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Question
Answer the following question:
If A = diag [2 –3 –5], B = diag [4 –6 –3] and C = diag [–3 4 1] then find B + C – A
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Solution
A = diag [2 –3 –5]
∴ A = `[(2, 0, 0),(0, -3, 0),(0, 0, -5)]`
B = diag [4 –6 –3]
∴ B = `[(4, 0, 0),(0, -6, 0),(0, 0, -3)]`
C = diag [–3 4 1]
∴ C = `[(-3, 0, 0),(0, 4, 0),(0, 0, 1)]`
B + C – A = `[(4, 0, 0),(0, -6, 0),(0, 0, -3)] + [(-3, 0, 0),(0, 4, 0),(0, 0, 1)] - [(2,0, 0),(0, -3, 0),(0, 0, -5)]`
= `[(4 - 3 - 2, 0, 0),(0, -6 + 4 + 3, 0),(0, 0, -3 + 1 + 5)]`
= `[(-1, 0, 0),(0, 1, 0),(0, 0, 3)]`
= diag [–1, 1, 3]
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