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Q26E

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Found in: Page 289

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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# Find the determinants of the linear transformations in Exercises 17 through 28.26. ${\mathbit{T}}\left(M\right){\mathbf{=}}\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]{\mathbit{M}}{\mathbf{+}}{\mathbit{M}}\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$from the space V of symmetric 2 × 2 matrices to V

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

$detT=detB=-16$

See the step by step solution

## Step 1: Definition.

A determinant is a unique number associated with 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.

## Step 2: Given.

Given linear transformation,

$T\left(M\right)=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]M+M\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

## Step 3: To find determinant.

For $M\in V$, we have

$M=\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]$

So we compute

$T\left(M\right)=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]M+M\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\phantom{\rule{0ex}{0ex}}T\left(M\right)=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]+\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\phantom{\rule{0ex}{0ex}}T\left(M\right)=\left[\begin{array}{cc}2a+4b& 2a+4b+2d\\ 2a+4b+2d& 4b+6d\end{array}\right]$

Since,

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

$B=\left[\begin{array}{ccc}2& 4& 0\\ 2& 4& 2\\ 0& 4& 6\end{array}\right]$

Therefore,

$detT=detB=-16$

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