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Show that the matrix $$A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$$ satisfies the equation $$A^2-(a+b)A+ (ad-bc) I =0$$

Please help me prove this. I have hard time doing it.

peaceful breeze
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Suan Suan
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  • Sum of the eigen value is $a+b$ and their product is $ad-bc$. Use Cayley-Hamilton Theorem. – Kavi Rama Murthy Mar 02 '21 at 12:02
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    $a+b$ should be $a+d$ (the trace of the matrix). You can calculate $A^2$ by matrix multiplication, then it should be easy to show that the expression is zero. – Blitzer Mar 02 '21 at 12:20

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$$A^2=\begin{bmatrix} a^2+bc & ab+bd\\ ac+cd & ad+bd \end{bmatrix}$$ $$-(a+d)A=\begin{bmatrix} -a^2-ad & -ab-bd\\ -ac-cd & -ad-d^2\end{bmatrix}$$ $$(bc-ad)I=\begin{bmatrix} ad-bc & 0 \\ 0 & bc-ad \end{bmatrix}$$ Add these three results to get $$A^2-(a+d)A+(bc-ad)I=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Note for OP: the middle co-efficiens has to be $(a+d)$ and not $(a+b)$ as written by you. This may be the reason why you could not get it.

Z Ahmed
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