If matrix A - a ij -3 - 3- matrix B - b ij -3 - 3 where a ij - a ji -0 and b ij - b ji -0- then A 4- B 3 isA- skew-symmetric matrixB- singularC- symmetricD- zero matrix

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Standard X

Mathematics

Transpose of a Matrix

If matrix A =...

Question

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

skew-symmetric matrix

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singular

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symmetric

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zero matrix

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Solution

The correct option is B singular
Since matrix A is skew-symmetric,
$∴ | A | = 0$
$∴ | A^{4} . B^{3} | = 0$

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Similar questions

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

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

and

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

then A is necessarily

Q. If A is a skew-symmetric matrix of order 3, then the matrix

A^{4}

Q. Assertion :The determinants of a matrix

A = [a_{i j}]_{5 \times 5}

where

a_{i j} + a_{j i} = 0

for each i and j is zero.
Because Reason: The determinant of a skew symmetric matrix of odd order is zero.

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If matrix A=[aij]3×3,matrix B=[bij]3×3 where aij+aji=0 and bij−bji=0,then A4.B3 is

The correct option is B singularSince matrix A is skew-symmetric, ∴|A|=0 ∴|A4.B3|=0

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

The correct option is B singular
Since matrix A is skew-symmetric,
$∴ | A | = 0$
$∴ | A^{4} . B^{3} | = 0$