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If A is a singular matrix, then adj A is
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a
non-singular
b
singular
c
symmetric
d
not defined
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Correct option is B
Let A be of order n × n.Since A × adj A = |A| I∴ |A × adj A| = |A|n⇒ |A| |adj A| = |A|n∴ |adj A| = |A|n–1Since A is singular,∴ |A| = 0.∴ |adj A| = 0.Hence, adj A is singular.Similar Questions
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