Question

Consider a linear transformation $\mathit{T}\left(\overrightarrow{x}\right)\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$ from ${\mathbf{ℝ}}^{\mathbf{2}}$to${\mathbf{ℝ}}^{\mathbf{2}}$ . Suppose for two vectors ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$ and ${\overrightarrow{\mathbf{v}}}_{\mathbf{2}}$in ${\mathbf{ℝ}}^{\mathbf{2}}$ we have $\mathit{T}\mathbf{\left(}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{3}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$androle="math" localid="1660719222222" $\mathit{T}\mathbf{\left(}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{4}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}$ . What can you say about det A ? Justify your answer carefully.

Short answer

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

$detA=detT=12$

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. 

Consider the linear transformation,

$T\left(\overrightarrow{x}\right)=A\overrightarrow{x}$.

To find A .

Since $A\left({\overrightarrow{v}}_1\right)=3{\overrightarrow{v}}_1$and $A\left({\overrightarrow{v}}_2\right)=4{\overrightarrow{v}}_2$, it means that the two vectors are linearly independent.

Therefore, $B=\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2\right\}$, is a basis for ${\mathrm{ℝ}}^2{\mathrm{ℝ}}^2$.

The matrix of corresponding to B is

$T=\left[\begin{array}{cc}3& 0\\ 0& 4\end{array}\right]$

Thus,

$detA=detT=12$.