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If A and B are symmetric matrices, then AB - BA is a
- a)symmetric matrix
- b)skew symmetric matrix
- c)diagonal matrix
- d)null matrix
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If A and B are symmetric matrices, then AB - BA is aa)symmetric matrix...
A and B are symmetric matrices, therefore, we have:
A' = A and B' = B ...(1)
Consider (AB - BA)' = (AB)' - (BA)' [(A - B)' = A' - B']
= B'A' - A'B' [(AB)' = B'A']
= BA - AB [by (1)]
= -(AB - BA)
∴ (AB - BA)' = -(AB - BA)
Thus, (AB - BA) is a skew-symmetric matrix.
A' = A and B' = B ...(1)
Consider (AB - BA)' = (AB)' - (BA)' [(A - B)' = A' - B']
= B'A' - A'B' [(AB)' = B'A']
= BA - AB [by (1)]
= -(AB - BA)
∴ (AB - BA)' = -(AB - BA)
Thus, (AB - BA) is a skew-symmetric matrix.
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Community Answer
If A and B are symmetric matrices, then AB - BA is aa)symmetric matrix...
Proof:
Let A and B be two symmetric matrices of the same order. Then,
AB - BA
= AB - BA (as A and B are symmetric)
= ABB^T - BAB^T (as B^T = B)
= AB(B^T - B^T) - (A - A)B^T (distributive property of matrices)
= AB(B^T - B^T) - null matrix (as A - A = null matrix)
= AB(null matrix) - null matrix
= null matrix - null matrix
= null matrix
Therefore, AB - BA is a skew symmetric matrix.
Let A and B be two symmetric matrices of the same order. Then,
AB - BA
= AB - BA (as A and B are symmetric)
= ABB^T - BAB^T (as B^T = B)
= AB(B^T - B^T) - (A - A)B^T (distributive property of matrices)
= AB(B^T - B^T) - null matrix (as A - A = null matrix)
= AB(null matrix) - null matrix
= null matrix - null matrix
= null matrix
Therefore, AB - BA is a skew symmetric matrix.
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If A and B are symmetric matrices, then AB - BA is aa)symmetric matrixb)skew symmetric matrixc)diagonal matrixd)null matrixCorrect answer is option 'B'. Can you explain this answer?
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