Find the matrix $A^{15}$.

Question

Let $I= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$ and $O=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$.

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1.Let $A=\begin{pmatrix} 1 & 3 \\ 3 & 5 \\ \end{pmatrix}$ and $ B=\begin{pmatrix} x & 3 \\ 3 & 6 \\ \end{pmatrix}$. Find the value of $x$ which satisfies $AB=BA$.

$AB=\begin{pmatrix} 1 & 3 \\ 3 & 5 \\ \end{pmatrix}\cdot\begin{pmatrix} x & 3 \\ 3 & 6 \\ \end{pmatrix}=\begin{pmatrix} x+9 & 21 \\ 3x+15 & 39 \\ \end{pmatrix}$

$BA=\begin{pmatrix} x & 3 \\ 3 & 6 \\ \end{pmatrix}\cdot\begin{pmatrix} 1 & 3 \\ 3 & 5 \\ \end{pmatrix}=\begin{pmatrix} x+9 & 3x+15 \\ 21 & 39 \\ \end{pmatrix}$ .So that we get $3x+15=21 \Rightarrow x=2$

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2.Let $A=\begin{pmatrix} 1 & 2 \\ 2 & 4 \\ \end{pmatrix}$ and $ B=\begin{pmatrix} -2 & x \\ 4 & y \\ \end{pmatrix}$.Find the values of x and y which satify $BA=O$. $BA=\begin{pmatrix} -2 & x \\ 4 & y \\ \end{pmatrix}\cdot\begin{pmatrix} 1 & 2 \\ 2 & 4 \\ \end{pmatrix}=\begin{pmatrix} -2+2x & 0 \\ 4+2y & 8+4y \\ \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$

S0 that we can get $x=1,y=-2$.

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3.Let $A$ satisfying $A^2=A-I$. Find $A^{15}$.

Please help to show me about this. Thank you in advance!

Answer 1

You may proceed as follows:

From the given condition we have

$$A-A^2 = A(I-A) =I$$

It follows

\begin{eqnarray} A^{15} & = & A(A^2)^7 \\ & = & A(A-I)^7 \\ & = & -(A-I)^6 \\ & = & -(A^2-2A + I)^3 \\ & = & -(-A)^3 \\ & = & A(A-I) \\ & = & -I \end{eqnarray}

Another way would be seeing

$$A^3+I = (A+I)(A^2-A+I)=O \Rightarrow A^3 = -I$$

Answer 2

If you recognize $x^2-x+1$ as a cyclotomic polynomial, then $$x^{15} + 1 = (x^3+1)a(x) = (x^2-x+1)b(x)$$ gives $A^{15}+I=0$.

The systematic way, which does not need insights, is to use polynomial division: $$ x^{15}=(x^2-x+1)q(x)-1 $$ where $q(x)=x^{13} + x^{12} - x^{10} - x^9 + x^7 + x^6 - x^4 - x^3 + x + 1$ is not really relevant. Only the remainder matters here.