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Hello all I learning about determinants and this problem currently has me completely stumped. I can not figure how the answer in the book was achieved.

Problem:

$\det(\lambda I_2 - A)$ where $$ A = \begin{bmatrix} 4 & 2 \\ -1 & 1\\ \end{bmatrix} $$

The book answer is: $ \lambda^2 - 5 \lambda + 6$

I could only figure out how the 6 was found. I would like to note we haven't gone over eigenvalues either.

Thanks.

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There's nothing about eigenvalues here. It's a simple calculation: $$\lambda I_2 - A = \left[\array{\lambda - 4 & -2 \\ 1 & \lambda - 1}\right],$$

so its determinant is $(\lambda - 4)(\lambda - 1) + 2 = \lambda^2 - 5\lambda + 6$ by the standard determinant formula for $2\times 2$ matrices.

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  • $\begingroup$ How did you know how use lambda on the diagnol, my book doesn't mention this notation I believe. $\endgroup$ – user1238097 Jan 9 at 0:23
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    $\begingroup$ Matrix should be $$\begin{bmatrix}\lambda-4&-2\\1&\lambda+1\end{bmatrix}$$ $\endgroup$ – Dave Jan 9 at 0:27
  • $\begingroup$ @Dave are you sure I user3482749 is correct. $\endgroup$ – user1238097 Jan 9 at 0:33
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    $\begingroup$ @user1238097 I am sure. Indeed, $$\lambda I-A=\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}-\begin{bmatrix}4&2\\-1&-1\end{bmatrix}$$ Notice the minus sign in front of $A$. $\endgroup$ – Dave Jan 9 at 1:12
  • $\begingroup$ @Dave you are definitely right. I am at a loss on the book answer still. I realized I added a negative a22 of the matrix on accident. $\endgroup$ – user1238097 Jan 9 at 2:03

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