Question

Solve the following system of equations using an inverse matrix. You must alsoindicate the inverse matrix, A-1, that was used to solvsystem. You mayoptionally write the inverse matrix with a scalar coefficient.- 2X-4Y=1-6X-9Y=-8

Answer 1

Given:

`\begin{gathered} -2x-4y=1 \\ -6x-9y=-8 \end{gathered}`

To find:

The solutions

Step-by-step explanation:

It can be written as,

`\begin{bmatrix}{-2} & {-4} \\ {-6} & {-9}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{1} & {} \\ {-8} & {}\end{bmatrix}`

It is of the form,

`\begin{gathered} AX=B \\ X=A^(-1)B \end{gathered}`

Let us find the inverse matrix of A,

`\begin{gathered} A^(-1)=(1)/(|A|)adjA \\ =(1)/([-2(-9)-(-4)(-6)])\begin{bmatrix}{-9} & 4 \\ {6} & {-2}\end{bmatrix}^ \\ A^(-1)=(1)/(-6)\begin{bmatrix}{-9} & 4{} \\ {6} & {-2}\end{bmatrix} \\ A^(-1)=(-1)/(6)\begin{bmatrix}{-9} & {4} \\ {6} & {-2}\end{bmatrix} \end{gathered}`

Then, find the matrix X,

`\begin{gathered} X=A^(-1)B \\ \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=-(1)/(6)\begin{bmatrix}{-9} & {4} \\ {6} & {-2}\end{bmatrix}\begin{bmatrix}{1} & {} \\ {-8} & {}\end{bmatrix} \\ =-(1)/(6)\begin{bmatrix}{-9-32} & {} \\ {6+16} & {}\end{bmatrix} \\ =-(1)/(6)\begin{bmatrix}{-41} & {} \\ {22} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{(41)/(6)} & {} \\ -{(11)/(3)} & {}\end{bmatrix} \end{gathered}`

Therefore, the solutions are,

`\begin{gathered} x=(41)/(6) \\ y=-(11)/(3) \end{gathered}`

Final answer:

The inverse matrix of A is,

`(1)/(A)=(-1)/(6)\begin{bmatrix}{-9} & {4} \\ {6} & {-2}\end{bmatrix}`

The solutions are,

`\begin{gathered} x=(41)/(6) \\ y=-(11)/(3) \end{gathered}`