Question
$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^2 - 5A + 7 I_2 = 0$.
✓
Answer
We have
$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right] $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\ -5 A =\left[\begin{array}{cc}(-5) \cdot 3 & (-5) \cdot 1 \\ (-5) \cdot(-1) & (-5) \cdot 2\end{array}\right] $
$ =\left[\begin{array}{cc}-5 & -5 \\ 5 & -10\end{array}\right]$
$ -5 A =\left[\begin{array}{cc} (-5) \cdot 3 & (-5) \cdot 1 \\ (-5) \cdot(-1) & (-5) \cdot 2
\end{array}\right] $
$=\left[\begin{array}{cc} -5 & -5 \\ 5 & -10 \end{array}\right] 7 I _2 =7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]$
So $A^2-5 A+71_2$
$=\left[\begin{array}{cr} 8 & 5 \\ -5 & 3 \end{array}\right]+\left[\begin{array}{cc} -15 & -5 \\ 5 & -10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] $
$=\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\
0 & 0 \end{array}\right]$
So $A^2 - 5A + 7I_2 = 0$.
Hence proved.
$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right] $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\ -5 A =\left[\begin{array}{cc}(-5) \cdot 3 & (-5) \cdot 1 \\ (-5) \cdot(-1) & (-5) \cdot 2\end{array}\right] $
$ =\left[\begin{array}{cc}-5 & -5 \\ 5 & -10\end{array}\right]$
$ -5 A =\left[\begin{array}{cc} (-5) \cdot 3 & (-5) \cdot 1 \\ (-5) \cdot(-1) & (-5) \cdot 2
\end{array}\right] $
$=\left[\begin{array}{cc} -5 & -5 \\ 5 & -10 \end{array}\right] 7 I _2 =7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]$
So $A^2-5 A+71_2$
$=\left[\begin{array}{cr} 8 & 5 \\ -5 & 3 \end{array}\right]+\left[\begin{array}{cc} -15 & -5 \\ 5 & -10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] $
$=\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\
0 & 0 \end{array}\right]$
So $A^2 - 5A + 7I_2 = 0$.
Hence proved.
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