The first step into solving for eigenvalues, is adding in a $\displaystyle -\lambda$ along the main diagonal. 

$\displaystyle A=\begin{bmatrix} 5-\lambda & -3 \\ 1& -4-\lambda \end{bmatrix}$

Now the next step to take the determinant.

$\displaystyle \det(A)=(5-\lambda)(-4-\lambda)-(-3)(1)$

$\displaystyle \det(A)=(5-\lambda)(-4-\lambda)+3$

Now lets FOIL, and solve for $\displaystyle \lambda$.

$\displaystyle (5-\lambda)(-4-\lambda)+3=-20-5\lambda+4\lambda+\lambda^2+3$

$\displaystyle =\lambda^2-\lambda-17$

Now lets use the quadratic equation to solve for $\displaystyle \lambda$.

$\displaystyle \lambda=\frac{1\pm\sqrt{(1)^2-4(1)(-17)}}{2(1)}$

$\displaystyle \lambda=\frac{1\pm\sqrt{1+68}}{2}$

$\displaystyle \lambda=\frac{1\pm\sqrt{69}}{2}$

So our eigen values are

$\displaystyle \lambda=\frac{1+ \sqrt{69}}{2}$

$\displaystyle \lambda=\frac{1- \sqrt{69}}{2}$

The eigenvalues, $\displaystyle \lambda$,  for the matrix are values for which the determinant of $\displaystyle \begin{bmatrix} 2- \lambda & 1\\ 3 & 4 - \lambda \end{bmatrix}$ is equal to zero. First, find the determinant:

$\displaystyle (2-\lambda)(4-\lambda) - (3)(1) = 8 - 4 \lambda - 2 \lambda + \lambda ^2 - 3 = \lambda ^2 - 6 \lambda + 5$

Now set the determinant equal to zero and solve this quadratic:

$\displaystyle \lambda^2 - 6 \lambda + 5 = 0$ this can be factored:

$\displaystyle (\lambda - 5 ) (\lambda - 1 ) = 0$

The eigenvalues are 5 and 1.

To determine if something is an eignevector, multiply times A:

$\displaystyle \begin{bmatrix} 2 &1 \\3 & 4\end{bmatrix} \begin{bmatrix} 1\\ 3\end{bmatrix} = \begin{bmatrix} 15\\ 5\end{bmatrix}$

Since this is equivalent to $\displaystyle 5\begin{bmatrix} 1\\ 3\end{bmatrix}$, $\displaystyle \begin{bmatrix} 1\\ 3\end{bmatrix}$ is an eigenvector (and 5 is an eigenvalue).

$\displaystyle \begin{bmatrix} 2 & 1\\ 3& 4\end{bmatrix} \begin{bmatrix} -3\\1 \end{bmatrix} = \begin{bmatrix} -5\\ -5\end{bmatrix}$

This cannot be re-written as $\displaystyle \mathbf{v}$ times a scalar, so this is not an eigenvector.

The eigenvalues are scalar quantities, $\displaystyle \lambda$, where the determinant of $\displaystyle \begin{bmatrix} -3-\lambda & 6 \\ 2& 1-\lambda \end{bmatrix}$ is equal to zero.

First, find an expression for the determinant:

$\displaystyle (-3-\lambda)(1-\lambda) - 12$

$\displaystyle =-3 - \lambda + 3 \lambda + \lambda ^2 - 12 = \lambda ^2 + 2 \lambda -15$

Now set this equal to zero, and solve:

$\displaystyle \lambda^2 + 2 \lambda -15 = 0$ this can be factored (or solved in another way)

$\displaystyle (\lambda+ 5) (\lambda-3) = 0$

The eigenvalues are -5 and 3.

To determine if something is an eigenvector, multiply by the matrix A: 

$\displaystyle \begin{bmatrix} -3 & 6\\ 2& 1\end{bmatrix} \begin{bmatrix} -3\\ 1\end{bmatrix} = \begin{bmatrix} 15\\ -5 \end{bmatrix}$

This is equivalent to $\displaystyle -5\begin{bmatrix} -3\\ 1\end{bmatrix}$ so this is an eigenvector.

$\displaystyle \begin{bmatrix} -3 & 6\\ 2& 1\end{bmatrix} \begin{bmatrix} 1\\ 1\end{bmatrix} = \begin{bmatrix} 3\\ 3 \end{bmatrix}$

This is equivalent to $\displaystyle 3\begin{bmatrix} 1\\ 1\end{bmatrix}$ so this is also an eigenvector.

The eigenvalues are scalar quantities $\displaystyle \lambda$ where the determinant of $\displaystyle \begin{bmatrix} -5-\lambda & 2\\ 3& -4-\lambda\end{bmatrix}$ is equal to zero. First, write an expression for the determinant:

$\displaystyle (-5-\lambda) (-4-\lambda) -6 = 0$

$\displaystyle 20 + 4\lambda+5 \lambda + \lambda^2 - 6 = 0$

$\displaystyle \lambda^2 + 9 \lambda +14 = 0$ this can be solved by factoring:

$\displaystyle (\lambda+2) (\lambda+7) = 0$

The solutions are -2 and -7

To determine if a vector is an eigenvector, multiply with A:

$\displaystyle \begin{bmatrix} -5 & 2\\ 3& -4 \end{bmatrix} \begin{bmatrix} -2\\ 3\end{bmatrix} = \begin{bmatrix} 16\\-18 \end{bmatrix}$. This cannot be expressed as an integer times $\displaystyle \mathbf{u}$, so $\displaystyle \mathbf{u}$ is not an eigenvector

$\displaystyle \begin{bmatrix} -5 & 2\\ 3& -4 \end{bmatrix} \begin{bmatrix} 2\\ 3\end{bmatrix} = \begin{bmatrix} -4\\-6 \end{bmatrix}$ This can be expressed as $\displaystyle -2\begin{bmatrix} 2\\ 3\end{bmatrix}$, so $\displaystyle \mathbf{v}$ is an eigenvector.

$\displaystyle \begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}6&-13\\8&-19\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}6 - \lambda &-13\\8&- \lambda - 19\end{bmatrix} \\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&13\lambda + \lambda ^{2} - 10\\&\lambda_{1}=0.73;\lambda_{2}=-13.73\end{align*}$

$\displaystyle \begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}0&-2\\6&9\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}-\lambda&-2\\6&9 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&\lambda ^{2} - 9\lambda + 12\\&\lambda_{1}=1.63;\lambda_{2}=7.37\end{align*}$

$\displaystyle \begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-7&3&-11\\10&-10&0\\8&16&19\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 7&3&-11\\10&- \lambda - 10&0\\8&16&19 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{From that, we can find the eigenvalues:}\\&(-1)\cdot (\lambda ^{3} - 2\lambda ^{2} - 195\lambda + 1880)\\&\lambda_{1}=9.29+5.19i;\lambda_{2}=9.29-5.19i;\lambda_{3}=-16.59\end{align*}$