umthumaL3e

2022-12-02

With T defined by T(x)=Ax, find a vector x such that T(x)=b

$A=\left(\begin{array}{ccc}1& -3& 2\\ 3& -8& 8\\ 0& 1& 2\\ 1& 0& 8\end{array}\right)\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}b=\left(\begin{array}{c}1\\ 6\\ 3\\ 10\end{array}\right)$

What I have done so far is thatI have merged the two matrices into a single augmented matrix. And row reduced it to get:

$\left(\begin{array}{cccc}1& -3& 2& 1\\ 0& 1& 2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

So does this just mean that the answer to the question is $\mathbf{x}=\left(\begin{array}{c}1\\ 3\\ 0\\ 0\end{array}\right)$

$A=\left(\begin{array}{ccc}1& -3& 2\\ 3& -8& 8\\ 0& 1& 2\\ 1& 0& 8\end{array}\right)\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}b=\left(\begin{array}{c}1\\ 6\\ 3\\ 10\end{array}\right)$

What I have done so far is thatI have merged the two matrices into a single augmented matrix. And row reduced it to get:

$\left(\begin{array}{cccc}1& -3& 2& 1\\ 0& 1& 2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

So does this just mean that the answer to the question is $\mathbf{x}=\left(\begin{array}{c}1\\ 3\\ 0\\ 0\end{array}\right)$

Madyson Hood

Beginner2022-12-03Added 10 answers

What you now have to do is solve the system of equations

${x}_{1}-3{x}_{2}+2{x}_{3}=1$

${x}_{2}+2{x}_{3}=3$

What happens when you solve for ${x}_{2}$ in the second equation? Hint: (use a parameter, like let ${x}_{3}=t$)

${x}_{1}-3{x}_{2}+2{x}_{3}=1$

${x}_{2}+2{x}_{3}=3$

What happens when you solve for ${x}_{2}$ in the second equation? Hint: (use a parameter, like let ${x}_{3}=t$)

Goundoubuf

Beginner2022-12-04Added 1 answers

Hint 1: If A has 3 columns, the dimension of x must be $3$

Hint 2: To check your result, compute Ax and see if you got $b$

Hint 2: To check your result, compute Ax and see if you got $b$

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$A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$T must be a linear transformation, we assume. Can u find the T standard matrix.$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{4},T\left({e}_{1}\right)=(3,1,3,1)\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}T\left({e}_{2}\right)=(-5,2,0,0),\text{}where\text{}{e}_{1}=(1,0)\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{e}_{2}=(0,1)$

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