Find all x in \mathbb{R}^{4} that are mapped into the zero vector by the t

kolonelyf4 2021-11-16 Answered
Find all x in \(\displaystyle{\mathbb{{{R}}}}^{{{4}}}\) that are mapped into the zero vector by the transformation \(\displaystyle{\mathbf{{{X}}}}\mapsto{A}{\mathbf{{{X}}}}\) for the given matrix A.
\[A=\begin{bmatrix}1 & 3 & 9 & 2 \\1 & 0 & 3 & -4\\0 & 1 & 2 & 3\\-2 & 3 & 0 & 5 \end{bmatrix}\]

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Expert Answer

May Dunn
Answered 2021-11-17 Author has 456 answers

Step 1 Solve Ax=0 to find all \(\displaystyle{x}\in{\mathbb{{{R}}}}^{{{4}}}\).The audmented matrix is
\[\begin{bmatrix}1 & 3 & 9 & 2 & 0\\1 & 0 & 3 & -4 & 0\\0 & 1 & 2 & 3 & 0\\-2 & 3 & 0 & 5 & 0\end{bmatrix}.\]
Reduce matrix to reduced row echelon from
\[\begin{bmatrix}1 & 3 & 9 & 2 & |0\\1 & 0 & 3 & -4 & |0\\0 & 1 & 2 & 3 & |0\\-2 & 3 & 0 & 5 & |0\end{bmatrix}\begin{array}{cc} R_{1}\leftrightarrow R_{4} \\ R_{2}+1/2R_{1}\rightarrow R_{2}\sim \\ R_{4}+1/2R_{1}\rightarrow R_{4} \end{array}\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{3}{2} & 3 & \frac{-3}{2} & |0\\0 & 1 & 2 & 3 & |0\\0 & \frac{9}{2} & 9 & \frac{9}{2} & |0\end{bmatrix}\]
\[\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{3}{2} & 3 & \frac{-3}{2} & |0\\0 & 1 & 2 & 3 & |0\\0 & \frac{9}{2} & 9 & \frac{9}{2} & |0\end{bmatrix}\begin{array}{cc} R_{2}\leftrightarrow R_{4} \\ R_{3}-2/9R_{2}\rightarrow R_{3}\sim \\ R_{4}+1/3R_{2}\rightarrow R_{4} \end{array}\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{9}{2} & 9 & \frac{9}{2} & |0\\0 & 0 & 0 & 2 & |0\\0 & 0 & 0 & -3 & |0\end{bmatrix}\]
\[\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{9}{2} & 9 & \frac{9}{2} & |0\\0 & 0 & 0 & 2 & |0\\0 & 0 & 0 & -3 & |0\end{bmatrix}\begin{array}{cc} -1/3R_{4}\rightarrow R_{4} \\ R_{3}-2R_{4}\rightarrow R_{3} \\ R_{2}-9/2R_{4}\rightarrow R_{2} \end{array}\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{9}{2} & 9 & 0 & |0\\0 & 0 & 0 & 0 & |0\\0 & 0 & 0 & 1 & 0\end{bmatrix}\]
\[\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{9}{2} & 9 & 0 & |0\\0 & 0 & 0 & 0 & |0\\0 & 0 & 0 & 1 & |0\end{bmatrix}\begin{array}{cc} R_{1}-5R_{4}\rightarrow R_{1} \\ R_{3}\leftrightarrow R_{4} \\ 2/9R_{2}\rightarrow R_{2} \end{array}\begin{bmatrix}-2 & 3 & 0 & 0 & |0\\0 & 1 & 2 & 0 & |0\\0 & 0 & 0 & 1 & |0\\0 & 0 & 0 & 0 & 0\end{bmatrix}\]

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memomzungup4
Answered 2021-11-18 Author has 1294 answers
Step 2
\[\begin{bmatrix}-2 & 3 & 0 & 5 & |0\\0 & \frac{9}{2} & 9 & 0 & |0\\0 & 0 & 0 & 0 & |0\\0 & 0 & 0 & 1 & |0\end{bmatrix}\begin{array}{cc} R_{1}-5R_{4}\rightarrow R_{1} \\ R_{3}\leftrightarrow R_{4} \\ 2/9R_{2}\rightarrow R_{2} \end{array}\begin{bmatrix}-2 & 3 & 0 & 0 & |0\\0 & 1 & 2 & 0 & |0\\0 & 0 & 0 & 1 & |0\\0 & 0 & 0 & 0 & 0\end{bmatrix}\]
\[\begin{bmatrix}-2 & 3 & 0 & 0 & |0\\0 & 1 & 2 & 0 & |0\\0 & 0 & 0 & 1 & |0\\0 & 0 & 0 & 0 & |0\end{bmatrix}\begin{array}{cc}R_{1}-3R_{2}\rightarrow R_{1}\\ -1/2R_{1}\rightarrow R_{1} \end{array}\begin{bmatrix}1 & 0 & 3 & 0 & |0\\0 & 1 & 2 & 0 & |0\\0 & 0 & 0 & 1 & |0\\0 & 0 & 0 & 0 & 0\end{bmatrix}\].
Terefore, we get
\(\displaystyle{x}_{{{1}}}+{3}{x}_{{{3}}}={0}\Rightarrow{x}_{{{1}}}=-{3}{x}_{{{3}}}\)
\(\displaystyle{x}_{{{1}}}+{2}{x}_{{{3}}}={0}\Rightarrow{x}_{{{1}}}=-{2}{x}_{{{3}}}\)
\(\displaystyle{x}_{{{3}}}={x}_{{{3}}}\)
\(\displaystyle{x}_{{{4}}}={0}\).
That is
\[x=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}=\begin{bmatrix}-3x_{3}\\-2x_{3}\\x_{3}\\0\end{bmatrix}=x_{3}\begin{bmatrix}-3\\-2\\1\\0\end{bmatrix}\]
Hense, the required vectors are
\[x=x_{3}\begin{bmatrix}-3\\-2\\1\\0\end{bmatrix}\]
where \(\displaystyle{x}_{{{3}}}\in{\mathbb{{{R}}}}^{{{4}}}\).
Result
\[x=x_{3}\begin{bmatrix}-3\\-2\\1\\0\end{bmatrix}\]
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