Linear algebra is the branch of mathematics dealing with vector spaces and linear mappings between those spaces. It is a powerful tool for solving simultaneous linear equations because it allows us to represent the problem as a matrix, and use matrix operations to solve it efficiently.
Let’s consider a system of linear equations with n variables and n equations. The general form of such a system can be written as:
$$\begin{aligned}
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\
&\vdots \\
a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n &= b_n\end{aligned}$$
We can use matrices to represent this system. Let A be the matrix of coefficients, X the matrix of variables, and B the matrix of constants:
$$A = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{bmatrix},
X = \begin{bmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{bmatrix},
B = \begin{bmatrix}
b_1 \\
b_2 \\
\vdots \\
b_n
\end{bmatrix}$$
With this representation, we can rewrite the system of linear equations as a matrix equation:
AX = B
We can then use methods like Gaussian elimination, LU-factorization, or QR-factorization to solve the matrix equation and find the variables. One common method is the Gaussian elimination which involves applying a series of row operations to bring the augmented matrix [A | B] into row-echelon form or reduced row-echelon form.
Let’s consider a simple example to illustrate the use of linear algebra in solving simultaneous linear equations. Suppose we have the following system of equations:
$$\begin{aligned}
2x_1 + 3x_2 &= 5 \\
4x_1 + 9x_2 &= 15\end{aligned}$$
We write this system in matrix form as follows:
$$A = \begin{bmatrix}
2 & 3 \\
4 & 9
\end{bmatrix},
X = \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix},
B = \begin{bmatrix}
5 \\
15
\end{bmatrix}$$
We form the augmented matrix [A | B]:
$$\begin{bmatrix}
2 & 3 & | & 5 \\
4 & 9 & | & 15
\end{bmatrix}$$
We apply Gaussian elimination by first dividing the first row by 2, and then using it to eliminate the 4 in the second row:
$$\begin{bmatrix}
1 & 1.5 & | & 2.5 \\
0 & 3 & | & 10
\end{bmatrix}$$
Finally, we divide the second row by 3, and back-substitute to find the solution:
$$\begin{bmatrix}
1 & 0 & | & 1 \\
0 & 1 & | & 10/3
\end{bmatrix}$$
$$X = \begin{bmatrix}
1 \\
10/3
\end{bmatrix}$$
Thus, the solution to the system of linear equations is x1 = 1 and x2 = 10/3.