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Exercise 7.1 · Q12

Q.If AA is a square matrix such that A2=AA^2=A, find the value of 7A−(I+A)37A-(I+A)^3.

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Since II commutes with every matrix, (I+A)3(I+A)^3 can be expanded exactly like an ordinary binomial cube; using A2=AA^2=A (so A3=AA^3=A too) collapses it to I+7AI+7A.

Step 1. Expand (I+A)3(I+A)^3 as a binomial. Because IA=AI=AIA=AI=A, II and AA commute, so the usual binomial expansion is valid:

(I+A)3=I3+3I2A+3IA2+A3=I+3A+3A2+A3.(I+A)^3=I^3+3I^2A+3IA^2+A^3=I+3A+3A^2+A^3. …

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