Question

Solve the following system of equations using an inverse matrix. You must alsoindicate the inverse matrix, A-1, that was used to solve the system. You mayoptionally write the inverse matrix with a scalar coefficient.2X-3Y=-103X-6Y=-8

Answer 1

Given:

The system of equations are

`\begin{gathered} 2x-3y=-10 \\ 3x-6y=-8 \end{gathered}`

Required:

To solve the given system of equations using an inverse matrix.

Step-by-step explanation:

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

`\begin{gathered} A=\begin{bmatrix}{2} & {-3} \\ {3} & {-6}\end{bmatrix} \\ X=\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix} \\ B=\begin{bmatrix}{-10} & {} \\ {-8} & {}\end{bmatrix} \end{gathered}`

Then

`\begin{bmatrix}{2} & {-3} \\ {3} & {-6}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-10} & {} \\ {-8} & {}\end{bmatrix}----(1)`

First, we need to calculate inverse of A. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

`\begin{gathered} A^(-1)=(1)/(|A|)\begin{bmatrix}{-6} & {3} \\ {-3} & {2}\end{bmatrix} \\ \\ =(1)/(-12-(-9))\begin{bmatrix}{-6} & {3} \\ {-3} & {2}\end{bmatrix} \\ \\ =(1)/(-3)\begin{bmatrix}{-6} & {3} \\ {-3} & {2}\end{bmatrix} \\ \\ =\begin{bmatrix}{2} & {-1} \\ {1} & {-(2)/(3)}\end{bmatrix} \end{gathered}`

Now we are ready to solve. Multiply both sides of the equation by inverse of A,

`\begin{bmatrix}{2} & {-1} \\ {1} & {-(2)/(3)}\end{bmatrix}\begin{bmatrix}{2} & {-3} \\ {3} & {-6}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{2} & {-1} \\ {1} & {-(2)/(3)}\end{bmatrix}\begin{bmatrix}{-10} & {} \\ {-8} & {}\end{bmatrix}`
`\begin{bmatrix}{4-3} & {-6+6} \\ {2-2} & {-3+4}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-20+8} & {} \\ {-10+(16)/(3)} & {}\end{bmatrix}`
`\begin{gathered} \begin{bmatrix}{1} & {0} \\ {0} & {1}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-12} & {} \\ {-(14)/(3)} & {}\end{bmatrix} \\ \\ \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-12} & {} \\ {-(14)/(3)} & {}\end{bmatrix} \end{gathered}`

Final Answer:

The solution is

`\begin{gathered} x=-12 \\ \\ y=-(14)/(3) \end{gathered}`