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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 2A + B – 5C
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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)]`
2A + B – 5C = 2 diag [2 – 3 – 5] + diag [4 – 6 – 3] – 5 diag [ –3 4 1]
`= 2[(2, 0, 0),(0, -3, 0),(0, 0, -5)] + [(4, 0, 0),(0, -6, 0),(0, 0, -3)] -5[(-3, 0, 0),(0, 4, 0),(0, 0, 1)]`
`= [(4, 0, 0),(0, -6, 0),(0, 0, -10)] + [(4, 0, 0),(0, -6, 0),(0, 0, -3)] - [(-15, 0, 0),(0, 20, 0),(0, 0, 5)]`
`= [(4 + 4 - (-15), 0, 0),(0, -6 - 6 - 20, 0),(0, 0, -10 - 3 - 5)]`
`= [(23, 0, 0),(0, -32, 0),(0, 0, -18)]`
= diag [23 – 32 – 18].
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