Q28. - Algebra 2

Question

Use Cramer’s Rule to solve each system of equations.

$r-2s-5t=-1$

$r+2s-2t=5$

$4r+s+t=-1$

Step-by-Step Solution

Answer

The required solution is $\left(-\frac{11}{19},\frac{39}{19},-\frac{14}{19}\right)$.

Step 1- Determine the value of r

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

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

Step 2 – Determine the value of s

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

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

Step 3 – Determine the value of t

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.

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

Step 4- Write the solution

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