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

Step-by-step 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*1)-(-2) \\ |A|=3 \\ A^(-1)=\frac{\begin{bmatrix}{1} & {1} \\ {-2} & {1}\end{bmatrix}}{3} \\ A^(-1)=\begin{bmatrix}{(1)/(3)} & {(1)/(3)} \\ {-(2)/(3)} & {(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}{(1)/(3)} & {(1)/(3)} \\ {-(2)/(3)} & {(1)/(3)}\end{bmatrix}\begin{bmatrix}{1} \\ {5}\end{bmatrix} \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{((1)/(3)*1)+((1)/(3)*5)} \\ {(-(2)/(3)*1)+((1)/(3)*5)}\end{bmatrix} \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{(6)/(3)} \\ {(3)/(3)}\end{bmatrix} \\ \begin{bmatrix}{x} \\ {y}\end{bmatrix}=\begin{bmatrix}{2} \\ {1}\end{bmatrix} \end{gathered}`

Therefore:

x = 2 and y = 1