Q9. - Algebra 2

Question

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

$r+4s+3t=10$

$2r-2s+t=15$

$r+2s-3t=-1$

Step-by-Step Solution

Answer

The required solution is $\left(6,-\frac{1}{2},2\right)$.

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.

$\begin{array}{c}r=\frac{\left|\begin{array}{ccc}10& 4& 3\\ 15& -2& 1\\ -1& 2& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 4& 3\\ 2& -2& 1\\ 1& 2& -3\end{array}\right|}\\ =\frac{300}{50}\\ =6\end{array}$

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.

$\begin{array}{c}s=\frac{\left|\begin{array}{ccc}1& 10& 3\\ 2& 15& 1\\ 1& -1& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 4& 3\\ 2& -2& 1\\ 1& 2& -3\end{array}\right|}\\ =\frac{-25}{50}\\ =-\frac{1}{2}\end{array}$

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.

$\begin{array}{c}s=\frac{\left|\begin{array}{ccc}1& 4& 10\\ 2& -2& 15\\ 1& 2& -1\end{array}\right|}{\left|\begin{array}{ccc}1& 4& 3\\ 2& -2& 1\\ 1& 2& -3\end{array}\right|}\\ =\frac{100}{50}\\ =2\end{array}$

Step 4- Write the solution

The obtained value of r is 6, s is $-\frac{1}{2}$ and t is 2 so the solution is $\left(6,-\frac{1}{2},2\right)$.