Q. 5

# If A = diag (2, -5, 9), B = diag (1, 1, -4) and C = diag (-6, 3, 4), find

i. A – 2B

ii. B + C – 2A

iii. 2A + 3B – 5C

Answer :

i. A-2B = diag -2diag

= diag-diag

= diag

= diag

Hence, A-2B=diag

ii. B+C-2A

=diag +diag-2diag

=diag+diag-diag

=diag

=diag

Hence, B+C-2A=diag

iii. 2A+3B-5C

=2diag+3diag-5diag

=diag+diag-diag

=diag

=diag

Hence, 2A+3B-5C=diag

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