91Ó°ÊÓ

We have given the system of the equation of linear transformation which is

$$\begin{gathered} y_1=x_1+3x_2+3x_3 \\ {y}_2=x_1+4x_2+8x_3 \\ {y}_3=2x_1+7x_2+12x_3 \end{gathered}$$

A function T from Xto Yis called invertible if the equation$T\left(x\right)=y$has a unique solution.

In this case, the inverse$T^{-1}$from Yto Xis defined by,

$T^{-1}\left(y\right)=\left(\text{The unique}x\text{in}X\text{such that}T\left(x\right)=y\right)$

The given system of linear equations is written as

$$\begin{gathered} \left|x_1+3x_2+3x_3=y_1 \\ {x}_1+4x_2+8x_3=y_2 \\ 2x_1+7x_2+12x_3=y_3\right| \end{gathered}$$

For solving the above linear equations, subtract equation 1 from equation2 and twice of equation 1 from equation 3 and we will get

$$\begin{gathered} \left|x_1+3x_2+3x_3=y_1 \\ {x}_2+5x_3=y_2-y_1 \\ {x}_2+6x_3=y_3-2y_1\right| \end{gathered}$$

Now subtract the equation 2 from the equation 3 we will get

$$\begin{gathered} \left|x_1+3x_2+3x_3=y_1 \\ {x}_2+5x_3=y_2-y_1 \\ {x}_3=-y_1-y_2+y_3\right| \end{gathered}$$

On solving the above equations, we get

$$\begin{gathered} \left|x_1=-8y_1-15y_2+12y_3 \\ {x}_2=4y_1+6y_2-5y_3 \\ {x}_3=-y_1-y_2+y_3\right| \end{gathered}$$

Thus, the inverse of linear transformation exits.

Hence, the inverse of linear transformation exits. The inverse transformation is

$$\begin{gathered} x_1=-8y_1-15y_2+12y_3 \\ {x}_2=4y_1+6y_2-5y_3 \\ {x}_3=-y_1-y_2+y_3 \end{gathered}$$