NDA Maths · Matrices & Determinants

Special Matrices and Their Tell-Tale Properties

Named matrix types — symmetric, skew-symmetric, diagonal, orthogonal, rotation, idempotent, involutory — each carry a defining equation that instantly fixes their determinant and inverse.

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Why this matters

Twenty-two PYQs, mostly EASY–MODERATE — high-yield because each type is recognised by one equation and then answers itself (a skew-symmetric matrix of odd order has determinant 0; an orthogonal matrix has inverse equal to its transpose). Learn the catalog and the recognition questions become instant. The determinant facts used here — odd-order skew-symmetric is 0, orthogonal is plus or minus 1, diagonal is the product of the diagonal — are proved in the next note, Determinants: Evaluation and Properties; here, use them as recognition shortcuts.

Test yourself — a quick 14-question Matrices & Determinants recall quiz, instant score.

Concept 1 of 6

The special-matrix catalog

Intuition

Each special matrix is defined by one equation relating

A

A^T

A^{-1}

, or a power of itself. Recognising that equation is the whole game — it pins down the determinant, the inverse, and the diagonal at a glance. For complex matrices, the conjugate-transpose

(\bar{A})^T

plays the role of

A^T

(Hermitian / skew-Hermitian).

Definition

Memorise the defining property of each type; the exam tests recognition more than computation.

Type	Defining property	Key consequence
Symmetric	$A^T = A$	$a_{ij} = a_{ji}$ ; inverse (if any) is symmetric
Skew-symmetric	$A^T = -A$	diagonal all 0; odd order $\Rightarrow \det = 0$ Odd-order skew-symmetric is ALWAYS singular (det 0). Even-order need not be.
Diagonal	off-diagonal all 0	$\det =$ product of diagonal entries
Orthogonal	$AA^T = I$	$A^{-1} = A^T$ ; $\det = \pm 1$
Idempotent	$A^2 = A$	$\det \in \{0, 1\}$
Involutory	$A^2 = I$	$A^{-1} = A$ ; $\det = \pm 1$
Hermitian	$(\bar{A})^T = A$	complex analogue of symmetric; diagonal entries are real; $A+(\bar{A})^T$ is always Hermitian For a REAL matrix, Hermitian $=$ symmetric. The conjugate-transpose $(\bar{A})^T$ is also written $A^{*}$ or $A^{\dagger}$ .
Skew-Hermitian	$(\bar{A})^T = -A$	complex analogue of skew-symmetric; diagonal entries are 0 or purely imaginary

Recognise the defining equation first; the determinant and inverse follow immediately.

Practice this conceptself-check · 6 quick reps

From the bank · past-year question

Example 1Matrices & DeterminantsEASY

A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

, then the matrix A is a/an

[Q9 · Apr · 2019]

Drill 5 more on the special-matrix catalog

Concept 2 of 6

Symmetric and skew-symmetric matrices

Intuition

Symmetric mirrors across the diagonal (

A^T = A

); skew-symmetric anti-mirrors (

A^T = -A

, forcing zero diagonal). Any square matrix splits uniquely into a symmetric part plus a skew-symmetric part — a decomposition the exam loves.

Definition

$A$ is symmetric if $A^T = A$ , skew-symmetric if $A^T = -A$ (so $a_{ii}=0$ ). Every square matrix decomposes as $A = \tfrac12(A + A^T) + \tfrac12(A - A^T)$ — symmetric part $P$ plus skew part $Q$ . Useful facts: $A + A^T$ is symmetric, $A - A^T$ is skew, $AA^T$ is symmetric; an odd-order skew-symmetric matrix has $\det = 0$ .

Symmetric + skew decomposition

A = \underbrace{\tfrac12(A + A^T)}_{\text{symmetric}} + \underbrace{\tfrac12(A - A^T)}_{\text{skew-symmetric}}

Worked example

Let

A

be a skew-symmetric matrix of order 3. Find

\det A

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Matrices & DeterminantsEASY

P

is a skew-symmetric matrix of order 3, then what is

\det(P)

equal to?

[Q20 · Sep · 2024]

Odd-order skew-symmetric $\Rightarrow \det = 0$ ; diagonal entries are 0

From

A^T = -A

: the diagonal satisfies

a_{ii} = -a_{ii}

, so every diagonal entry is 0. And for ODD order,

\det A = (-1)^n\det A = -\det A

, forcing

\det A = 0

. Don't assume the determinant is unknown — for odd order it is always 0. (Even order need not be.)

Drill 7 more on symmetric and skew-symmetric matrices

Concept 3 of 6

Diagonal, scalar, and identity matrices

Intuition

Diagonal matrices are the friendliest: their determinant is just the product of the diagonal, their powers raise each diagonal entry to that power, and their inverse reciprocates each diagonal entry. A scalar matrix is

kI

; the identity is

1\cdot I

Definition

For a diagonal matrix $D = \text{diag}(d_1,\dots,d_n)$ : $\det D = \prod d_i$ , $D^k = \text{diag}(d_1^k,\dots,d_n^k)$ , and $D^{-1} = \text{diag}(1/d_1,\dots,1/d_n)$ (when all $d_i \neq 0$ ). A scalar matrix is $kI$ ; $\det(kI_n) = k^n$ .

Diagonal determinant and inverse

\det D = \prod_i d_i, \qquad D^{-1} = \operatorname{diag}\!\left(\tfrac{1}{d_1}, \dots, \tfrac{1}{d_n}\right) \qquad D^k = \operatorname{diag}(d_1^k, \dots, d_n^k) \qquad \det(kI_n) = k^n

Worked example

For

A = \begin{pmatrix}3&0&0\\0&2&0\\0&0&5\end{pmatrix}

, find

\det A

and

\det(A^2)

Practice this concept4 quick reps

From the bank · past-year question

Example 3Matrices & DeterminantsEASY

A = \begin{pmatrix}2&0&0\\0&3&0\\0&0&4\end{pmatrix}

, then which of the following statements are correct? (1)

A^n

will always be singular for any positive integer

n

. (2)

A^n

will always be a diagonal matrix for any positive integer

n

. (3)

A^n

will always be a symmetric matrix for any positive integer

n

[Q9 · Sep · 2023]

Concept 4 of 6

Orthogonal matrices

Intuition

An orthogonal matrix has perpendicular unit columns, captured by

AA^T = I

. The huge payoff: its inverse is simply its transpose (no adjoint needed), and its determinant is

\pm 1

Definition

$A$ is orthogonal if $AA^T = A^T A = I$ . Then $A^{-1} = A^T$ and $\det A = \pm 1$ (since $\det(AA^T) = (\det A)^2 = 1$ ). Rotation matrices are the standard example.

Orthogonality

AA^T = I \;\Longrightarrow\; A^{-1} = A^T,\quad \det A = \pm 1 \qquad AA^T = A^T A = I

Worked example

A square matrix

A

is called orthogonal under which condition?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Matrices & DeterminantsEASY

A square matrix A is called orthogonal if (where A' is the transpose of A)

[Q4 · Sep · 2018]

Drill 3 more on orthogonal matrices

Concept 5 of 6

Rotation matrices

Intuition

The rotation matrix

R(\theta) = \begin{pmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{pmatrix}

rotates the plane. Composing rotations adds angles, so

R(\theta)R(\phi) = R(\theta+\phi)

and

R(\theta)^n = R(n\theta)

— no matrix grinding needed.

Definition

$R(\theta) = \begin{pmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{pmatrix}$ is orthogonal with $\det = 1$ . Composition law: $R(\theta)R(\phi) = R(\theta+\phi)$ ; hence $R(\theta)^n = R(n\theta)$ and $R(\theta)^{-1} = R(-\theta) = R(\theta)^T$ .

Rotation composition

R(\theta)\,R(\phi) = R(\theta + \phi), \qquad R(\theta)^n = R(n\theta)

Worked example

A = \begin{pmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{pmatrix}

, what is

A^2

Practice this concept4 quick reps

From the bank · past-year question

Example 5Matrices & DeterminantsEASY

A = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix}

, then what is

AA^T

equal to (where

A^T

is the transpose of

A

[Q30 · Apr · 2017]

Drill 2 more on rotation matrices

Concept 6 of 6

Idempotent and involutory matrices

Intuition

An idempotent matrix is unchanged when squared (

A^2 = A

— like a projection); an involutory matrix is its own inverse (

A^2 = I

). Spot the defining square and the powers collapse instantly.

Definition

Idempotent: $A^2 = A$ (so $A^n = A$ for all $n \ge 1$ ; $\det A \in \{0,1\}$ ). Involutory: $A^2 = I$ (so $A^{-1} = A$ ; even powers are $I$ , odd powers are $A$ ). A common test matrix is the all-ones matrix $J$ , where $J^2 = nJ$ for order $n$ .

Idempotent, involutory, all-ones, nilpotent

A^2 = A \Rightarrow A^n = A \qquad A^2 = I \Rightarrow A^{-1} = A \qquad J_n^2 = nJ_n \qquad A^k = O \text{ (nilpotent)}

Worked example

For

A = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}

, express

A^2

in terms of

A

, and state whether

A

is invertible.

Practice this concept4 quick reps

From the bank · past-year question

Example 6Matrices & DeterminantsMODERATE

Consider the following in respect of the matrix

A=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}

: 1. Inverse of

A

does not exist 2.

A^{3}=A

3A=A^{2}

Which of the above are correct?

[Q50 · Sep · 2021]

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Symmetric and skew-symmetric matrices
Symmetric + skew decomposition
$A = \underbrace{\tfrac12(A + A^T)}_{\text{symmetric}} + \underbrace{\tfrac12(A - A^T)}_{\text{skew-symmetric}}$
Diagonal, scalar, and identity matrices
Diagonal determinant and inverse
$\det D = \prod_i d_i, \qquad D^{-1} = \operatorname{diag}\!\left(\tfrac{1}{d_1}, \dots, \tfrac{1}{d_n}\right) \qquad D^k = \operatorname{diag}(d_1^k, \dots, d_n^k) \qquad \det(kI_n) = k^n$
Orthogonal matrices
Orthogonality
$AA^T = I \;\Longrightarrow\; A^{-1} = A^T,\quad \det A = \pm 1 \qquad AA^T = A^T A = I$
Rotation matrices
Rotation composition
$R(\theta)\,R(\phi) = R(\theta + \phi), \qquad R(\theta)^n = R(n\theta)$
Idempotent and involutory matrices
Idempotent, involutory, all-ones, nilpotent
$A^2 = A \Rightarrow A^n = A \qquad A^2 = I \Rightarrow A^{-1} = A \qquad J_n^2 = nJ_n \qquad A^k = O \text{ (nilpotent)}$

Reference tables (1)

The special-matrix catalog8 rows

Type	Defining property	Key consequence
Symmetric	$A^T = A$	$a_{ij} = a_{ji}$ ; inverse (if any) is symmetric
Skew-symmetric	$A^T = -A$	diagonal all 0; odd order $\Rightarrow \det = 0$ Odd-order skew-symmetric is ALWAYS singular (det 0). Even-order need not be.
Diagonal	off-diagonal all 0	$\det =$ product of diagonal entries
Orthogonal	$AA^T = I$	$A^{-1} = A^T$ ; $\det = \pm 1$
Idempotent	$A^2 = A$	$\det \in \{0, 1\}$
Involutory	$A^2 = I$	$A^{-1} = A$ ; $\det = \pm 1$
Hermitian	$(\bar{A})^T = A$	complex analogue of symmetric; diagonal entries are real; $A+(\bar{A})^T$ is always Hermitian For a REAL matrix, Hermitian $=$ symmetric. The conjugate-transpose $(\bar{A})^T$ is also written $A^{*}$ or $A^{\dagger}$ .
Skew-Hermitian	$(\bar{A})^T = -A$	complex analogue of skew-symmetric; diagonal entries are 0 or purely imaginary

Recognise the defining equation first; the determinant and inverse follow immediately.

Watch out for (1)

Odd-order skew-symmetric $\Rightarrow \det = 0$ ; diagonal entries are 0
→ Symmetric and skew-symmetric matrices

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