📂Matrix Algebra

Augmented Matrices and Elementary Row Operations

Definition¹

Let’s assume a linear system is given as follows.

$$ \begin{equation} \begin{aligned} a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} &= b_{1}\\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} &= b_{2}\\ &\vdots\\ a_{m1}x_{1} + a_{m2}x_{2} + \cdots + a_{mn}x_{n} &= b_{m} \end{aligned} \label{linsys2} \end{equation} $$

The representation of constants of a linear system in a matrix is called an augmented matrix.

$$ \begin{equation} \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} & b_{1} \\ a_{21} & a_{22} & \cdots & a_{2n} & b_{2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} & b_{m} \end{bmatrix} \label{augmented} \end{equation} $$

Description

Matrices were designed to solve systems of linear equations easily. For example, one could extract only the constants of $\eqref{linsys2}$ and represent them like $\eqref{augmented}$ or as follows.

$$ \begin{align*} \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{bmatrix} &= \begin{bmatrix} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{bmatrix} \\ A\mathbf{x} &= \mathbf{b} \end{align*} $$

In particular, when represented in an augmented matrix, the linear system can be solved through elementary row operations, which essentially are the same as the method of addition and subtraction taught in middle school.

Definition

The following three operations are considered elementary row operations.

• Multiply a row by a constant that is not $0$.

• Swap the positions of two rows.

• Add a multiple of one row to another row.

Example

Solving a system of equations through the method of addition and subtraction is akin to modifying the augmented matrix with elementary row operations to leave only one element in the last column of each row. In the language of linear algebra, this is ’transforming the augmented matrix into a reduced row echelon form through elementary row operations'.

$$ \begin{array}{rcrcrcr} x &+& y &+& 2z &=&\ 9 \\ 2x &+& 4y &-& 3z &=&\ 1 \\ 3x &+& 6y &-& 5z &=&\ 0 \end{array} \quad \begin{bmatrix} 1 & 1 & 2 & 9 \\ 2 & 4 & -3 & 1 \\ 3 & 6 & -5 & 0 \end{bmatrix} $$

Multiply the first row by $-2$ and add it to the second row.

$$ \begin{array}{rcrcrcr} x &+& y &+& 2z &=&\ 9 \\ && 2y &-& 7z &=&\ -17 \\ 3x &+& 6y &-& 5z &=&\ 0 \end{array} \quad \begin{bmatrix} 1 & 1 & 2 & 9 \\ 0 & 2 & -7 & -17 \\ 3 & 6 & -5 & 0 \end{bmatrix} $$

Multiply the first row by $-3$ and add it to the third row.

$$ \begin{array}{rcrcrcr} x &+& y &+& 2z &=&\ 9 \\ && 2y &-& 7z &=&\ -17 \\ && 3y &-&11z &=&\ -27 \end{array} \quad \begin{bmatrix} 1 & 1 & 2 & 9 \\ 0 & 2 & -7 & -17 \\ 0 & 3 & -11 & -27 \end{bmatrix} $$

Multiply the second row by $\dfrac{1}{2}$.

$$ \begin{array}{rcrcrcr} x &+& y &+& 2z &=&\ 9 \\ && y &-& \dfrac{7}{2} z &=&\ -\dfrac{17}{2} \\ && 3y &-&11z &=&\ -27 \end{array} \quad \begin{bmatrix} 1 & 1 & 2 & 9 \\ 0 & 1 & -\dfrac{7}{2} & -\dfrac{17}{2} \\ 0 & 3 & -11 & -27 \end{bmatrix} $$

Multiply the second row by $-3$ and add it to the third row.

$$ \begin{array}{rcrcrcr} x &+& y &+& 2z &=&\ 9 \\ && y &-& \dfrac{7}{2} z &=&\ -\dfrac{17}{2} \\ && &-&\dfrac{1}{2}z &=&\ -\dfrac{3}{2} \end{array} \quad \begin{bmatrix} 1 & 1 & 2 & 9 \\ 0 & 1 & -\dfrac{7}{2} & -\dfrac{17}{2} \\ 0 & 0 & -\dfrac{1}{2} & -\dfrac{3}{2} \end{bmatrix} $$

Multiply the third row by $-2$.

$$ \begin{array}{rcrcrcr} x &+& y &+& 2z &=&\ 9 \\ && y &-& \dfrac{7}{2} z &=&\ -\dfrac{17}{2} \\ && &&z &=&\ 3 \end{array} \quad \begin{bmatrix} 1 & 1 & 2 & 9 \\ 0 & 1 & -\dfrac{7}{2} & -\dfrac{17}{2} \\ 0 & 0 & 1 & 3 \end{bmatrix} $$

Multiply the second row by $-1$ and add it to the first row.

$$ \begin{array}{rcrcrcr} x && &+& \dfrac{11}{2}z &=&\ \dfrac{35}{2} \\ && y &-& \dfrac{7}{2} z &=&\ -\dfrac{17}{2} \\ && &&z &=&\ 3 \end{array} \quad \begin{bmatrix} 1 & 0 & \dfrac{11}{2} & \dfrac{35}{2} \\ 0 & 1 & -\dfrac{7}{2} & -\dfrac{17}{2} \\ 0 & 0 & 1 & 3 \end{bmatrix} $$

Multiply the third row by $-\dfrac{11}{2}$ and add it to the first row, then multiply the third row by $\dfrac{7}{2}$ and add it to the first row.

$$ \begin{array}{rcrcrcr} x && && &=&\ 1 \\ && y && &=&\ 2 \\ && &&z &=&\ 3 \end{array} \quad \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{bmatrix} $$

Therefore, the solution of the given linear system is $x=1$, $y=2$, $z=3$.