# Evaluate the determinant with Identity matrix

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.

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.
• Matrix should be $$\begin{bmatrix}\lambda-4&-2\\1&\lambda+1\end{bmatrix}$$ – Dave Jan 9 at 0:27
• @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 $A$. – Dave Jan 9 at 1:12