A - a ij - n - n and a ij - i 2- j 2 then A is necessarilyA- symmetric matrixB- skew symmetric matrixC- zero matrixD- a unit matrix

The correct option is B skew symmetric matrixa_ji = j^2 i^2 = (i^2 j^2) = a_ij ...

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Transpose of a Matrix

A =[ a ij ] n...

Question

$A = [a_{i j}]_{n \times n}$ and $a_{i j} = i^{2} - j^{2}$ then A is necessarily

a unit matrix

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symmetric matrix

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skew symmetric matrix

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zero matrix

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Similar questions

A = [a_{i j}]_{n \times n}

a_{i j} = i^{2} - j^{2}

A = [a_{i j}]_{n \times n}

a_{i j} = i^{2} - j^{2}

A = [a_{i j}]_{n \times n}

a_{i j} = I^{2} - j^{2}

A = [a_{i j}]_{n \times n}

a_{i j} = i^{2} - j^{2}

A

A + A^{T}

A - A^{T} =

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