T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)
A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.
Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)
and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)
So, the matrix of the transformation is
[tex]\left[\begin{array}{ccc}7&6\\4&5\end{array}\right][/tex]
The inverse of the matrix is
[tex]\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right][/tex]
So, the Inverse Transformation is given by
[tex]T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})[/tex]
So, no option is correct. And the answer is
[tex]T^{-1}(x,y)=(\frac{5x-6y}{11}, \frac{-4x+7y}{11})[/tex]
Learn more about linear transformations here-
brainly.com/question/13005179
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