Chapter 8: Problem 14
\(\operatorname{det} \mathbf{D}=5\)
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Chapter 8: Problem 14
\(\operatorname{det} \mathbf{D}=5\)
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$$\left(\begin{array}{ccccc} 1 & 3 & 5 & -1 & 1 \\ 0 & 1 & 1 & -1 & 4 \\ 1 & 2 & 5 & -4 & -2 \\ 1 & 4 & 6 & -2 & 6 \end{array}\right) \quad \frac{\operatorname{row}}{\text { operations }}\left(\begin{array}{rrrr|r} 1 & 3 & 5 & -1 & 1 \\ 0 & 1 & 1 & -1 & 4 \\ 0 & 0 & 1 & -4 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right)$$ Since the bottom row implies \(0=1,\) the system is inconsistent.
$$\left(\begin{array}{rrrrr} 1 & 0 & 1 & -1 & 1 \\ 0 & 2 & 1 & 1 & 3 \\ 1 & -1 & 0 & 1 & -1 \\ 1 & 1 & 1 & 1 & 2 \end{array}\right) \quad \frac{\operatorname{row}}{\text { operations }}\left(\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 1 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 1 & -5 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ The solution is \(x_{1}=0, x_{2}=1, x_{3}=1, x_{4}=0\).
(a) \(-(b)\) We compute $$\begin{aligned} &\mathbf{A} \mathbf{K}_{1}=\left(\begin{array}{ccc} 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \end{array}\right)\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right)=\left(\begin{array}{r} -2 \\ 2 \\ 0 \end{array}\right)=-2\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right)=-2 \mathbf{K}_{1}\\\ &\mathbf{A} \mathbf{K}_{2}=\left(\begin{array}{rrr} 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \end{array}\right)\left(\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right)=\left(\begin{array}{r} -2 \\ 0 \\ 2 \end{array}\right)=-2\left(\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right)=-2 \mathbf{K}_{2}\\\ &\mathbf{A} \mathbf{K}_{3}=\left(\begin{array}{lll} 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 4 \\ 4 \\ 4 \end{array}\right)=4\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=4 \mathbf{K}_{3} \end{aligned}$$ and observe that \(\mathbf{K}_{1}\) is an eigenvector with corresponding eigenvalue \(-2, \mathbf{K}_{2}\) is an eigenvector with corre- sponding eigenvalue \(-2,\) and \(\mathbf{K}_{3}\) is an eigenvector with corresponding eigenvalue 4 (c) since \(\mathbf{K}_{1} \cdot \mathbf{K}_{2}=1 \neq 0, \mathbf{K}_{1}\) and \(\mathbf{K}_{2}\) are not orthogonal, while \(\mathbf{K}_{1} \cdot \mathbf{K}_{3}=0\) and \(\mathbf{K}_{2} \cdot \mathbf{K}_{3}=0\) so \(\mathbf{K}_{3}\) is orthogonal to both \(\mathbf{K}_{1}\) and \(\mathbf{K}_{2},\) To transform \(\left\\{\mathbf{K}_{1}, \mathbf{K}_{2}\right\\}\) into an orthogonal set we let \(\mathbf{V}_{1}=\mathbf{K}_{1}\) and compute \(\mathbf{K}_{2} \cdot \mathbf{V}_{1}=1\) and \(\mathbf{V}_{1} \cdot \mathbf{V}_{1}=2 .\) Then $$\mathbf{V}_{2}=\mathbf{K}_{2}-\frac{\mathbf{K}_{2} \cdot \mathbf{V}_{1}}{\mathbf{V}_{1} \cdot \mathbf{V}_{1}} \mathbf{V}_{1}=\left(\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right)-\frac{1}{2}\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right)=\left(\begin{array}{r} \frac{1}{2} \\ \frac{1}{2} \\ -1 \end{array}\right)$$ Now, \(\left\\{\mathbf{V}_{1}, \mathbf{V}_{2}, \mathbf{K}_{3}\right\\}\) is an orthogonal set of eigenvectors with $$\left\|\mathbf{V}_{1}\right\|=\sqrt{2}, \quad\left\|\mathbf{V}_{2}\right\|=\frac{3}{\sqrt{6}}, \quad \text { and } \quad\left\|\mathbf{K}_{3}\right\|=\sqrt{3}$$ An orthonormal set of vectors is $$\left(\begin{array}{c} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{array}\right), \quad\left(\begin{array}{r} \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \end{array}\right), \quad \text { and } \quad\left(\begin{array}{c} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array}\right)$$ and so the matrix $$\mathbf{P}=\left(\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{array}\right)$$
(a) \(\mathbf{M}^{T}=\left(\begin{array}{ccccccccccc}22 & 8 & 19 & 27 & 21 & 3 & 3 & 27 & 21 & 18 & 21 \\ 13 & 3 & 21 & 22 & 3 & 25 & 27 & 6 & 7 & 14 & 23 \\\ 2 & 27 & 21 & 7 & 27 & 5 & 21 & 17 & 2 & 25 & 7\end{array}\right)\) (b) \(\mathbf{B}^{T}=\mathbf{M}=\left(\begin{array}{rrr}1 & 1 & 0 \\ 1 & 0 & 1 \\\ 1 & 1 & -1\end{array}\right)=\left(\begin{array}{rrrrrrrrrr}37 & 38 & 61 & 56 & 51 & 33 & 51 & 50 & 30 & 57 & 51 \\ 24 & 35 & 40 & 34 & 48 & 8 & 24 & 44 & 23 & 43 & 28 \\ 11 & -24 & 0 & 15 & -24 & 20 & 6 & -11 & 5 & -11 & 16\end{array}\right)\) (c) \(\mathbf{B A}^{-1}=\mathbf{B}\left(\begin{array}{rrr}-1 & 1 & 1 \\ 2 & -1 & -1 \\ 1 & 0 & -1\end{array}\right)=\mathbf{M}\)
If we take \(\mathbf{K}_{1}=\left(\begin{array}{l}0 \\ 1 \\\ 1\end{array}\right)\) as in Example 4 in the text then we look for a vector \(\mathbf{K}_{2}=\left(\begin{array}{l}a \\ b \\ c\end{array}\right)\) such that \(1(a)+\) \(\frac{1}{4} b-\frac{1}{4} c=0\) and \(\mathbf{K}_{1} \cdot \mathbf{K}_{2}=0\) or \(b+c=0 .\) The last equation implies \(c=-b\) so \(a+\frac{1}{4} b-\frac{1}{4}(-b)=a+\frac{1}{2} b=0\) If we let \(b=-2,\) then \(a=1\) and \(c=2,\) so a second eigenvector with eigenvalue -9 and orthogonal to \(\mathbf{K}_{1}\) is \(\mathbf{K}_{2}=\left(\begin{array}{r}1 \\ -2 \\ 2\end{array}\right)\)
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