Introduction to Matrices: Gauss–Jordan Elimination

So far I’ve spoken extensively about calculus, both integration and differentiation, and their respective applications. This post is going to introduce a different field of mathematics called linear algebra. Specifically, this post is going to introduce the use of matrices. In the future we will see how important matrices are in quantum mechanics for complex higher-dimensional abstractions and concept that are impossible to visualize.

I will begin now by introducing a basic example of the intersection of two lines.

Recall in elementary school the following equation for a line.

     \[  y = mx + b \]

For this example we have two lines represented by the following two equations.

     \[  y = 3x + 7 \\ \] \[  y = -2x - 3 \\ \]

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In order to find the point of intersection of these two lines we can perform a simple substitution and solve for x.

     \begin{align*}     3x + 7 &= -2x - 3 \\ 5x + 7 &= -3 \\ 5x &= -10 \\ x &= \frac{-10}{5}\\ x &= -2\\ \end{align*}

Now we can substitute the value of x back into one of the equations in order to solve for y.

     \begin{align*}     y &= 3x + 7  \\  y &= 3(-2) + 7  \\ y &= 1 \\ \end{align*}

Therefore the point of intersection of the two lines is (-2, 1).

As mentioned in linear algebra we often use matrices for a better mathematical representation of systems. A typical example of a matrix is as follows.

     \[      A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}  \]

For this example we can represent this system of equations as a matrix. To begin we must rearrange each equation to the following form.

     \begin{align*}     y &= 3x + 7  \\  - 3x + y &=  7    \end{align*}

     \begin{align*}    y &=  -2x - 3 \\  2x + y &= -3  \\ \end{align*}

Now we can represent the system of equations as the following set of matrices and create an augmented matrix.

     \[  A = \begin{bmatrix}  -3 & 1 \\  2 & 1  \end{bmatrix}  , \quad B =  \begin{bmatrix}  7 \\ -3   \end{bmatrix}  \]

     \[  (A|B)= \left[ \begin{array}{cc|c}  -3 & 1   & 7 \\ 2  & 1  & -3 \end{array} \right] \]

Now that we’ve created the augmented matrix we need to get the matrix into a reduced row echelon form using elementary row operations.

     \begin{align*}   M &= \left[ \begin{array}{cc|c}  -3 & 1  & 7 \\ 2  & 1  & -3 \end{array} \right] \end{align*}

First we subtract row 2 from row 1.

     \begin{align*}    &= \left[ \begin{array}{cc|c}  -3 - (2) & 1 - (1)  & 7 - (-3) \\ 2  & 1  & -3  \end{array} \right] \\  &= \left[ \begin{array}{cc|c}  -5  & 0  & 10 \\ 2  & 1  & -3 \end{array} \right] \end{align*}

Now we divide row 1 by -5.

     \begin{align*}  &= \left[ \begin{array}{cc|c}  \frac{-5}{-5}  & \frac{0}{-5}  & \frac{10}{-5} \\ 2  & 1  & -3 \end{array} \right] \\  &= \left[ \begin{array}{cc|c}  1  & 0  & -2 \\ 2  & 1  & -3 \end{array} \right] \end{align*}

Finally to achieve reduced row echelon form we multiply row 1 by 2 and subtract it from row 2.

     \begin{align*}    &= \left[ \begin{array}{cc|c}  1  & 0  & -2 \\ 2 - (2(1)) & 1 - (2(0))  & -3 - (2(-2)) \end{array} \right] \\  &= \left[ \begin{array}{cc|c}  1  & 0  & -2 \\ 0 & 1  & 1 \end{array} \right] \end{align*}

From looking at this matrix we now have the two equations x = -2 and y = 1 representing the point of intersection of the two lines that we found previously.

For this example using the Guass-Jordan elimination method doesn’t seem practical. But for more complicated systems of equations especially those involving higher dimensions we find that matrices offer a much more practical abstraction.

In the next posts I will discuss matrix operations and their importance into quantum mechanics as well as introduce the concept of eigenvalues and eigenvectors.

Dan Roy, Founder and Creator of Universal Physics.

Category(s): Mathematics
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