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Question
If A = diag (2 − 59), B = diag (11 − 4) and C = diag (−6 3 4), find
2A + 3B − 5C
Solution
Here,
`A = [[2 0 0],[0 -5 0],[0 0 0]]`
`B=[[1 0 0],[0 1 0],[0 0 -4]]`
and C = `[[-6 0 0],[0 3 0],[0 0 4]]`
`2A+3B-5C`
⇒2A+3B−5C=2 `[[2 0 0],[0 -5 0],[0 0 9]]` +3`[[1 0 0],[0 1 0],[0 0 -4]]` - 5`[[-6 0 0],[0 3 0],[0 0 4]]`
⇒2A+3B−5C= `[[4 0 0],[0 -10 0],[0 0 18]]` + `[[3 0 0],[0 3 0],[0 0 -12]]` - `[[-30 0 0],[0 15 0],[0 0 20]]`
⇒2A+3B−5C= `[[4 +3+30 0+0-0 0+0-0],[0+0-0 -10+3-15 0+0-0],[0+0-0 0+0-0 18-12-20]]`
⇒2A+3B−5C=`[[37 0 0],[0 -22 0],[0 0 -14 ]]`
=diag(37 −22 −14)
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