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 localid="1647259000209" width="90">r=jbckeflhiabcdefghiafter it perform simplification of determinants to write the value of r.

$\begin{array}{c}r=\frac{\left|\begin{array}{ccc}40& 25& 0\\ −2& 12& 6\\ −10& −25& 50\end{array}\right|}{\left|\begin{array}{ccc}2& 25& 0\\ 10& 12& 6\\ 36& −25& 50\end{array}\right|}\\ =\frac{31000}{−5600}\\ =−\frac{155}{28}\end{array}$

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="1647259251174" $s=\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of s.

$\begin{array}{c}s=\frac{\left|\begin{array}{ccc}2& 40& 0\\ 10& −2& 6\\ 36& −10& 50\end{array}\right|}{\left|\begin{array}{ccc}2& 25& 0\\ 10& 12& 6\\ 36& −25& 50\end{array}\right|}\\ =\frac{−11440}{−5600}\\ =\frac{143}{70}\end{array}$

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 $t=\frac{\left|\begin{array}{ccc}a& b& j\\ d& e& k\\ g& h& l\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of t.

$\begin{array}{c}s=\frac{\left|\begin{array}{ccc}2& 25& 40\\ 10& 12& −2\\ 36& −25& −10\end{array}\right|}{\left|\begin{array}{ccc}2& 25& 0\\ 10& 12& 6\\ 36& −25& 50\end{array}\right|}\\ =\frac{−26920}{−5600}\\ =\frac{673}{140}\end{array}$

The obtained value of r is $-\frac{155}{28}$, s is $\frac{143}{70}$ and t is $\frac{673}{140}$ so the solution is $\left(-\frac{155}{28},\frac{143}{70},\frac{673}{140}\right)$.