Question

Please show all steps/work. I have attached the problem as well as a similar example problem that was completed to show the necessary steps. For step 2 please reduce using row echelon form (as seen in the example).

Answer 1

Answer:

x = 2, y = 1

Explanation:

The given system of equations is:

x - y = 1

2x + y = 5

Write the system of equations in the form AX = B

$\begin{bmatrix}{1} & {-1} \\ {2} & {1}\end{bmatrix}\begin{bmatrix}{x} \\ {y}\end{bmatrix}=\text{ }\begin{bmatrix}{1} \\ {5}\end{bmatrix}$$\begin{gathered} \text{where:} \\ A=\begin{bmatrix}{1} & {-1} \\ {2} & {1}\end{bmatrix} \\ |A|=(1\times1)-(-2) \\ |A|=3 \\ A^{-1}=\frac{\begin{bmatrix}{1} & {1} \\ {-2} & {1}\end{bmatrix}}{3} \\ A^{-1}=\begin{bmatrix}{\frac{1}{3}} & {\frac{1}{3}} \\ {-\frac{2}{3}} & {\frac{1}{3}}\end{bmatrix} \end{gathered}$$\text{Multiply both sides of the matrix equation by A}^{-1}$$\begin{gathered} AA^{-1}X=A^{-1}B \\ X=A^{-1}B \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{\frac{1}{3}} & {\frac{1}{3}} \\ {-\frac{2}{3}} & {\frac{1}{3}}\end{bmatrix}\begin{bmatrix}{1} \\ {5}\end{bmatrix} \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{(\frac{1}{3}\times1)+(\frac{1}{3}\times5)} \\ {(-\frac{2}{3}\times1)+(\frac{1}{3}\times5)}\end{bmatrix} \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{\frac{6}{3}} \\ {\frac{3}{3}}\end{bmatrix} \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{2} \\ {1}\end{bmatrix} \end{gathered}$

Therefore:

x = 2 and y = 1