Definitions, examples, and properties of special matrices. A = Aᵀ for symmetric.

Types of Matrices — Symmetric, Skew-Symmetric, Orthogonal

Question

Identify and verify the type of each matrix below:

A = \begin{pmatrix} 2 & 3 & -1 \\ 3 & 0 & 4 \\ -1 & 4 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & -3 & 1 \\ 3 & 0 & -4 \\ -1 & 4 & 0 \end{pmatrix}

Also, state the defining condition for an orthogonal matrix.

Solution — Step by Step

Write $A^T$ by swapping rows and columns — the $(i,j)$ entry of $A^T$ is the $(j,i)$ entry of $A$ .

A^T = \begin{pmatrix} 2 & 3 & -1 \\ 3 & 0 & 4 \\ -1 & 4 & 5 \end{pmatrix}

We see $A^T = A$ , so $A$ is a symmetric matrix.

B^T = \begin{pmatrix} 0 & 3 & -1 \\ -3 & 0 & 4 \\ 1 & -4 & 0 \end{pmatrix}

Compare with $B$ : every entry of $B^T$ equals $-1$ times the corresponding entry of $B$ . That is, $B^T = -B$ . So $B$ is a skew-symmetric matrix.

For a skew-symmetric matrix, the diagonal entries satisfy $a_{ii} = -a_{ii}$ , which forces $a_{ii} = 0$ . Check $B$ : all three diagonal entries are indeed $0$ . This is a mandatory property — if even one diagonal entry is non-zero, the matrix cannot be skew-symmetric.

A square matrix $P$ is orthogonal if:

P^T P = P P^T = I

Equivalently, $P^T = P^{-1}$ . The rows (and columns) of $P$ form an orthonormal set — mutually perpendicular unit vectors. Neither $A$ nor $B$ above is orthogonal.

Why This Works

The key is the transpose operation. Symmetric means the matrix is a mirror image of itself across the main diagonal — so $a_{ij} = a_{ji}$ for all $i, j$ . Geometrically, symmetric matrices represent self-adjoint transformations, which is why they appear so often in physics and engineering (moment of inertia tensors, covariance matrices).

Skew-symmetric means the matrix is the negative mirror image: $a_{ij} = -a_{ji}$ . This forces every diagonal entry to be zero because $a_{ii} = -a_{ii}$ has only one solution. In CBSE 12 and JEE, skew-symmetric matrices almost always appear with this zero-diagonal check as a quick verification step.

Orthogonal matrices are a different beast — they preserve lengths and angles. The condition $P^TP = I$ is the algebraic way of saying “columns are orthonormal.” In JEE, orthogonal matrices mostly appear in rotation problems and the spectral theorem.

Alternative Method

Decomposition approach — any square matrix $M$ can be split uniquely into a symmetric part and a skew-symmetric part:

M = \underbrace{\frac{M + M^T}{2}}_{\text{symmetric}} + \underbrace{\frac{M - M^T}{2}}_{\text{skew-symmetric}}

If $M$ itself equals $\frac{M+M^T}{2}$ , it is symmetric. If it equals $\frac{M-M^T}{2}$ , it is skew-symmetric.

This decomposition is a regular CBSE 12 long-answer question (3–4 marks). Given any matrix, write it as sum of symmetric and skew-symmetric — it’s mechanical once you know the formula.

For matrix $B$ above, compute $\frac{B - B^T}{2}$ :

\frac{B - B^T}{2} = \frac{1}{2}\begin{pmatrix} 0 & -6 & 2 \\ 6 & 0 & -8 \\ -2 & 8 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -3 & 1 \\ 3 & 0 & -4 \\ -1 & 4 & 0 \end{pmatrix} = B

Since $B = \frac{B - B^T}{2}$ , confirmed skew-symmetric. Final answers: $A$ is symmetric, $B$ is skew-symmetric.

Common Mistake

Students often write $A^T = A$ correctly but then forget to actually verify entry by entry — they just say “it looks symmetric” from the diagonal. In exams, the verification step carries marks. Write out $A^T$ explicitly and show the equality. Also, many students confuse the orthogonal condition: they write $P^T = P$ (symmetric) instead of $P^T = P^{-1}$ . Symmetric $\neq$ orthogonal. A symmetric matrix can be orthogonal only if it is also an involution ( $P^2 = I$ ), which is a rare special case.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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