# find a vector $x \in \mathbb{R}^5$ (expressed by the unknowns $y_1, y_2, y_3, y_4$) that fulfills $T(x) = y$

Info given:

There is a linear transformation $$T$$: $$\mathbb{R}^5 \rightarrow \mathbb{R}^4$$ given by $$T\left(\begin{array}{r} \mathbf{x} \end{array}\right) = \left[\begin{array}{r} 2x_1-4x_2-x_3-3x_4+2x_5\\ -x_1+2x_2+x_3+x_5\\ x_1-2_x2-x_3-3x_4-x_5\\ -x_1+4x_2-x_3+5x_5 \end{array}\right] \quad \text{for} \quad \mathbf{x} = \left[\begin{array}{r} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{array}\right] \in \mathbb{R}^5$$

First I've found the matrix $$A$$ that fulfulls $$T(x) = Ax$$ for all $$x \in \mathbb{R}^5$$.

After that, I am told that we should let $$y = (y_1$$ $$y_2$$ $$y_3$$ $$y_4$$)$$^T$$ $$\in \mathbb{R}^4$$ be an arbitrary (but unknown) vector.

My task is then to find a vector $$x \in \mathbb{R}^5$$ (expressed by the unknowns $$y_1, y_2, y_3, y_4$$) that fulfills $$T(x) = y$$

My try:

I have tried myself and end up with the following: $$x = \frac{y}{A}$$

$$= \frac{\left(\begin{matrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{matrix}\right)}{\left(\begin{matrix} 2 & -4 & -1 & -3 & 2 \\ -1 & 2 & 1 & 0 & 1 \\ 1 & -2 & -1 & -3 & -1 \\ -1 & 4 & -1 & 0 & 1 \end{matrix}\right)}$$

$$=\left(\begin{matrix} \frac{y_1}{2-4-1-3+2} \\ \frac{y_2}{-1+2+1+1} \\ \frac{y_3}{1-2-1-3-1} \\ \frac{y_4}{-1+4-1+1} \end{matrix}\right)$$

I'm not sure if this is correct or if the approach is correct. Any help would be so appreciated!

• Dividing by a matrix? That doesn't make sense. – mattos May 15 '19 at 10:56

With all due respect, but $$x = \frac{y}{A}$$ is non-sense !
To find $$x$$, you have to solve the linear system $$Ax=y$$.
• Can this answer be corret? $$\left(\begin{matrix} 2*y-11*x_5 \\ y-4*x_5 \\ y-4*x_5 \\ \frac{-2*y}{3} \\ x_5 \end{matrix}\right)$$ – jubibanna May 15 '19 at 14:07