Explain how the proof is done

Question

A solution of matrix problem appears to be as follows

some one explain the following in the solution

why is A cube is eliminated and fourth power of A is obtained?

In the seventh line

In the third term why the power of A is not expressed in variable but in a constant?

Why it is again multiplied by square of A and fourth power of A is eliminated?

someone help me please

Answer 1

following from what @bob.sacamento wrote you will notice that: $$ \begin{align} A^2(A^2-abI) &= A^4-abA^2 \\ &= abA^2+A^3-abA -abA^2\\ &= abA+A^2-abI-abA \\ &=A^2 -abI \end{align} $$ with the help of this identity the inductive step follows easily. thus if $$ A^p-A^{p-2}-A^2+abI =0 $$ we may write this as: $$ A^{p-2}(A^2-I) = A^2 - abI $$ multiplying by $A^2$ and using the result above gives: $$ A^p(A^2-I) = A^2 - abI $$

Answer 2

Not sure of what formula has to be proved, but here's how it can be proved:

First question:

Multiply the first equality: $$A^3=ab A+A^2-ab I$$ by $A$: $$A^4=ab A^2+A^3-abA,$$ and add, simplifying what has to be simplified: $$A^4=abA^2+A^2-abI.$$

Second question:

Likewise, multiply: $$A^p-abA^{p-2}-A^2+abI=0$$ by $A^2$: $$A^{p+2}-abA^p-A^4+abA^2=0$$ and add $A^4-abA^2-A^2+abI=0$ ($1$st equation).