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Expert-verified Found in: Page 401 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # For the matrix ${\mathbit{A}}{\mathbf{=}}\mathbf{\left[}\begin{array}{cc}\mathbf{8}& \mathbf{-}\mathbf{2}\\ \mathbf{-}\mathbf{2}& \mathbf{5}\end{array}\mathbf{\right]}{\mathbf{,}}$ write ${\mathbit{A}}{\mathbf{=}}{{\mathbit{B}}}^{{\mathbf{2}}}$ as discussed in Exercise 30. See Example 1.

$\mathrm{A}={\mathrm{B}}^{2}\mathrm{where},\mathrm{B}=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}14& -2\\ -2& 11\end{array}\right]$

See the step by step solution

## Step 1: Given Information:

$A=\left[\begin{array}{cc}8& -2\\ -2& 5\end{array}\right]$

## Step 2: Estimating S:

Consider the matrix below:

$A=\left[\begin{array}{cc}8& -2\\ -2& 5\end{array}\right]$

Now, as shown below, calculate the eigen values:

$\mathrm{det}\left(\mathrm{A}-\mathrm{\lambda l}\right)=0$

$⇒\left|\begin{array}{cc}8-\lambda & -2\\ -2& 5-\lambda \end{array}\right|=0$

$⇒{\lambda }^{2}-13\lambda +36=0\phantom{\rule{0ex}{0ex}}⇒\left(\lambda -4\right)·\left(\mathrm{\lambda }-9\right)=0$

As a result, the eigenvalues are 9 and 4, yielding

$D=\left[\begin{array}{cc}9& 0\\ 0& 4\end{array}\right]$

So,

${\mathrm{D}}_{1}=\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]$

The eigen vectors are as follows:

$\left[\begin{array}{c}-2\\ 1\end{array}]\mathrm{and}\left[\begin{array}{c}1\\ 2\end{array}\right\right]$

Find S by combining the eigen vectors after they have been converted to unit vectors, as shown below.

$\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right]\mathrm{and}\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]$

As a result, we get S, as shown below:

$S=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]$

## Step 3: Estimating B:

Now, as shown below, evaluate B:

$\mathrm{B}={\mathrm{SD}}_{1}\mathrm{S}$

$=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]$

$=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]$

$=\frac{1}{5}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}-6& 3\\ 2& 4\end{array}\right]$

$⇒B=\frac{1}{5}\left[\begin{array}{cc}14& -2\\ -2& 11\end{array}\right]$

As a result, we can write A as:

$A={B}^{2}$

## Step 4: Determining the Result:

$\mathrm{A}={\mathrm{B}}^{2}\mathrm{where},\mathrm{B}=\frac{1}{5}\left[\begin{array}{cc}14& -2\\ -2& 11\end{array}\right]$ ### Want to see more solutions like these? 