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Expert-verified Found in: Page 199 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # True or False.The linear transformation ${\mathbf{T}}\left(M\right){\mathbf{=}}\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]{\mathbf{M}}{\mathbf{}}{\mathbf{from}}{\mathbf{}}{{\mathbf{R}}}^{\mathbf{2}\mathbf{×}\mathbf{2}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{{\mathbf{R}}}^{\mathbf{2}\mathbf{×}\mathbf{2}}{\mathbf{}}{\mathbf{}}{\mathbf{have}}{\mathbf{}}{\mathbf{rank}}{\mathbf{}}{\mathbf{1}}{\mathbf{.}}$

The given statement is False.

See the step by step solution

## Step 1: Determine the matrix B .

Consider the function $\mathrm{T}\left(\mathrm{M}\right)=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\mathrm{M}\mathrm{from}{\mathrm{R}}^{2×2}\mathrm{to}{\mathrm{R}}^{2×2}$ .

As localid="1659426855825" $\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right\}$is the basis element of ${R}^{2×2}$, the matrix B is defined as follows.

$T\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right\}=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}1& 0\\ 3& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+3\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

$T\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right\}=1.{u}_{1}+0.{u}_{2}+3.{u}_{3}+0.{u}_{4}$

The image of T at point $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ is defined as follows.

$T\left\{\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right\}=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}0& 1\\ 0& 3\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]+3\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=0.{u}_{1}+1.{u}_{2}+0.{u}_{3}+3.{u}_{4}$

The image of T at point $\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$ is defined as follows.

$T\left\{\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right\}=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2& 0\\ 6& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=2\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+6\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=2.{u}_{1}+0.{u}_{2}+6.{u}_{3}+0.{u}_{4}$

The image of T at point $\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$ is defined as follows.

$T\left\{\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right\}=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}0& 2\\ 0& 6\end{array}\right]\phantom{\rule{0ex}{0ex}}=2\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]+6\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}T\left\{\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right\}=0.{u}_{1}+2.{u}_{2}+0.{u}_{3}+6.{u}_{4}$

Therefore, the matrix B= $\left[\begin{array}{cccc}1& 0& 2& 0\\ 0& 1& 0& 2\\ 3& 0& 6& 0\\ 0& 3& 0& 6\end{array}\right]$.

## Step 2: Determine the rank of T .

As two rows are linearly dependent implies rank (T)=2

Hence, the statement is false. ### Want to see more solutions like these? 