Question

Determine a matriz X, tal que

$\left[\begin{array}{ccc}1&1&0\\3&-2&0\\5&0&1\end{array}\right]$ × X = $\left[\begin{array}{ccc}5\\0\\8\end{array}\right]$

Answer 1

$\mathrm{A_{m\times p}\times B_{p\times n}=(AB)_{m\times n}}\\\\ \mathrm{\left[\begin{array}{ccc}1&1&0\\3&-2&0\\5&0&1\end{array}\right]\times X=\left[\begin{array}{ccc}5\\0\\8\end{array}\right]\ \to\ X_{3\times 1}=\left[\begin{array}{ccc}a\\b\\c\end{array}\right]}\\\\\\ \mathrm{\left[\begin{array}{ccc}1&1&0\\3&-2&0\\5&0&1\end{array}\right]\times\left[\begin{array}{ccc}a\\b\\c\end{array}\right]=\left[\begin{array}{ccc}5\\0\\8\end{array}\right]}$

$\mathrm{1.a+1.b+0.c=5\ \to\ a+b=5}\\ \mathrm{3.a+(-2).b+0.c=0\ \to\ 3a-2b=0}\\ \mathrm{5.a+0.b+1.c=8\ \to\ 5a+c=8}\\\\ \mathrm{\begin{cases} \mathrm{a+b=5\ \mathbf{(I)}} \\ \mathrm{3a-2b=0\ \mathbf{(II)}} \\ \mathrm{5a+c=8\ \mathbf{(III)}} \end{cases}}\\\\\\ \mathrm{Somando\ 2.I+II:}\\\\ \mathrm{2a+2b+3a-2b=10+0\ \to\ 5a=10\ \to\ a=2}\\\\ \mathrm{Substituindo\ o\ valor\ de\ a\ em\ I\ e\ III:}\\\\ \mathrm{2+b=5\ \to\ b=5-2\ \to\ b=3}\\ \mathrm{5.2+c=8\ \to\ c=8-10\ \to\ c=-2}}$

$\textrm{Portanto:}\\\\ \mathrm{X=\left[\begin{array}{ccc}a\\b\\c\end{array}\right]\ \to\ \mathbf{X=\left[\begin{array}{ccc}2\\3\\-2\end{array}\right]}}$