Question
If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then find k so that $A^2= 8A + kI.$
✓
Answer
We have
$\begin{aligned} & A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] \end{aligned} $
$ A^2=A A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] $
$ =\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]$
And $\begin{aligned} & 8 A+k l=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]+k\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \end{aligned} $
$ =\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] $
$ =\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
Thus $A^2=8 A+k l$
$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]=\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
$\Rightarrow 1 = 8 + k$
$\Rightarrow k = -7$
Also $56 + k = 49$
$\Rightarrow k = -7.$
$\begin{aligned} & A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] \end{aligned} $
$ A^2=A A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] $
$ =\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]$
And $\begin{aligned} & 8 A+k l=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]+k\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \end{aligned} $
$ =\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] $
$ =\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
Thus $A^2=8 A+k l$
$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]=\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
$\Rightarrow 1 = 8 + k$
$\Rightarrow k = -7$
Also $56 + k = 49$
$\Rightarrow k = -7.$
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