# Find a single vector x whose image under t is b

Transformation is defined as T(x)=Ax, find whether x is unique or not.

$A=\begin{bmatrix} 1 & -5 & -7\\ 3 & 7 & 5\end{bmatrix}$

$B=\begin{bmatrix} 2\\ 2\end{bmatrix}$

This question aims to find the uniqueness of vector $x$ with the help of linear transformation.

This question uses the concept of Linear transformation with reduced row echelon form. Reduced row echelon form helps in solving the linear matrices. In reduced row echelon form, we apply different row operations using the properties of linear transformation.

To solve for $x$, we have $T(x)=b$ which is to solve $Ax=b$ in order to solve for $x$. The augmented matrix is given as:

$A \begin{bmatrix} A & B \end{bmatrix}$

$=\begin{bmatrix} 1 & -5 & -7 & |-2\\ -3 & 7 & 5 & |-2 \end{bmatrix}$

Applying row operations to get the reduced echelon form.

$\begin{bmatrix} 1 & -5 & -7 & |-2\\ -3 & 7 & 5 & |-2 \end{bmatrix}$

$R_1 \leftrightarrow R_2 ,R_2 + \frac {1}{3} R_1 \rightarrow R_2$

By using the above row operations, we get:

$\begin{bmatrix} -3 & 7 & 5 & -2\\ 0 & -\frac{8}{3} & – \frac{16}{3} & -\frac{8}{3} \end{bmatrix}$

$-\frac{3}{8}R_2 \rightarrow R_2 ,R_1 – 7R_2 \ \rightarrow R_1$

$\begin{bmatrix} -3 & 0 & -9 & -9\\ 0 & 1 & 2 & 1 \end{bmatrix}$

$-\frac{1}{3}R_1 \rightarrow R_1$

The above operations results in the following matrix:

$\begin{bmatrix} 1 & 0 & 3 & 3\\ 0 & 1 & 2 & 1 \end{bmatrix}$

We get:

$x_1+3x_3 = 3$

$x_1 = 3 – 3x_3$

$x_2 + 2x_3 = 1$

$x_2 = 1 -2x_3$

Now:

$x= \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 3 – x_3\\ 1 – 2x_3\\ x_3 \end{bmatrix}$

$=\begin{bmatrix} 3 \\ 1\\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -3 \\ -2\\ -1 \end{bmatrix}$

## Numerical Result

By applying a linear transformation of given matrices, it shows that $x$ does not have a unique solution.

## Example

Two matrices are given below. Find the unique vector x with the help of transformation $T(x)=Ax$

$A=\begin{bmatrix} 1 & -5 & -7\\ -3 & 7 & 5\end{bmatrix}$

$B=\begin{bmatrix} 4\\ 4\end{bmatrix}$

To solve for $x$, we have $T(x)=b$ which is to solve $Ax=b$ in order to solve for $x$. The augmented matrix is given as:

$A \begin{bmatrix} A & B \end{bmatrix}$

$R_2 + 3R_1$

$\begin{bmatrix} 1 & -5 & -7 & 4 \\ 0 & -8 & -16 & 16 \end{bmatrix}$

$-\frac{R_2}{8}$

$\begin{bmatrix} 1 & -5 & -7 & 4 \\ 0 & 1 & 2 & -2 \end{bmatrix}$

$R_1 + 5R_2$

$\begin{bmatrix} 1 & 0 & 3 & -6 \\ 0 & 1 & 2 & -2 \end{bmatrix}$

$x_1+3x_3 = -6$

$x_1 = -6 – 3x_3$

$x_2 + 2x_3 = -2$

$x_2 = -2 -2x_3$

$x= \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} -6 – 3x_3\\ -2 – 2x_3\\ x_3 \end{bmatrix}$

The above equation shows that $x$ does not have a unique solution.

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