$\left(\begin{array}{ccc}1& -3& 2\\ 2& -4& 4\\ 3& -8& 6\end{array}\right)=A$

a) $\left(\begin{array}{c}4\\ 6\\ 11\end{array}\right)=B$

b) $\left(\begin{array}{c}-3\\ -2\\ -7\end{array}\right)=B$

Addison Trujillo
2022-07-15
Answered

Let A be the matrix below and define a transformation $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$ by $T(U)=AU.$. For each of the vectors $B$ below, find a vector $U$ such that $T$ maps $U$ to $B$, if possible. Otherwise state that there is no such $U$.

$\left(\begin{array}{ccc}1& -3& 2\\ 2& -4& 4\\ 3& -8& 6\end{array}\right)=A$

a) $\left(\begin{array}{c}4\\ 6\\ 11\end{array}\right)=B$

b) $\left(\begin{array}{c}-3\\ -2\\ -7\end{array}\right)=B$

$\left(\begin{array}{ccc}1& -3& 2\\ 2& -4& 4\\ 3& -8& 6\end{array}\right)=A$

a) $\left(\begin{array}{c}4\\ 6\\ 11\end{array}\right)=B$

b) $\left(\begin{array}{c}-3\\ -2\\ -7\end{array}\right)=B$

You can still ask an expert for help

Zichetti4b

Answered 2022-07-16
Author has **13** answers

The reason you found no inverse is because A is not invertible. This can be seen by showing that detA=0

aggierabz2006zw

Answered 2022-07-17
Author has **5** answers

Hint: write $U=({u}_{1},{u}_{2},{u}_{3})$ and then work out $AU$ in terms of these components. You now have a system of linear equations to solve in the form $AU=B$.

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