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  • If A is square matrix such that A^2 = A, show that (I + A)^3 = 7A + I.

If A is square matrix such that A^2 = A, show that (I + A)^3 = 7A + I.

Answers (1)

We are given that,

\\A\textsuperscript{2} = A \\\because (a+b)\textsuperscript{3} = a\textsuperscript{3} + b\textsuperscript{3} + 3a\textsuperscript{2}b + 3ab\textsuperscript{2} \\As, (I + A)\textsuperscript{3} = I\textsuperscript{3} + A\textsuperscript{3} + 3I\textsuperscript{2}A + 3IA\textsuperscript{2} \\\ \text{because I is an identity matrix} \\ \therefore I\textsuperscript{3} = I\textsuperscript{2} = I \\ \therefore (I + A)\textsuperscript{3} = I + A\textsuperscript{3} + 3IA + 3IA

As, I is an identity matrix.

\\ \therefore IA = AI = A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A\textsuperscript{3} + 6IA \\\because A\textsuperscript{2} = A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A\textsuperscript{2}.A + 6A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A.A + 6A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A\textsuperscript{2} + 6A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A + 6A = I + 7A

Hence proved,

(I + A)\textsuperscript{3} = I + 7A

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