Linear Algebra & Geometry Notes
Determinant of a matrix
2$\times$2 matrices
If \(A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\)
Then the determinant of the matrix is equal to $ad - bc$
$det(A) = \begin{vmatrix} A \end{vmatrix} = \begin{vmatrix} a & b\\ c & d \end{vmatrix} = a \times d - b \times c$
3$\times$3 matrices
Laplace Expansion along any of the rows or columns
If $A = \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix} $
e.g.
$det(A) = \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f\\ h & i \end{vmatrix} + b \begin{vmatrix} d & f\\ g & i \end{vmatrix} + c \begin{vmatrix} d & e\\ g & h \end{vmatrix} $
Note:
This is just an example where the Laplace expansion was done alone the first row. But, the Laplace exapansion can be done along any of the rows or the columns. It is generally better to do it along the row or column which has the most zeros, since that will yield the easiest calculation
How to invert a matrix
2$\times$2 matrices
$A = \begin{bmatrix} a & b\\ c & d \end{bmatrix} $
The inverse is calculated with the following formula:
$ A^{-1} = \dfrac{1}{\begin{vmatrix} det(A) \end{vmatrix} } \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} $
3$\times$3 matrices
$A = \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix} $
The inverse is calculated with the following formula:
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Calculate the matrix of co-factors
Matrix of co-factors \(= \begin{bmatrix} \begin{vmatrix} e & f\\\ h & i \end{vmatrix} & \begin{vmatrix} d & f\\\ g & i \end{vmatrix} & \begin{vmatrix} d & e\\\ g & h \end{vmatrix}\\ \\ \begin{vmatrix} b & c\\\ h & i \end{vmatrix} & \begin{vmatrix} a & c\\\ g & i \end{vmatrix} & \begin{vmatrix} a & b\\\ g & h \end{vmatrix}\\ \\ \begin{vmatrix} b & c\\\ e & f \end{vmatrix} & \begin{vmatrix} a & c\\\ d & f \end{vmatrix} & \begin{vmatrix} a & b\\\ d & e \end{vmatrix}\\ \end{bmatrix}\) -
Get the adjugate or adjoint matrix from the matrix of co-factors
$adj(A) = A^T$ -
Multiply it by the signs corresponding to the $i, j$ positions
$a_{i,j} = -1^{i + j}$
- Lastly multiply by the reciprocal of the determinant
$A^{-1} = \dfrac{1}{\begin{vmatrix} det(A) \end{vmatrix} } \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix} $
Dot product
Cross product
Wikipedia Lovely tedious calculations that abuses standard math notation
Box/Mixed product
Wikipedia Half of Box prod will give the area of the tetahedron made by joining the three point. Basically half the area of the parallelopiped.
Cauchy-Schawartz inequality
$\begin{Vmatrix} \vec{x} . \vec{y} \end{Vmatrix} \le \begin{Vmatrix} \vec{x} \end{Vmatrix} + \begin{Vmatrix} \vec{y} \end{Vmatrix} $
The double pipes stand for the magnitude of the vectors which is calculated with pythagoras’ theorem in $\R^3$
note: The double pipes should not be confused with the single pipes of a matrix which stands for the determinant of a matrix
This is kindof like a triangle inequation and it is rearranged and used to calculate angles
Ortholinear projection
Solivng for intersections
Find line from two planes
Steps:
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Find the normals to the two planes
For this you just have to take a look at the coefficients of the cartesian equations of the planes
e.g.:
Plane:$ \pi_1 = x + 2y - 3z = 1,$ the corresponding normal vector is $ \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} $
The normal vector is usual denoted by a $\vec{n}$ -
After you have both normal vectors you have to cross product them to get the vector of the line
This works because the vector of the line is perpendicular to the normals of both planes -
Lastly you have to find a point which lies in the intersection of both planes (same as saying on the line), so that you can anchor the line vector to a point in 3d space.
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Now you have everything to compose the equation of the line. Just convert the point on the line to a vector coming from the origin and make the line vector parametric to give you result like the following:
Line: $l = \vec{x_0} + t \vec{x}$
Youtube video going through the steps
Eigen Values
It shows a trick to solve for eigen values