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Using $\det(A-\lambda{}I)=0$ Find the eigenvalues for the given matrix: $ A=\begin{bmatrix} 1&-1&0&0\\ 3&5&0&0\\ 0&0&1&5\\ 0&0&-1&1\\ \end{bmatrix} $

The patterns in this matrix are obvious, so I am assuming there is a way to simplify this problem without expanding by a row/column, which could become messy really fast (although the abundance of zeros should help.) I just need a little direction.

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    Subtract $\lambda$ from all the diagonal elements in the matrix you have written. There should be a simplification for calculating that determinant staring right back at you.2012-10-21

1 Answers 1

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Hint:

$(1)\;\;\text{When}\,\,X\,,\,Y\,\,\text{square matrices, we have}\;\;\det\begin{pmatrix} X&0\\0&Y\end{pmatrix}=\det X\cdot \det Y$

$(2)\;\;\text{In our particular case}\,\,\,\det(A-\lambda I)=\det\begin{pmatrix}\lambda-1&1&0&0\\-3&\lambda-5&0&0\\0&0&\lambda-1&-5\\0&0&1&\lambda-1\end{pmatrix}$