bmatrix} = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix}$. Then $$A^2 – 3A – 7I = \begin{bmatrix} 22 – 15 – 7 & 9 – 9 \\ -3 + 3 & 1 + 6 – 7 \end{bmatrix} = O.$$

For the inverse, multiply $A^2 – 3A – 7I = O$ by $A^{-1}$: $A – 3I – 7A^{-1} = O$, so $A^{-1} = \tfrac17(A – 3I)$. Hence $$A^{-1} = \frac17\begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix}.$$

Check: $A \cdot \begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = 7I$, so $AA^{-1} = I$.

Question 12

Find the matrix $A$ satisfying the matrix equation $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Solution. Write the equation as $PAQ = I$. Multiply on the left by $P^{-1}$ and on the right by $Q^{-1}$ (the order matters, because matrices do not commute): $A = P^{-1}Q^{-1}$.

$|P| = 4 – 3 = 1$, so $P^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$. $|Q| = 9 – 10 = -1$, so $Q^{-1} = -\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}$.

$$A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 6 – 5 & 4 – 3 \\ -9 + 10 & -6 + 6 \end{bmatrix}.$$

Question 13

Find $A$, if $\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} A = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$.

Solution. The left factor is $3 \times 1$ and the product is $3 \times 3$, so $A$ must be $1 \times 3$; write $A = [p\ q\ r]$. Row 2 of the product is $1 \cdot A = [p\ q\ r]$, and it reads $[-1\ 2\ 1]$. Check the other rows: $4A = [-4\ 8\ 4]$ and $3A = [-3\ 6\ 3]$, which match.

Question 14

If $A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$, then verify $(BA)^2 \neq B^2A^2$.

Solution. $BA$ is $2 \times 2$: $$BA = \begin{bmatrix} 6 + 1 + 4 & -8 + 1 + 0 \\ 3 + 2 + 8 & -4 + 2 + 0 \end{bmatrix} = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix},$$ so $$(BA)^2 = \begin{bmatrix} 121 – 91 & -77 + 14 \\ 143 – 26 & -91 + 4 \end{bmatrix} = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}.$$

Now $B$ is $2 \times 3$, which is not square, so $B^2 = B \cdot B$ is not defined (columns of the first factor, 3, differ from rows of the second, 2). Likewise $A$ is $3 \times 2$, so $A^2$ is not defined. Hence $B^2A^2$ does not exist, and $(BA)^2 \neq B^2A^2$ because the left side is a matrix and the right side is not defined.

Question 15

If possible, find $BA$ and $AB$, where $A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{bmatrix}$.

Solution. $A$ is $2 \times 3$ and $B$ is $3 \times 2$. Both products exist: $BA$ is $3 \times 3$ and $AB$ is $2 \times 2$.

$$BA = \begin{bmatrix} 8 + 1 & 4 + 2 & 8 + 4 \\ 4 + 3 & 2 + 6 & 4 + 12 \\ 2 + 2 & 1 + 4 & 2 + 8 \end{bmatrix} = \begin{bmatrix} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{bmatrix},$$ $$AB = \begin{bmatrix} 8 + 2 + 2 & 2 + 3 + 4 \\ 4 + 4 + 4 & 1 + 6 + 8 \end{bmatrix} = \begin{bmatrix} 12 & 9 \\ 12 & 15 \end{bmatrix}.$$

Question 16

Show by an example that for $A \neq O$, $B \neq O$, $AB = O$.

Solution. Take $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, both non-zero. Row 1 of $A$ is $[0\ 1]$; against the columns of $B$ it gives $0 \cdot 1 + 1 \cdot 0 = 0$ and $0 \cdot 0 + 1 \cdot 0 = 0$. Row 2 of $A$ is zero. So $AB = O$.

This shows that, unlike real numbers, a product of matrices can be zero without either factor being zero.

Question 17

Given $A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix}$. Is $(AB)’ = B’A’$?

Solution. $AB = \begin{bmatrix} 2 + 8 + 0 & 8 + 32 + 0 \\ 3 + 18 + 6 & 12 + 72 + 18 \end{bmatrix} = \begin{bmatrix} 10 & 40 \\ 27 & 102 \end{bmatrix}$, so $(AB)’ = \begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}$.

Now $B’ = \begin{bmatrix} 1 & 2 & 1 \\ 4 & 8 & 3 \end{bmatrix}$ and $A’ = \begin{bmatrix} 2 & 3 \\ 4 & 9 \\ 0 & 6 \end{bmatrix}$, so $$B’A’ = \begin{bmatrix} 2 + 8 + 0 & 3 + 18 + 6 \\ 8 + 32 + 0 & 12 + 72 + 18 \end{bmatrix} = \begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}.$$

They agree, as the reversal law $(AB)’ = B’A’$ guarantees.

Question 18

Solve for $x$ and $y$: $x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = O$.

Solution. Combining into one column and equating each entry to zero gives two linear equations: $$2x + 3y = 8, \qquad x + 5y = 11.$$ From the second, $x = 11 – 5y$. Substitute: $22 – 10y + 3y = 8$, so $7y = 14$ and $y = 2$, then $x = 1$. Check: $2 + 6 = 8$ and $1 + 10 = 11$.

Question 19

If $X$ and $Y$ are $2 \times 2$ matrices, solve $2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$, $3X + 2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}$ for $X$ and $Y$.

Solution. Treat the two matrix equations like simultaneous equations. Call the right sides $P$ and $Q$. Multiply the first by 3 and the second by 2: $6X + 9Y = 3P$ and $6X + 4Y = 2Q$. Subtracting gives $5Y = 3P – 2Q$. Multiplying the first by 2 and the second by 3 and subtracting the other way gives $5X = 3Q – 2P$.

$$X = \frac15\left(\begin{bmatrix} -6 & 6 \\ 3 & -15 \end{bmatrix} – \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix}\right) = \frac15\begin{bmatrix} -10 & 0 \\ -5 & -15 \end{bmatrix},$$ $$Y = \frac15\left(\begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} – \begin{bmatrix} -4 & 4 \\ 2 & -10 \end{bmatrix}\right) = \frac15\begin{bmatrix} 10 & 5 \\ 10 & 10 \end{bmatrix}.$$

Question 20

If $A = \begin{bmatrix} 3 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 7 & 3 \end{bmatrix}$, then find a non-zero matrix $C$ such that $AC = BC$.

Solution. $AC = BC$ means $(A – B)C = O$. Here $A – B = [-4\ \ 2]$ is $1 \times 2$, so $C$ must have two rows. Take $C = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$; then $-4c_1 + 2c_2 = 0$, so $c_2 = 2c_1$. Any non-zero $c_1 = k$ gives a valid $C$; the simplest is $k = 1$.

Check: $AC = 3 + 10 = 13$ and $BC = 7 + 6 = 13$.

Question 21

Give an example of matrices $A$, $B$ and $C$ such that $AB = AC$, where $A$ is a non-zero matrix, but $B \neq C$.

Solution. Take $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, $C = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. Row 1 of $A$ is $[1\ 0]$, which picks out row 1 of the other factor, and row 1 of both $B$ and $C$ is zero. Row 2 of $A$ is zero. So $AB = AC = O$, yet $B \neq C$.

This is why “cancelling” $A$ from $AB = AC$ is not allowed for matrices.

Question 22

If $A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}$, verify: (i) $(AB)C = A(BC)$ (ii) $A(B + C) = AB + AC$.

Solution. (i) $AB = \begin{bmatrix} 2 + 6 & 3 – 8 \\ -4 + 3 & -6 – 4 \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix}$, so $(AB)C = \begin{bmatrix} 8 + 5 & 0 \\ -1 + 10 & 0 \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}$.

$BC = \begin{bmatrix} 2 – 3 & 0 \\ 3 + 4 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix}$, so $A(BC) = \begin{bmatrix} -1 + 14 & 0 \\ 2 + 7 & 0 \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}$. The two agree.

(ii) $AC = \begin{bmatrix} 1 – 2 & 0 \\ -2 – 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ -3 & 0 \end{bmatrix}$, so $AB + AC = \begin{bmatrix} 7 & -5 \\ -4 & -10 \end{bmatrix}$.

Also $B + C = \begin{bmatrix} 3 & 3 \\ 2 & -4 \end{bmatrix}$ and $A(B + C) = \begin{bmatrix} 3 + 4 & 3 – 8 \\ -6 + 2 & -6 – 4 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ -4 & -10 \end{bmatrix}$. The two agree.

Question 23

If $P = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ and $Q = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$, prove that $PQ = \begin{bmatrix} xa & 0 & 0 \\ 0 & yb & 0 \\ 0 & 0 & zc \end{bmatrix} = QP$.

Solution. The $(i, j)$ entry of $PQ$ is the sum of (row $i$ of $P$) times (column $j$ of $Q$). Row $i$ of $P$ has a single non-zero entry, in position $i$, and column $j$ of $Q$ has a single non-zero entry, in position $j$. The two can meet only when $i = j$, so off-diagonal entries are $0$. On the diagonal: $$PQ = \begin{bmatrix} xa & 0 & 0 \\ 0 & yb & 0 \\ 0 & 0 & zc \end{bmatrix}.$$ The same argument for $QP$ gives $ax,\ by,\ cz$ on the diagonal, which equal $xa,\ yb,\ zc$ because ordinary numbers commute. Hence $PQ = QP$.

Question 24

If $\begin{bmatrix} 2 & 1 & 3 \end{bmatrix}\begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A$, find $A$.

Solution. Multiply the $3 \times 3$ matrix by the column first: $\begin{bmatrix} -1 + 0 + 1 \\ -1 + 0 + 0 \\ 0 + 0 – 1 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\ -1 \end{bmatrix}$.

Then the row $[2\ 1\ 3]$ times this column: $0 – 1 – 3 = -4$. The result is a $1 \times 1$ matrix.

Common mistakes

• Multiplying in the wrong order. Matrix multiplication is not commutative. In Question 12 the left inverse must go on the left and the right inverse on the right; swapping them gives a different matrix.
• Cancelling a matrix factor. $AB = AC$ does not give $B = C$ (Question 21), and $AB = O$ does not force $A = O$ or $B = O$ (Question 16). Cancellation needs $A$ to be invertible.
• Expanding $(A+B)(A-B)$ as $A^2 – B^2$. The cross terms $-AB + BA$ vanish only when $AB = BA$ (Question 9).
• Equating entries from only one equation. In Question 5 each of $b^2 – 3b + 2 = 0$ and $b^2 – 5b + 6 = 0$ has two roots; only the shared root satisfies both.
• Forgetting that an absolute value is non-negative when building a matrix from $|{-2i + 3j}|$ (Question 3).

Practise next

• The remaining Short Answer questions, with transpose, symmetric and skew-symmetric matrices: Questions 25 to 52.
• Objective and fill-in-the-blank questions on the same chapter: Questions 53 to 81.
• True or false statements about matrix algebra: Questions 82 to 101.
• Textbook practice on order, construction and equality: Exercise 3.1; operations and multiplication: Exercise 3.2; inverses: Exercise 3.4.