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Solve the following system of equations using an inverse matrix. You must alsoindicate the inverse matrix, A-?, that was used to solve the system. You mayoptionally write the inverse matrix with a scalar coefficient.94:0+9y = -1-33-9y = 2A-12II1y =-Color9Submit Answerattempt 1 out of

Solve The Following System Of Equations Using An Inverse Matrix You Must Alsoindicate The Inverse Matrix A That Was Used To Solve The System You Mayoptionally W class=

Sagot :

Remember the following rule to find the inverse of a 2x2 matrix:

[tex]\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}[/tex]

And it is well defined whenever ad-bc is different from 0.

The system of equations can be written as a product of a matrix A and a vector (x,y) as:

[tex]\begin{pmatrix}4&9\\-3&-9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-1\\2\end{pmatrix}[/tex]

The system can be solved by multiplying both members by the inverse of the matrix (4 9 \\ -3 -9):

[tex]\begin{gathered} \begin{pmatrix}4 & 9 \\ -3 & -9\end{pmatrix}^{-1}\begin{pmatrix}4 & 9 \\ -3 & -9\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}4 & 9 \\ -3 & -9\end{pmatrix}^{-1}\begin{pmatrix}-1 \\ 2\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}4 & 9 \\ -3 & -9\end{pmatrix}^{-1}\begin{pmatrix}-1 \\ 2\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{(4)(-9)-(-3)(9)}\begin{pmatrix}-9 & -9 \\ 3 & 4\end{pmatrix}\begin{pmatrix}-1 \\ 2\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{-36+27}\begin{pmatrix}-9 & -9 \\ 3 & 4\end{pmatrix}\begin{pmatrix}-1 \\ 2\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{-9}\begin{pmatrix}-9 & -9 \\ 3 & 4\end{pmatrix}\begin{pmatrix}-1 \\ 2\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=-\frac{1}{9}\begin{pmatrix}-1(-9)+2(-9) \\ -1(3)+2(4)\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=-\frac{1}{9}\begin{pmatrix}9-18 \\ -3+8\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=-\frac{1}{9}\begin{pmatrix}-9 \\ 5\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}-9\times-\frac{1}{9} \\ 5\times-\frac{1}{9}\end{pmatrix} \\ \\ \Rightarrow\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}1 \\ -\frac{5}{9}\end{pmatrix} \\ \\ \Rightarrow x=1,y=-\frac{5}{9} \end{gathered}[/tex]

Therefore, the inverse matrix used to solve the system is:

[tex]A^{-1}=-\frac{1}{9}\begin{pmatrix}-9 & -9 \\ 3 & 4\end{pmatrix}[/tex]

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