If in a square matrix A-aij- we find that aij-aji- i-j then A is a-

$I f m a t r i x A = [a_{i j}]_{3 \times 3}, m a t r i x B = [b_{i j}]_{3 \times 3} w h e r e a_{i j} + a_{j i} = 0 a n d b_{i j} - b_{j i} = 0, t h e n | A^{4} . B^{3} | i s$

I f m a t r i x A = [a_{i j}]_{3 \times 3}, m a t r i x B = [b_{i j}]_{3 \times 3} w h e r e a_{i j} + a_{j i} = 0 a n d b_{i j} - b_{j i} = 0, t h e n A^{4} . B^{3} i s

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If in a square matrix A=[aij], we find that aij=aji∀i,j then A is a:

The correct option is A Symmetric matrixBy changing aij→aji weare exchanging rows and columns⇒ Transposing the matrix, givenaij=aji⇒A=AT⇒ A is a symmetric matrix ⇒(A)

If in a square matrix $A = [a_{i j}]$ , we find that $a_{i j} = a_{j i} \forall i, j$ then $A$ is a:

The correct option is A Symmetric matrix
By changing $a_{i j} \to a_{j i}$ we

are exchanging rows and columns

$\Rightarrow$ Transposing the matrix, given

$a_{i j} = a_{j i} \Rightarrow A = A^{T}$

$\Rightarrow$ A is a symmetric

matrix $\Rightarrow (A)$