NCERT Class 12 Mathematics — Matrices, Exercise 3.2. All 22 questions solved.

Exercise 3.2 is the longest in the chapter and introduces the operations everything later depends on. Two rules govern when an operation is even allowed.

Addition and subtraction need matrices of identical order, and work entry by entry. Multiplication needs the inner dimensions to agree — an $m \times n$ times an $n \times p$ gives an $m \times p$ — and each entry of the product is a row of the first dotted with a column of the second.

Key insight. Matrix multiplication is not commutative: $\mathrm{AB}$ and $\mathrm{BA}$ are usually different matrices, and sometimes only one of them is even defined. Question 1 asks for both so you can see it, and question 14 is devoted entirely to it. Once that lands, most of the algebra in the rest of the chapter stops being surprising.

Question 1

Let $\mathrm{A} = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$, $\mathrm{B} = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$, $\mathrm{C} = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$. Find each of the following.

Solution. Addition and subtraction are entry by entry; $3\mathrm{A}$ multiplies every entry of A by 3.

(i) $\mathrm{A} + \mathrm{B} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}$

(ii) $\mathrm{A} – \mathrm{B} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}$

(iii) $3\mathrm{A} – \mathrm{C} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} – \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

(iv) For $\mathrm{AB}$, each entry is a row of A dotted with a column of B:

$$\mathrm{AB} = \begin{bmatrix} 2(1) + 4(-2) & 2(3) + 4(5) \\ 3(1) + 2(-2) & 3(3) + 2(5) \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$$

(v) Now the other order:

$$\mathrm{BA} = \begin{bmatrix} 1(2) + 3(3) & 1(4) + 3(2) \\ -2(2) + 5(3) & -2(4) + 5(2) \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}$$

$\mathrm{AB} \neq \mathrm{BA}$ — not by a sign or a factor, but completely.

Question 2

Compute the following.

Solution.

(i) $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}$ — the $-b$ and $b$ cancel.

(ii) Adding entry by entry gives perfect squares in every position:

$$\begin{bmatrix} a^2 + b^2 + 2ab & b^2 + c^2 + 2bc \\ a^2 + c^2 – 2ac & a^2 + b^2 – 2ab \end{bmatrix} = \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}$$

(iii) $\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$

(iv) Every entry becomes $\sin^2 x + \cos^2 x = 1$:

$$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

Question 3

Compute the indicated products.

Solution. Check the orders first — the number of columns in the left matrix must equal the number of rows in the right.

(i) $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix}$

(ii) A $3\times1$ times a $1\times3$ gives a $3\times3$ — the orders make this much bigger than it looks:

$$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$$

(iii) $\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

(iv) $\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$

(v) $\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}$

(vi) $\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

Question 4

Compute $(\mathrm{A}+\mathrm{B})$ and $(\mathrm{B}-\mathrm{C})$, and verify that $\mathrm{A} + (\mathrm{B} – \mathrm{C}) = (\mathrm{A} + \mathrm{B}) – \mathrm{C}$.

Solution.

$$\mathrm{A} + \mathrm{B} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}, \qquad \mathrm{B} – \mathrm{C} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}$$

Both sides of the identity come to

$$\begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$$

confirming that matrix addition is associative — the brackets can be moved freely, exactly as with numbers.

Question 5

Compute $3\mathrm{A} – 5\mathrm{B}$.

Solution. Multiplying A by 3 clears its denominators of 3, and multiplying B by 5 clears its denominators of 5:

$$3\mathrm{A} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}, \qquad 5\mathrm{B} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$$

The two are identical, so the difference is the zero matrix.

Question 6

Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}$.

Solution. Multiply each matrix by its scalar, then add entry by entry.

$$= \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}$$

The off-diagonal terms cancel and the diagonal ones give $\sin^2\theta + \cos^2\theta = 1$:

$$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Question 7

Find X and Y.

Question 7 (i)

$\mathrm{X} + \mathrm{Y} = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ and $\mathrm{X} – \mathrm{Y} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

Solution. Treat them exactly as simultaneous equations in numbers. Adding the two equations eliminates Y:

$$2\mathrm{X} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \quad\Rightarrow\quad \mathrm{X} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$$

Subtracting instead eliminates X:

$$2\mathrm{Y} = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} \quad\Rightarrow\quad \mathrm{Y} = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$$

Question 7 (ii)

$2\mathrm{X} + 3\mathrm{Y} = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$ and $3\mathrm{X} + 2\mathrm{Y} = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$

Solution. Call the two right-hand sides P and Q. Eliminating by the usual method — multiply the first by 3, the second by 2, subtract:

$$5\mathrm{Y} = 3\mathrm{P} – 2\mathrm{Q} \quad\Rightarrow\quad \mathrm{Y} = \frac{1}{5}\left(\begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} – \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix}\right) = \frac{1}{5}\begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$$

$$5\mathrm{X} = 3\mathrm{Q} – 2\mathrm{P} \quad\Rightarrow\quad \mathrm{X} = \frac{1}{5}\begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix}$$

Question 8

Find X, if $\mathrm{Y} = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $2\mathrm{X} + \mathrm{Y} = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$.

Solution.

$$2\mathrm{X} = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} – \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$$

$$\mathrm{X} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$$

Question 9

Find $x$ and $y$, if $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$.

Solution. Carry out the scalar multiplication and addition first:

$$\begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$

Equating the entries that contain unknowns:

$$2 + y = 5 \Rightarrow y = 3, \qquad 2x + 2 = 8 \Rightarrow x = 3$$

Question 10

Solve the equation for $x$, $y$, $z$ and $t$, if $2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$.

Solution.

$$\begin{bmatrix} 2x + 3 & 2z – 3 \\ 2y & 2t + 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$$

Four equations, each with one unknown:

$$2x + 3 = 9 \Rightarrow x = 3, \qquad 2z – 3 = 15 \Rightarrow z = 9$$

$$2y = 12 \Rightarrow y = 6, \qquad 2t + 6 = 18 \Rightarrow t = 6$$

Question 11

If $x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$, find the values of $x$ and $y$.

Solution.

$$\begin{bmatrix} 2x – y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$$

Adding the two equations eliminates $y$ immediately:

$$5x = 15 \Rightarrow x = 3, \qquad y = 5 – 3(3) = -4$$

Question 12

Given $3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix}$, find the values of $x$, $y$, $z$ and $w$.

Solution. Equating the four positions:

$$3x = x + 4 \Rightarrow x = 2$$

$$3y = 6 + x + y \Rightarrow 2y = 8 \Rightarrow y = 4$$

$$3w = 2w + 3 \Rightarrow w = 3$$

$$3z = -1 + z + w \Rightarrow 2z = -1 + 3 = 2 \Rightarrow z = 1$$

The order matters: $x$ and $w$ have to be found first, since the equations for $y$ and $z$ depend on them.

Question 13

If $\mathrm{F}(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, show that $\mathrm{F}(x)\,\mathrm{F}(y) = \mathrm{F}(x+y)$.

Solution. Multiplying out, the third row and column stay clear of everything, so only the top-left $2\times2$ block matters. Its $(1,1)$ entry is

$$\cos x\cos y – \sin x\sin y = \cos(x+y)$$

and the $(1,2)$ entry is

$$-\cos x\sin y – \sin x\cos y = -\sin(x+y)$$

Similarly the $(2,1)$ entry gives $\sin x\cos y + \cos x\sin y = \sin(x+y)$ and the $(2,2)$ entry gives $\cos(x+y)$. So

$$\mathrm{F}(x)\mathrm{F}(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \mathrm{F}(x+y)$$

The compound angle formulae fall out of the multiplication itself. Geometrically $\mathrm{F}(x)$ rotates space about the $z$-axis by $x$, and doing two rotations in succession adds the angles.

Question 14

Show that (i) $\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$ and (ii) the corresponding $3\times3$ statement.

Solution.

(i) Computing both orders:

$$\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}, \qquad \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$$

These differ in every entry.

(ii) For the $3\times3$ pair:

$$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$$

$$\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$$

Again completely different. One counterexample would suffice to show that matrix multiplication is not commutative in general; this question gives two.

Question 15

Find $\mathrm{A}^2 – 5\mathrm{A} + 6\mathrm{I}$, if $\mathrm{A} = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$.

Solution. First the square, computed as $\mathrm{A}\times\mathrm{A}$:

$$\mathrm{A}^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$$

Then subtract $5\mathrm{A}$ and add $6\mathrm{I}$. Note that $6\mathrm{I}$ adds 6 only along the diagonal, not to every entry:

$$\mathrm{A}^2 – 5\mathrm{A} + 6\mathrm{I} = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} – \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$$

$$= \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$$

Question 16

If $\mathrm{A} = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, prove that $\mathrm{A}^3 – 6\mathrm{A}^2 + 7\mathrm{A} + 2\mathrm{I} = 0$.

Solution. Building the powers one at a time:

$$\mathrm{A}^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}, \qquad \mathrm{A}^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$$

Substituting:

$$\mathrm{A}^3 – 6\mathrm{A}^2 + 7\mathrm{A} + 2\mathrm{I} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

Compute $\mathrm{A}^3$ as $\mathrm{A}^2 \times \mathrm{A}$ rather than starting again — it halves the work and the chances of an error.

Question 17

If $\mathrm{A} = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $\mathrm{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find $k$ so that $\mathrm{A}^2 = k\mathrm{A} – 2\mathrm{I}$.

Solution.

$$\mathrm{A}^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$$

$$k\mathrm{A} – 2\mathrm{I} = \begin{bmatrix} 3k – 2 & -2k \\ 4k & -2k – 2 \end{bmatrix}$$

Equating any one entry gives $k$; the simplest is the $(1,2)$ position:

$$-2k = -2 \quad\Rightarrow\quad k = 1$$

Checking the others: $3(1) – 2 = 1$ ✓, $4(1) = 4$ ✓, $-2(1) – 2 = -4$ ✓. All four agree, so $k = 1$ genuinely works.

Question 18

If $\mathrm{A} = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix}$ and I is the identity matrix of order 2, show that $\mathrm{I} + \mathrm{A} = (\mathrm{I} – \mathrm{A})\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.

Solution. Write $t = \tan\frac{\alpha}{2}$ to keep things readable. Then

$$\mathrm{I} + \mathrm{A} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}, \qquad \mathrm{I} – \mathrm{A} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$$

Multiplying out the right-hand side:

$$(\mathrm{I} – \mathrm{A})\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \cos\alpha + t\sin\alpha & -\sin\alpha + t\cos\alpha \\ -t\cos\alpha + \sin\alpha & t\sin\alpha + \cos\alpha \end{bmatrix}$$

Now use the half-angle forms, which is what makes this work:

$$\cos\alpha = \frac{1 – t^2}{1 + t^2}, \qquad \sin\alpha = \frac{2t}{1 + t^2}$$

For the $(1,1)$ entry:

$$\cos\alpha + t\sin\alpha = \frac{1 – t^2}{1 + t^2} + \frac{2t^2}{1 + t^2} = \frac{1 + t^2}{1 + t^2} = 1$$

For the $(1,2)$ entry:

$$-\sin\alpha + t\cos\alpha = \frac{-2t + t – t^3}{1 + t^2} = \frac{-t(1 + t^2)}{1 + t^2} = -t$$

The other two entries follow by the same substitution, giving $\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \mathrm{I} + \mathrm{A}$.

Question 19

A trust fund has ₹30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) ₹1800 (b) ₹2000.

Solution. Let ₹$x$ go into the 5% bond, so ₹$(30000 – x)$ goes into the 7% bond. As a matrix product, the total interest is

$$\begin{bmatrix} x & 30000 – x \end{bmatrix}\begin{bmatrix} 5/100 \\ 7/100 \end{bmatrix} = \frac{5x + 7(30000 – x)}{100} = \frac{210000 – 2x}{100}$$

(a) Total interest ₹1800:

$$\frac{210000 – 2x}{100} = 1800 \Rightarrow 210000 – 2x = 180000 \Rightarrow x = 15000$$

(b) Total interest ₹2000:

$$210000 – 2x = 200000 \Rightarrow x = 5000$$

Question 20

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Solution. Convert dozens to units first — this is the step the question is really testing.

$$10 \text{ dozen} = 120, \qquad 8 \text{ dozen} = 96, \qquad 10 \text{ dozen} = 120$$

Now the total is a row of quantities times a column of prices:

$$\begin{bmatrix} 120 & 96 & 120 \end{bmatrix}\begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix} = 9600 + 5760 + 4800 = 20160$$

Question 21

Assume X, Y, Z, W and P are matrices of order $2\times n$, $3\times k$, $2\times p$, $n\times 3$ and $p\times k$ respectively. The restriction on $n$, $k$ and $p$ so that $\mathrm{PY} + \mathrm{WY}$ will be defined are:

Solution. Two conditions must hold: each product must be defined, and the two results must have the same order so they can be added.

$\mathrm{PY}$ is $(p\times k)(3\times k)$, which needs $k = 3$; it is then $p \times k$.

$\mathrm{WY}$ is $(n\times 3)(3\times k)$, which is always defined; it is $n \times k$.

For the sum, $p\times k$ must equal $n\times k$, so $p = n$.

Question 22

If $n = p$, then the order of the matrix $7\mathrm{X} – 5\mathrm{Z}$ is:

Solution. Multiplying by a scalar never changes a matrix’s order, so $7\mathrm{X}$ is $2\times n$ and $5\mathrm{Z}$ is $2\times p$. Since $n = p$, these are the same order and the subtraction is defined, giving a matrix of order

$$2 \times n$$

Common mistakes

• Assuming $\mathrm{AB} = \mathrm{BA}$. They are usually different and sometimes only one is defined. Questions 1 and 14 exist to make this concrete.
• Multiplying entry by entry. Matrix multiplication is row-times-column, not a term-by-term product. Only addition and scalar multiplication work entrywise.
• Adding a scalar to every entry instead of adding $k\mathrm{I}$. In question 15, $6\mathrm{I}$ adds 6 along the diagonal only.
• Ignoring the orders before multiplying. In question 3(ii) a $3\times1$ times a $1\times3$ gives a $3\times3$, not a single number — the reverse order is what gives a number.
• Solving question 12’s equations in the wrong order. $y$ and $z$ depend on $x$ and $w$, so those two have to come first.
• Forgetting to convert dozens in question 20. The prices are per book, so the quantities must be in books: 120, 96, 120.

Practise next

• Exercise 3.3 — transpose, symmetric and skew-symmetric matrices, which build directly on the multiplication practised here.
• Exercise 3.4 — invertible matrices, and the link to Chapter 4.