91Ó°ÊÓ

Apply the Cramer’s rule for the value of r.

$r=\frac{\left|\begin{array}{ccc}j& b& c\\ k& e& f\\ l& h& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$

Use the provided equations and substitute the value of a, b, c, d, e, f, g, h, i, j, k and l into role="math" localid="1647334847725" $r=\frac{\left|\begin{array}{ccc}j& b& c\\ k& e& f\\ l& h& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of r.

$\begin{array}{c}r=\frac{\left|\begin{array}{ccc}−1& −2& −5\\ 5& 2& −2\\ −1& 1& 1\end{array}\right|}{\left|\begin{array}{ccc}1& −2& −5\\ 1& 2& −2\\ 4& 1& 1\end{array}\right|}\\ =\frac{−33}{57}\\ =−\frac{11}{19}\end{array}$

Apply the Cramer’s rule for the value of s.

$s=\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$

Use the provided equations and substitute the value of a, b, c, d, e, f, g, h, i, j, k and l into role="math" localid="1647335016076" $s=\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of s.

$\begin{array}{c}s=\frac{\left|\begin{array}{ccc}1& −1& −5\\ 1& 5& −2\\ 4& −1& 1\end{array}\right|}{\left|\begin{array}{ccc}1& −2& −5\\ 1& 2& −2\\ 4& 1& 1\end{array}\right|}\\ =\frac{117}{57}\\ =\frac{39}{19}\end{array}$

Apply the Cramer’s rule for the value of t.

$t=\frac{\left|\begin{array}{ccc}a& b& j\\ d& e& k\\ g& h& l\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$

Use the provided equations and substitute the value of a, b, c, d, e, f, g, h, i, j, k and l into role="math" localid="1647335183063" $t=\frac{\left|\begin{array}{ccc}a& b& j\\ d& e& k\\ g& h& l\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of t.

$\begin{array}{c}t=\frac{\left|\begin{array}{ccc}1& −2& −1\\ 1& 2& 5\\ 4& 1& −1\end{array}\right|}{\left|\begin{array}{ccc}1& −2& −5\\ 1& 2& −2\\ 4& 1& 1\end{array}\right|}\\ =\frac{−42}{57}\\ =−\frac{14}{19}\end{array}$

The obtained value of r is $-\frac{11}{19}$, s is $\frac{39}{19}$ and t is $-\frac{14}{19}$ so the solution is $\left(-\frac{11}{19},\frac{39}{19},-\frac{14}{19}\right)$.