, where I is the identity matrix (1s on the diagonal, 0s elsewhere)Compute the determinant of A'Set the determinant of A' equal to zero and solve for lambda.![A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]](/tpl/images/0225/7250/9c721.png)
What eigen value for this matix
(1 -2)
(-2 0)...
Answers: 1
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