Linear Algebra Lecture 3: Multiplication and inverse matrices
- Matrix multiplication: (4 ways!)
- Inverse of
$ A, AB, A^T $ - Gaiss-Hordan / Find $A^{-1}$
Given $ A_{m \times n} B_{n \times p} = C_{m \times p}$
Method 1: element view
$=a_{31}b_{14}+a_{32}b_{24}+...
=\sum_{k=1}^n a_{3k}b_{k4}$
Method 2: vector view
$C_{34} = (row\,3\,of\,A) \cdot (column\,4\,of\,B)$
Method 3:
rows of C are combinations of rows of B
columns of C are combinations of columns of A
Method 4:
AB = sum of (columns of A ) x (rows of B)
$$
\begin{bmatrix}
2 & 7 \\
3 & 8 \\
4 & 9
\end{bmatrix}
\begin{bmatrix}
1 & 6 \\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
2 \\
3 \\
4
\end{bmatrix}
\begin{bmatrix}
1 \\
6
\end{bmatrix}
+
\begin{bmatrix}
7 \\
8 \\
9
\end{bmatrix}
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
Block:
Same as plutiplications of elements
$$
\begin{bmatrix}
A_1 & A_2 \\
A_3 & A_4
\end{bmatrix}
\begin{bmatrix}
B_1 & B_2 \\
B_3 & B_4
\end{bmatrix}
=
\begin{bmatrix}
A_1B_1+A_2B_3 & A_1B_2+A_2B_4 \\
A_3B_1+A_4B_2 & A_3B_2+A_4B_4
\end{bmatrix}
$$
Inverse (Square matrices)
$$A^{-1}A = I = A A^{-1}$$
non-invertable and why: e.g.
$$
A =
\begin{bmatrix}
1 & 3 \\
2 & 6
\end{bmatrix}
$$
because I can’t find vector x with Ax = 0.
Gauss-Jordan: find `$A^{-1}###
$$
\begin{bmatrix}
1 & 3 \\
2 & 7
\end{bmatrix}
\begin{bmatrix}
a & c \\
b & d
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
$$
A \qquad A^{-1} \quad =\quad I
$$
We can write this as :
$$
\begin{bmatrix}
1 & 3 \\
2 & 7
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix}
=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & 3 \\
2 & 7
\end{bmatrix}
\begin{bmatrix}
c \\
d
\end{bmatrix}
=
\begin{bmatrix}
0 \\
1
\end{bmatrix}
$$
Let’s bring them together and do the elimations:
$$ \left[
\begin{array}{cc|cc}
1 & 3 & 1 & 0 \\
2 & 7 & 0 & 1
\end{array}
\right]
-> \left[
\begin{array}{cc|cc}
1 & 3 & 1 & 0 \\
0 & 1 & -2 & 2
\end{array}
\right]
->
\left[
\begin{array}{cc|cc}
1 & 0 & 7 & -3 \\
0 & 1 & -2 &1
\end{array}
\right]
$$
$$ A \qquad I \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad I \qquad\qquad A $$
EA = I tells us $ E = A^{-1}$
$$E[A\;I] = [I\;A^{-1}]$$
