ka1leE

2021-09-20

Use the definition of Ax to write the matrix equation as a vector equation, or vice versa.
$\left[\begin{array}{cc}7& -3\\ 2& 1\\ 9& -6\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ -5\end{array}\right]=\left[\begin{array}{c}1\\ -9\\ 12\\ -4\end{array}\right]$

hesgidiauE

From the Theorem 3 the matrix equation is
$Ax=b$
where A in $m×n$ matrix $b\in {R}^{m}$.
We have
${x}_{1}\left[\begin{array}{c}4\\ -1\\ 7\\ -4\end{array}\right]+{x}_{2}\left[\begin{array}{c}-5\\ 3\\ -5\\ 1\end{array}\right]+{x}_{3}\left[\begin{array}{c}7\\ -8\\ 0\\ 2\end{array}\right]=\left[\begin{array}{c}6\\ -8\\ 0\\ -7\end{array}\right]$
then the matrix equation of the given vector equation is
$\left[\begin{array}{c}{x}_{1}\cdot 4\\ -{x}_{1}\\ {x}_{1}\cdot 7\\ -{x}_{1}\end{array}\right]+\left[\begin{array}{c}-5{x}_{2}\\ 3\cdot {x}_{2}\\ -5{x}_{2}\\ {x}_{2}\end{array}\right]+\left[\begin{array}{c}7{x}_{3}\\ -8{x}_{3}\\ 0\\ 2\cdot {x}_{3}\end{array}\right]=\left[\begin{array}{c}6\\ -8\\ 0\\ -7\end{array}\right]$
$\left[\begin{array}{ccc}4& -5& 7\\ -1& 3& -8\\ 7& -5& 0\\ -4& 1& 2\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}6\\ -8\\ 0\\ 7\end{array}\right]$
Hence, the matrix equation is
$\left[\begin{array}{ccc}4& -5& 7\\ -1& 3& -8\\ 7& -5& 0\\ -4& 1& 2\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}6\\ -8\\ 0\\ 7\end{array}\right]$
Result: $\left[\begin{array}{ccc}4& -5& 7\\ -1& 3& -8\\ 7& -5& 0\\ -4& 1& 2\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}6\\ -8\\ 0\\ 7\end{array}\right]$

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