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
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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