Question

Find the determinants of the linear transformations in Exercises$\mathbf{17}$through$\mathbf{28}$.

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

Short answer

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

$\det T=\det B=14$

Step-by-step solution

Step by Step Solution: Step 1: 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,

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

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 role="math" localid="1660311515849" $V$is a subspace of role="math" localid="1660311511892" ${\mathrm{ℝ}}^3$, spanned by $(1,1,-1)$and $(3,0,-1)$.

Now, we compute:

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

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

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

Substitute $\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right]$for role="math" localid="1660311504999" $\overrightarrow{v}$and simplify as:

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

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

Therefore, $\det T=\det B=14$.