First, it must be established that $\displaystyle A- B$ is defined. This is the case if and only $\displaystyle A$ and $\displaystyle B$ have the same number of rows, which is true, and they have the same number of columns, which is also true. $\displaystyle A -B$ is therefore defined.

Subtraction of two matrices is performed by subtracting corresponding elements, so

$\displaystyle A-B= \begin{bmatrix}- \frac{4}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& \frac{3}{11} \end{bmatrix}- \begin{bmatrix} \frac{5}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& -\frac{8}{11} \end{bmatrix}$

$\displaystyle = \begin{bmatrix}- \frac{4}{7}-\frac{5}{7} & \frac{2}{9} - \frac{2}{9} \\ \\ -\frac{5}{6} - \left ( -\frac{5}{6} \right )& \frac{3}{11} -\left ( -\frac{8}{11} \right )\end{bmatrix}$

$\displaystyle = \begin{bmatrix} {\color{Red} -1} & 0 \\ 0 & 1 \end{bmatrix}$

However, $\displaystyle I = \begin{bmatrix} {\color{Red} 1 }& 0 \\ 0 & 1 \end{bmatrix}$.

Therefore, $\displaystyle A-B \ne I$.

By Cramer's rule, the value of $\displaystyle x$ is equal to $\displaystyle x= \frac{D_{x}}{D}$, where $\displaystyle D$ is the determinant of the matrix of coefficients

$\displaystyle D =\begin{vmatrix} 2 & 5 \\ 4 & -1 \end{vmatrix}$,

and $\displaystyle D_{x}$ is the same matrix with the x-coefficients replaced by the constants:

$\displaystyle D_{x}= \begin{vmatrix} 17 & 5 \\ 13 & -1 \end{vmatrix}$

The determinant of a two-by-two matrix is equal to the product of the entries in the main diagonal minus the product of the other two entries. Therefore,

$\displaystyle D_{x}= \begin{vmatrix} 17 & 5 \\ 13 & -1 \end{vmatrix} = 17(-1)-5(13)= -17 -65 = -82$

$\displaystyle D= \begin{vmatrix} 2 & 5 \\ 4 & -1 \end{vmatrix} = 2(-1) - 5(4) = -2 - 20 = -22$

and

$\displaystyle x= \frac{D_{x}}{D} = \frac{-82}{-22} = \frac{41}{11}$.

First, the pivot element must be identified. Select the pivot column by spotting the "most negative" element in the bottom, or objective, row. This is the $\displaystyle -2$ in Column 3, making that the pivot column.

$\displaystyle \begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ 1 & 2 & 4 & 0 & 1 & 0 & 600\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & {\color{Red} \textbf{-2}} & 0&0&0 & 0 \end{bmatrix}$

Select the pivot row by dividing the rightmost element in each other row by the pivot-column element in that row. The "least positive" quotient determines the pivot row:

$\displaystyle \begin{bmatrix} 2 & 3 & {\color{Blue} \textbf{2}} &1 & 0 & 0 & {\color{Blue} \textbf{1,200}} \\ 1 & 2 & {\color{Blue} \textbf{4}} & 0 & 1 & 0 & {\color{Blue} \textbf{600}}\\ 0 & 4 & {\color{Blue} \textbf{3}} & 0 & 0 & 1 & {\color{Blue} \textbf{900}} \\ -1 & -1 & {\color{Red} \textbf{-2}} & 0&0&0 & 0 \end{bmatrix}\begin{matrix} 1,200 \div 2 = 600 \\ 600 \div 4 = {\color{Red} \textbf{150}} \\ 900 \div 3 = 300\\\; \end{matrix}$

The "least positive" of the three quotients is 150, in Row 2. The pivot element is the 4 in Row 2, Column 3.

Perform the pivot by first multiplying each entry in Row 2 by $\displaystyle \frac{1}{4}$ in order to make the pivot element a 1:

$\displaystyle \frac{1}{4}R2\rightarrow R2$

$\displaystyle \begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ \frac{1}{4} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 150\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & -2 & 0&0&0 & 0 \end{bmatrix}$

Make the other elements in that column zeroes by adding multiples of the pivot row to the other rows, as follows:

$\displaystyle -2R2 + R 1 \rightarrow R1$

$\displaystyle -3R2 + R3 \rightarrow R3$

$\displaystyle 2R2 + R4 \rightarrow R4$

$\displaystyle \begin{bmatrix} \frac{3}{2} & 2 & 0 &1 & -\frac{1}{2} & 0 & 900 \\ \frac{1}{4} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 150\\ -\frac{3}{4} & \frac{5}{2} & 0 & 0 & -\frac{3}{4} & 1 & 450 \\ -\frac{1}{2} & 0& 0 & 0&\frac{1}{2}&0 & {\color{Red} \textbf{300}} \end{bmatrix}$

The entry at bottom right is 300.