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Let $$a$$ denote the element of the $${i^{th}}$$ row and $${j^{th}}$$ column in a $$3 \times 3$$ matrix and let $${a_{ij}} = \, - {a_{ji}}$$ for every i and j then this matrix is an -


A
Orthogonal matrix
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B
singular matrix
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C
matrix whose principal diagonal elements are all zero
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D
skew-symmetric matrix
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Solution

The correct option is B skew-symmetric matrix
If $$a_{ij}$$ is the element of $$i^{th}$$ row and $$j^{th}$$ column and $$a_{ij}=-a_{ji}$$
then lets assume
$$A=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{bmatrix}$$

Where, $$a_{ij}=-a_{ji}$$

$$a_{11}=-a_{11}\implies a_{11}=0$$, similarly $$a_{22}=a_{33}=0$$

and $$a_{21}=-a_{12}, a_{31}=-a_{13}, a_{32}=-a_{23}$$

Putting the values
$$A=\begin{bmatrix} { 0 } & { a }_{ 12 } & { a }_{ 13 } \\ { -a }_{ 12 } & { 0 } & { a }_{ 23 } \\ { -a }_{ 13 } & { -a }_{ 23 } & { 0 } \end{bmatrix}$$

$$ { A }^{ T }=\begin{bmatrix} { 0 } & { -a }_{ 12 } & { -a }_{ 13 } \\ { a }_{ 12 } & { 0 } & { -a }_{ 23 } \\ { a }_{ 13 } & { a }_{ 23 } & {0 } \end{bmatrix}\\$$
and
$$-A=\begin{bmatrix} { 0 } & -{ a }_{ 12 } & -{ a }_{ 13 } \\ { a }_{ 12 } & { 0 } & -{ a }_{ 23 } \\ { a }_{ 13 } & { a }_{ 23 } & { 0 } \end{bmatrix}$$
thus
$$A^T=-A$$

$$\therefore$$ The matrix is a skew-symmetric matrix.

Mathematics

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