Question

Find the determinants of the linear transformations in Exercises $\mathbf{17}$through .

28.$\mathit{T}\mathbf{\left(}\overrightarrow{\mathbf{v}}\mathbf{\right)}\mathbf{=}\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\mathbf{\times }\overrightarrow{\mathbf{v}}$ from the plane$\mathit{V}$ given by ${\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$to $\mathbf{V}$.

Short answer

Therefore, the determinant of the linear transformations is given by,

$\mathrm{detT}=\mathrm{detB}=14$.

Step-by-step solution

Definition

A determinant is a unique number associatedwith a square matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given

Given linear transformation,

$\mathrm{T}\left(\overrightarrow{\mathrm{v}}\right)=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\times \overrightarrow{\mathrm{v}}$

To find determinant

First, we notice that the plane$V$ contains$(0,0,0)$, as well as$(1,1,-1)$ and$(3,0,-1)$. This means that$V$ is a subspace of${ℝ}^3$, spanned by$(1,1,-1)$ and$(3,0,-1)$.

Now, we compute:

$\mathrm{T}\left(\overrightarrow{\mathrm{v}}\right)=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\times \overrightarrow{\mathrm{v}}$

Substitute $\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$ for $\overrightarrow{v}$ and simplify as:

$\begin{array}{l}T\left(\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\right)=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\times \left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\\ T\left(\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\right)=\left[\begin{array}{c}-5\\ 4\\ -1\end{array}\right]\\ T\left(\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\right)=4\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]-3\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]\end{array}$

Substitute $\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]$ for $\overrightarrow{v}$ and simplify as:

$\begin{array}{l}T\left(\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]\right)=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\times \left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]\\ T\left(\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]\right)=\left[\begin{array}{c}-2\\ 10\\ -6\end{array}\right]\\ T\left(\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]\right)=10\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]-4\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]\end{array}$

Since$\mathcal{B}=\left\{\right(1,1,-1),(3,0,-1\left)\right\}$is a basis for$V$, this means that the matrix of$\mathrm{T}$corresponding to$\mathcal{B}$is$B=\left[\begin{array}{cc}4& -3\\ 10& -4\end{array}\right]$

Therefore, $\mathrm{detT}=\mathrm{detB}=14$.