JEE Mains Maths · Matrices

Symmetric, Skew-Symmetric and Orthogonal Matrices

Three matrix families defined by how A relates to its transpose — symmetric (A = Aᵀ), skew-symmetric (A = −Aᵀ, forcing a zero diagonal), and orthogonal (AAᵀ = I, so A⁻¹ = Aᵀ) — each carrying tell-tale determinant and structure facts JEE tests relentlessly.

All Matrices notes JEE Mains Maths strategy

Why this matters

Fifteen PYQs, every one MODERATE — this is one of the most reliably recurring Matrices themes in JEE Mains. The questions cluster into a few fixed shapes: counting symmetric/skew matrices over an entry set, splitting A into its symmetric and skew parts, tracking whether a product like ABᵀ or A¹³B²⁶ comes out symmetric or skew, the XᵀAX = 0 characterisation of skew matrices, and orthogonal/rotation matrices where the inverse is just the transpose. Learn the five recognition patterns below and most of these answer themselves with a one-line transpose argument rather than grinding entries.

Concept 1 of 5: Definitions, entry patterns, and counting

A symmetric matrix mirrors across its main diagonal (

a_{ij}=a_{ji}

); a skew-symmetric matrix anti-mirrors (

a_{ij}=-a_{ji}

), which forces every diagonal entry to be 0. Because the lower triangle is fully determined by the upper, you only get to choose the diagonal plus the upper off-diagonal entries — that count is what JEE's 'how many such matrices?' problems hinge on.

Definition

$A$ is symmetric if $A^T = A$ and skew-symmetric if $A^T = -A$ (so $a_{ii}=0$ ). For an $n\times n$ matrix the number of free entries is:

symmetric: the $n$ diagonal entries $+$ the $\tfrac{n(n-1)}{2}$ upper off-diagonal entries $= \tfrac{n(n+1)}{2}$ ;
skew-symmetric: only the $\tfrac{n(n-1)}{2}$ upper off-diagonal entries (diagonal is forced to 0).

So if each free entry is chosen from a set of $k$ values, there are $k^{\,n(n+1)/2}$ symmetric matrices — and 0 skew-symmetric matrices unless $0$ is in the value set (the diagonal must be 0).

Free-entry counts for order n

\#\{\text{symmetric}\} = k^{\,n(n+1)/2}, \qquad \#\{\text{skew}\} = \begin{cases} k^{\,n(n-1)/2} & 0 \in \text{value set}\\[2pt] 0 & 0 \notin \text{value set}\end{cases}

norder of the square matrix
knumber of allowed values per entry

Worked example

How many symmetric

2\times 2

matrices have all entries drawn from

\{1,2,3\}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q71Moderate

Example 1 · Matrices · Symmetric, Skew-Symmetric and Orthogonal Matrices

The number of symmetric matrices of order 3 , with all the entries from the set

\mathbf{\{ 0,1,2,3,4,5,6,7,8,9\}}

, is :

A symmetric matrix is NOT determined by $n^2$ free choices

For a symmetric

3\times3

matrix you choose only

\tfrac{3\cdot4}{2}=6

entries (3 diagonal + 3 upper), not

9

. Counting all

9

entries freely over-counts massively — the lower triangle is a mirror, not a free choice.

Skew-symmetric over a set missing $0$ gives ZERO matrices

The diagonal of a skew matrix is forced to

0

. If the allowed value set does not contain

0

(e.g.

\{-3,-2,-1,1,2\}

), the skew count collapses to

0

— a favourite JEE curveball hidden inside a union count.

Drill 3 more on definitions, entry patterns, and counting

Concept 2 of 5: The symmetric-plus-skew decomposition

Every square matrix splits uniquely into a symmetric part plus a skew-symmetric part. The symmetric part is

P=\tfrac12(A+A^T)

and the skew part is

Q=\tfrac12(A-A^T)

. JEE loves asking for

\det P

\det Q

, which for a

2\times2

come out as clean expressions in the entries.

Definition

For any square $A$ , $A = P + Q$ with $P=\tfrac12(A+A^T)$ symmetric and $Q=\tfrac12(A-A^T)$ skew-symmetric; this splitting is unique. For a $2\times2$ $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ : $P=\begin{pmatrix}a&\frac{b+c}{2}\\\frac{b+c}{2}&d\end{pmatrix}$ , $Q=\begin{pmatrix}0&\frac{b-c}{2}\\-\frac{b-c}{2}&0\end{pmatrix}$ , so $\det Q=\left(\tfrac{b-c}{2}\right)^2$ and $\det P = ad-\left(\tfrac{b+c}{2}\right)^2$ .

Unique symmetric + skew split

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

Worked example

Write

A=\begin{pmatrix}4&2\\0&6\end{pmatrix}

P+Q

with

P

symmetric and

Q

skew-symmetric.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 11 · Q47Moderate

Example 2 · Matrices · Symmetric, Skew-Symmetric and Orthogonal Matrices

Let

A =\begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix},a \in R

be written as

P + Q

where

P

is a symmetric matrix and

Q

is skew symmetric matrix. If

det(Q) = 9

, then the modulus of the sum of all possible values of determinant of

P

is equal to:

Concept 3 of 5: Transposing products of symmetric and skew matrices

Almost every 'is this expression symmetric or skew?' question is solved by one move: take the transpose using

(XY)^T=Y^TX^T

, replace

A^T

A

(symmetric) and

B^T

-B

(skew), and read off the overall sign. A crucial parity fact drives it: for skew

B

(B^n)^T=(-B)^n

, so

B^n

is skew when $n$ is odd, symmetric when $n$ is even.

Definition

Use $(XY)^T=Y^TX^T$ and $(X\pm Y)^T=X^T\pm Y^T$ . With $A^T=A$ (symmetric), $B^T=-B$ (skew): powers obey $(A^n)^T=A^n$ (always symmetric) and $(B^n)^T=(-1)^n B^n$ . A result is symmetric if its transpose equals itself, skew if its transpose equals its negative. Handy commutator facts: for symmetric $A$ and skew $B$ , $AB-BA$ is symmetric while $AB+BA$ is skew.

Transpose rules and skew-power parity

(XY)^T = Y^T X^T, \qquad (B^n)^T = (-1)^n B^n \ \ (B\ \text{skew})

Worked example

A

is symmetric and

B

is skew-symmetric (same order), determine whether

AB+BA

is symmetric or skew-symmetric.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q153Moderate

Example 3 · Matrices · Symmetric, Skew-Symmetric and Orthogonal Matrices

Let

A

and

B

be any two

3 \times 3

symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?

$B^n$ flips type with the parity of $n$

For skew

B

B^{26}

is symmetric but

B^{5}

is skew. Students who assume every power of a skew matrix stays skew mis-answer questions like

B^{5}-A^{5}

: since

(B^5)^T=-B^5

and

(A^5)^T=A^5

, the difference is neither symmetric nor skew in general.

A product of symmetric matrices need not be symmetric

(AB)^T=B^TA^T=BA

for symmetric

A,B

— which equals

AB

only if they commute. Don't declare

AB

symmetric automatically; it is symmetric iff

AB=BA

Drill 2 more on transposing products of symmetric and skew matrices

Concept 4 of 5: The quadratic form test for skew-symmetry

If a matrix

A

satisfies

X^TAX=0

for every column vector

X

, that is a strong structural clue:

A

must be skew-symmetric (in particular its diagonal is all zeros). It does NOT mean

A=0

— the vanishing quadratic form only kills the symmetric part of

A

Definition

The scalar $X^TAX$ equals its own transpose $X^TA^TX$ , so $X^T(A+A^T)X=0$ for all $X$ . Since $A+A^T$ is symmetric, its quadratic form vanishing for all $X$ forces $A+A^T=O$ , i.e. $A^T=-A$ . Taking $X=e_i$ (a standard basis vector) directly gives $a_{ii}=0$ .

Quadratic-form characterisation

X^TAX = 0 \ \ \forall X \;\Longrightarrow\; A^T = -A \ \ (A\ \text{skew-symmetric})

Worked example

Verify that

A=\begin{pmatrix}0&2\\-2&0\end{pmatrix}

gives

X^TAX=0

for every

X=\begin{pmatrix}x\\y\end{pmatrix}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q74Moderate

Example 4 · Matrices · Symmetric, Skew-Symmetric and Orthogonal Matrices

Let be a

3 \times 3

matrix such that

X^{T}AX = O

for all nonzero

3 \times 1

matrices

X =\begin{bmatrix} x \\ y \\ z \end{bmatrix}

. If

A\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} 1 \\ 4 \\ - 5 \end{bmatrix},A\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=\begin{bmatrix} 0 \\ 4 \\ - 8 \end{bmatrix}

, and

det(adj(2(\text{ }A + I))) =2^{\alpha}3^{\beta}5^{\gamma},\alpha,\beta,\gamma, \in N

, then

\alpha^{2}+\beta^{2}+\gamma^{2}

$X^TAX=0$ does not mean $A=O$

The vanishing quadratic form only annihilates the symmetric part

A+A^T

. A non-zero skew matrix such as

\begin{pmatrix}0&2\\-2&0\end{pmatrix}

satisfies

X^TAX=0

identically while being far from the zero matrix. Read the condition as 'skew', not 'zero'.

Concept 5 of 5: Orthogonal, rotation, and Cayley-transform matrices

An orthogonal matrix has orthonormal rows/columns, packaged as

AA^T=I

. Two huge payoffs: the inverse is just the transpose (no adjoint needed), and the determinant is

\pm1

. Rotation matrices are the standard example, they compose by adding angles, and a skew matrix run through the Cayley transform

(I+A)(I-A)^{-1}

comes out orthogonal.

Definition

$A$ is orthogonal if $AA^T=A^TA=I$ ; then $A^{-1}=A^T$ and $\det A=\pm1$ (from $(\det A)^2=\det(AA^T)=1$ ). The rotation matrix $f(x)=\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}$ is orthogonal with $\det=1$ , satisfies $f(x)f(y)=f(x+y)$ , and $f(-x)=f(x)^{-1}=f(x)^T$ . If $A$ is skew-symmetric, the Cayley transform $(I+A)(I-A)^{-1}$ is orthogonal; and a symmetric $A$ with $A^2=I$ has orthonormal rows (each row is a unit vector, distinct rows are orthogonal).

Orthogonality and rotation composition

AA^T=I \;\Rightarrow\; A^{-1}=A^T,\ \det A=\pm1 \qquad f(x)f(y)=f(x+y)

Worked example

Show that

A=\begin{pmatrix}\tfrac{3}{5}&\tfrac{4}{5}\\-\tfrac{4}{5}&\tfrac{3}{5}\end{pmatrix}

is orthogonal, and write

A^{-1}

Practice this conceptself-check · 6 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 4 · Q65Moderate

Example 5 · Matrices · Symmetric, Skew-Symmetric and Orthogonal Matrices

If for the matrix,

A=\begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix},AA^{\top}=I_{2}

, then the value of

\alpha^{4}+\beta^{4}

is:

$\det A=\pm1$ , not always $+1$

Orthogonality only pins

(\det A)^2=1

. Proper rotations have

\det=+1

, but reflections are orthogonal with

\det=-1

. Never assume the determinant is

1

unless the matrix is specifically a rotation.

Orthogonal is about $A^T=A^{-1}$ , not $A^T=A$

Don't confuse orthogonal (

AA^T=I

, so

A^T=A^{-1}

) with symmetric (

A^T=A

). A symmetric matrix with

A^2=I

happens to be orthogonal too, but symmetry alone never implies orthogonality.

Drill 5 more on orthogonal, rotation, and cayley-transform matrices

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)

Definitions, entry patterns, and counting
Free-entry counts for order n
$\#\{\text{symmetric}\} = k^{\,n(n+1)/2}, \qquad \#\{\text{skew}\} = \begin{cases} k^{\,n(n-1)/2} & 0 \in \text{value set}\\[2pt] 0 & 0 \notin \text{value set}\end{cases}$
The symmetric-plus-skew decomposition
Unique symmetric + skew split
$A = \underbrace{\tfrac12(A+A^T)}_{P\ \text{symmetric}} + \underbrace{\tfrac12(A-A^T)}_{Q\ \text{skew}}$
Transposing products of symmetric and skew matrices
Transpose rules and skew-power parity
$(XY)^T = Y^T X^T, \qquad (B^n)^T = (-1)^n B^n \ \ (B\ \text{skew})$
The quadratic form test for skew-symmetry
Quadratic-form characterisation
$X^TAX = 0 \ \ \forall X \;\Longrightarrow\; A^T = -A \ \ (A\ \text{skew-symmetric})$
Orthogonal, rotation, and Cayley-transform matrices
Orthogonality and rotation composition
$AA^T=I \;\Rightarrow\; A^{-1}=A^T,\ \det A=\pm1 \qquad f(x)f(y)=f(x+y)$

Watch out for (7)

A symmetric matrix is NOT determined by $n^2$ free choices
→ Definitions, entry patterns, and counting
Skew-symmetric over a set missing $0$ gives ZERO matrices
→ Definitions, entry patterns, and counting
$B^n$ flips type with the parity of $n$
→ Transposing products of symmetric and skew matrices
A product of symmetric matrices need not be symmetric
→ Transposing products of symmetric and skew matrices
$X^TAX=0$ does not mean $A=O$
→ The quadratic form test for skew-symmetry
$\det A=\pm1$ , not always $+1$
→ Orthogonal, rotation, and Cayley-transform matrices
Orthogonal is about $A^T=A^{-1}$ , not $A^T=A$
→ Orthogonal, rotation, and Cayley-transform matrices

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