Answer
\[\lambda^2-10\lambda+25=0\]
Work Step by Step
\[A=\left[\begin{array}{cc}7&-2\\
2&3\\
\end{array}\right]\]
Characteristic polynomial of $A$ is given by \[|A-\lambda I|=0\]
\[A-\lambda I=\left[\begin{array}{cc}7&-2\\
2&3\\
\end{array}\right]-\lambda\left[\begin{array}{cc}1&0\\
0&1\\
\end{array}\right]\]
\[A-\lambda I=\left[\begin{array}{cc}7-\lambda&-2\\
2&3-\lambda\\
\end{array}\right]\]
\[|A-\lambda I|=\left|\begin{array}{cc}7-\lambda&-2\\
2&3-\lambda\\
\end{array}\right|\]
\[|A-\lambda I|=(7-\lambda)(3-\lambda)+4\]
\[|A-\lambda I|=21-7\lambda-3\lambda+\lambda^2+4\]
\[|A-\lambda I|=\lambda^2-10\lambda+25\]
Ao characteristic equation of $A$ is
\[\lambda^2-10\lambda+25=0\]
Hence \[\lambda^2-10\lambda+25=0\]
