Linear equations and Matrix

One reason for matrix notation is to provide a shorthand way of writing linear equations.

\(
\begin{align*}
\begin{cases}
3x+2y = 14 \\
5x+3y = 22
\end{cases}
\end{align*}
\)

\( \begin{bmatrix}
3 &2 \\
5 &3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} =
\begin{bmatrix}
14 \\
22
\end{bmatrix} \)

\( AX=B \)

We have a linear map from \( \mathbf{\mathbb {R^2}} \) to \( \mathbf{\mathbb{R^2}} \) and we want to know which vectors x, if any, map to the vector b.

for 2×2 matrices \( A = \begin{bmatrix}a&b \\ c&d\end{bmatrix} \),

The determinant of the matrix \(\det A = ad - bc. \)

If the determinant is zero, the matrix is not invertible.

A matrix with determinant 0 is not invertible. The converse of this turns out to be true as well: a matrix with nonzero determinant is invertible. Thus the determinant gives us a way of finding out whether a matrix can be inverted. (The Princeton Companion to Mathematics - Part III Mathematical Concepts - III.15 Determinants)

\( A^{-1} = \frac {1}{\det A} \begin{bmatrix}
d &-b \\
-c &a
\end{bmatrix} \)

\( X = A^{-1}B \)

\(
X = \frac {1}{-1} \begin{bmatrix}3&-2 \\ -5&3\end{bmatrix}
\begin{bmatrix} 14 \\ 22 \end{bmatrix}
= \begin{bmatrix}-3&2 \\ 5&-3\end{bmatrix}
\begin{bmatrix} 14 \\ 22 \end{bmatrix}
= \begin{bmatrix} 2 \\ 4 \end{bmatrix}
\)

\(
\begin{align*}
\begin{cases}
x = 2 \\
y = 4
\end{cases}
\end{align*}
\)

Row reduction provides a method to compute the inverse of an invertible matrix.

\(
\displaystyle [A|I]=
\left[
{\begin{array}{cc|cc}
3 & 2 & 1 & 0 \\
5 & 3 & 0 & 1
\end{array}}
\right] \longrightarrow
\left[
{\begin{array}{cc|cc}
1 & \frac{2}{3} & \frac{1}{3} & 0 \\
5 & 3 & 0 & 1
\end{array}}
\right] \longrightarrow
\left[
{\begin{array}{cc|cc}
1 & \frac{2}{3} & \frac{1}{3} & 0 \\
0 & -\frac{1}{3} & -\frac{5}{3} & 1
\end{array}}
\right] \longrightarrow
\)
\(
\> \> \> \>
\left[
{\begin{array}{cc|cc}
1 & \frac{2}{3} & \frac{1}{3} & 0 \\
0 & 1 & 5 & -3
\end{array}}
\right] \longrightarrow
\left[
{\begin{array}{cc|cc}
1 & 0 & -3 & 2 \\
0 & 1 & 5 & -3
\end{array}}
\right]
= [I|A^{-1}]
\)