NCERT Class 12 Mathematics — Matrices, Exercise 3.3. All 12 questions solved.

Exercise 3.3 introduces the transpose, written $\mathrm{A}’$ or $\mathrm{A}^{T}$: interchange the rows and columns, so an $m\times n$ matrix becomes $n\times m$ and the entry $a_{ij}$ moves to position $(j, i)$.

Four properties do all the work:

$$(\mathrm{A}’)’ = \mathrm{A}, \qquad (k\mathrm{A})’ = k\mathrm{A}’,$$ $$(\mathrm{A}+\mathrm{B})’ = \mathrm{A}’ + \mathrm{B}’, \qquad (\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’$$

A matrix is symmetric if $\mathrm{A}’ = \mathrm{A}$ and skew symmetric if $\mathrm{A}’ = -\mathrm{A}$.

Key insight. Notice which property reverses the order. Addition does not — $(\mathrm{A}+\mathrm{B})’ = \mathrm{A}’+\mathrm{B}’$ — but multiplication does: $(\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’$. Every proof in this exercise and the next turns on that single reversal, and writing $\mathrm{A}’\mathrm{B}’$ by reflex is the most expensive habit you can bring to this chapter.

One more fact worth carrying: a skew symmetric matrix has zeros all down its diagonal, because $a_{ii} = -a_{ii}$ forces $a_{ii} = 0$.

Question 1

Find the transpose of each of the following matrices.

Solution. Rows become columns.

(i) A $3\times1$ column becomes a $1\times3$ row: $\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}’ = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}$

(ii) $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}’ = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

(iii) $\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}’ = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$

Note that the diagonal never moves — only the off-diagonal entries are reflected across it.

Question 2

If $\mathrm{A} = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}$ and $\mathrm{B} = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$, verify that (i) $(\mathrm{A}+\mathrm{B})’ = \mathrm{A}’+\mathrm{B}’$, (ii) $(\mathrm{A}-\mathrm{B})’ = \mathrm{A}’-\mathrm{B}’$.

Solution.

(i) $\mathrm{A}+\mathrm{B} = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$, so

$$(\mathrm{A}+\mathrm{B})’ = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$$

Transposing A and B separately and adding gives the same matrix.

(ii) $\mathrm{A}-\mathrm{B} = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$, so

$$(\mathrm{A}-\mathrm{B})’ = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} = \mathrm{A}’ – \mathrm{B}’$$

Question 3

If $\mathrm{A}’ = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$ and $\mathrm{B} = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$, verify that (i) $(\mathrm{A}+\mathrm{B})’ = \mathrm{A}’+\mathrm{B}’$, (ii) $(\mathrm{A}-\mathrm{B})’ = \mathrm{A}’-\mathrm{B}’$.

Solution. Note that $\mathrm{A}’$ is given, not A. Since $(\mathrm{A}’)’ = \mathrm{A}$:

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

(i) $\mathrm{A}+\mathrm{B} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$, so

$$(\mathrm{A}+\mathrm{B})’ = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix} = \mathrm{A}’ + \mathrm{B}’$$

(ii) $\mathrm{A}-\mathrm{B} = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}$, so

$$(\mathrm{A}-\mathrm{B})’ = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix} = \mathrm{A}’ – \mathrm{B}’$$

Question 4

If $\mathrm{A}’ = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $\mathrm{B} = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, then find $(\mathrm{A}+2\mathrm{B})’$.

Solution. Again $\mathrm{A}’$ is given, so transpose it back first:

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

$$\mathrm{A} + 2\mathrm{B} = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$$

$$(\mathrm{A}+2\mathrm{B})’ = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$$

Question 5

For the matrices A and B, verify that $(\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’$.

Question 5 (i)

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

Solution. A is $3\times1$ and B is $1\times3$, so AB is $3\times3$:

$$\mathrm{AB} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}, \qquad (\mathrm{AB})’ = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$$

For the other side, $\mathrm{B}’$ is $3\times1$ and $\mathrm{A}’$ is $1\times3$, so $\mathrm{B}’\mathrm{A}’$ is $3\times3$ — and equals the same matrix. Note that $\mathrm{A}’\mathrm{B}’$ would be a $1\times3$ times a $3\times1$, a single number: the reversed order is not optional.

Question 5 (ii)

$\mathrm{A} = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$, $\mathrm{B} = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$

Solution.

$$\mathrm{AB} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}, \qquad (\mathrm{AB})’ = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix} = \mathrm{B}’\mathrm{A}’$$

Question 6

Verify that $\mathrm{A}’\mathrm{A} = \mathrm{I}$.

Question 6 (i)

$\mathrm{A} = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

Solution.

$$\mathrm{A}’\mathrm{A} = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$$

$$= \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha – \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha – \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Question 6 (ii)

$\mathrm{A} = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

Solution. The same computation with $\sin$ and $\cos$ exchanged gives $\mathrm{A}’\mathrm{A} = \mathrm{I}$ again.

A matrix with $\mathrm{A}’\mathrm{A} = \mathrm{I}$ is called orthogonal, and its transpose is its inverse. Both matrices here are rotations, which is exactly what you would expect to be undone by rotating back.

Question 7

(i) Show that $\mathrm{A} = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.

Solution. Transposing reflects entries across the diagonal. Here $a_{12} = a_{21} = -1$, $a_{13} = a_{31} = 5$ and $a_{23} = a_{32} = 1$, so nothing moves:

$$\mathrm{A}’ = \mathrm{A}$$

(ii) Show that $\mathrm{A} = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$ is a skew symmetric matrix.

Solution. Transposing gives

$$\mathrm{A}’ = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = -\mathrm{A}$$

The diagonal is all zeros, as it must be for any skew symmetric matrix.

Question 8

For the matrix $\mathrm{A} = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that (i) $(\mathrm{A}+\mathrm{A}’)$ is a symmetric matrix, (ii) $(\mathrm{A}-\mathrm{A}’)$ is a skew symmetric matrix.

Solution. $\mathrm{A}’ = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.

(i) $\mathrm{A}+\mathrm{A}’ = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$ — the off-diagonal entries are equal, so it is symmetric.

(ii) $\mathrm{A}-\mathrm{A}’ = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ — zero diagonal and opposite off-diagonal entries, so it is skew symmetric.

These are not coincidences. In general $(\mathrm{A}+\mathrm{A}’)’ = \mathrm{A}’+\mathrm{A}$, which is the same matrix, and $(\mathrm{A}-\mathrm{A}’)’ = \mathrm{A}’-\mathrm{A}$, which is its negative. Question 10 uses both facts.

Question 9

Find $\frac{1}{2}(\mathrm{A}+\mathrm{A}’)$ and $\frac{1}{2}(\mathrm{A}-\mathrm{A}’)$, when $\mathrm{A} = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.

Solution. Look at A before computing: its diagonal is zero and every off-diagonal pair is opposite, so A is already skew symmetric and $\mathrm{A}’ = -\mathrm{A}$.

$$\frac{1}{2}(\mathrm{A}+\mathrm{A}’) = \frac{1}{2}(\mathrm{A}-\mathrm{A}) = \mathrm{O}$$

$$\frac{1}{2}(\mathrm{A}-\mathrm{A}’) = \frac{1}{2}(\mathrm{A}+\mathrm{A}) = \mathrm{A}$$

This is the extreme case of question 10’s decomposition: a skew symmetric matrix splits into a zero symmetric part and itself.

Question 10

Express the following matrices as the sum of a symmetric and a skew symmetric matrix.

Solution. Every square matrix splits as

$$\mathrm{A} = \underbrace{\tfrac{1}{2}(\mathrm{A}+\mathrm{A}’)}_{\text{symmetric}} + \underbrace{\tfrac{1}{2}(\mathrm{A}-\mathrm{A}’)}_{\text{skew symmetric}}$$

which is question 8’s observation turned into a method. Applying it to each:

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

(ii) $\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$ is already symmetric, so its skew part is zero:

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

(iii) $\begin{bmatrix} 3 & 3 & -1 \ -2 & -2 & 1 \ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \[3pt] \frac{1}{2} & -2 & -2 \[3pt] \frac{-5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \[3pt] \frac{-5}{2} & 0 & 3 \[3pt] \frac{-3}{2} & -3 & 0 \end{bmatrix}$

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

Question 11

If A, B are symmetric matrices of same order, then $\mathrm{AB} – \mathrm{BA}$ is a

Solution. Transpose the whole expression and see what comes back. Use $(\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’$, and then $\mathrm{A}’ = \mathrm{A}$, $\mathrm{B}’ = \mathrm{B}$ since both are symmetric.

$$(\mathrm{AB} – \mathrm{BA})’ = (\mathrm{AB})’ – (\mathrm{BA})’ = \mathrm{B}’\mathrm{A}’ – \mathrm{A}’\mathrm{B}’ = \mathrm{BA} – \mathrm{AB}$$

$$= -(\mathrm{AB} – \mathrm{BA})$$

The transpose is the negative of the original, which is the definition of skew symmetric. Option (C) would require $\mathrm{AB} = \mathrm{BA}$, which symmetric matrices need not satisfy.

Question 12

If $\mathrm{A} = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ and $\mathrm{A} + \mathrm{A}’ = \mathrm{I}$, then the value of $\alpha$ is

Solution.

$$\mathrm{A}’ = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$$

Adding, the sine terms cancel and only the diagonal survives:

$$\mathrm{A} + \mathrm{A}’ = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}$$

Setting this equal to I:

$$2\cos\alpha = 1 \quad\Rightarrow\quad \cos\alpha = \frac{1}{2} \quad\Rightarrow\quad \alpha = \frac{\pi}{3}$$

Checking the alternatives: $\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$, $\cos\pi = -1$ and $\cos\tfrac{3\pi}{2} = 0$, none of which give $\tfrac12$.

Common mistakes

• Writing $(\mathrm{AB})’ = \mathrm{A}’\mathrm{B}’$. It is $\mathrm{B}’\mathrm{A}’$. In question 5(i) the wrong order is not merely a different answer — it is a $1\times1$ where a $3\times3$ was wanted.
• Using the given $\mathrm{A}’$ as though it were A. Questions 3 and 4 both supply the transpose, so the first step is to transpose it back.
• Forgetting that a skew symmetric matrix has zero diagonal. It follows from $a_{ii} = -a_{ii}$, and it is a fast way to check your answer in question 10.
• Dropping the $\tfrac12$ in the decomposition. $\mathrm{A} + \mathrm{A}’$ is symmetric, but the symmetric part of A is $\tfrac12(\mathrm{A}+\mathrm{A}’)$. Without the halves the two pieces sum to $2\mathrm{A}$.
• Assuming symmetric matrices commute. They generally do not, which is exactly why $\mathrm{AB} – \mathrm{BA}$ in question 11 is not the zero matrix.

Practise next

• Exercise 3.4 — invertible matrices, and the condition two matrices must satisfy to be inverses of each other.
• Miscellaneous Exercise on Chapter 3 — proofs combining the transpose rules with symmetry.