Q8. - Algebra 2

Question

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

$a+9b-2c=2$

$-a-3b+4c=1$

$2a+3b-6c=-5$

Step-by-Step Solution

Answer

The required solution is $\left(-5,\frac{2}{3},-9\right)$.

Step 1- Determine the value of a

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

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

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

$\begin{array}{c}a=\frac{\left|\begin{array}{ccc}2& 9& -2\\ 1& -3& 4\\ -5& 3& -6\end{array}\right|}{\left|\begin{array}{ccc}1& 9& -2\\ -1& -3& 4\\ 2& 3& -6\end{array}\right|}\\ =\frac{-90}{18}\\ =-5\end{array}$

Step 2 – Determine the value of b

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

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

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

$\begin{array}{c}b=\frac{\left|\begin{array}{ccc}1& 2& -2\\ -1& 1& 4\\ 2& -5& -6\end{array}\right|}{\left|\begin{array}{ccc}1& 9& -2\\ -1& -3& 4\\ 2& 3& -6\end{array}\right|}\\ =\frac{12}{18}\\ =\frac{2}{3}\end{array}$

Step 3 – Determine the value of c

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

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

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

$\begin{array}{c}b=\frac{\left|\begin{array}{ccc}1& 9& 2\\ -1& -3& 1\\ 2& 3& -5\end{array}\right|}{\left|\begin{array}{ccc}1& 9& -2\\ -1& -3& 4\\ 2& 3& -6\end{array}\right|}\\ =\frac{-9}{18}\\ =-\frac{1}{2}\end{array}$

Step 4- Write the solution

The obtained value of a is $-5$, b is $\frac{2}{3}$ and c is $-9$ so the solution is $\left(-5,\frac{2}{3},-9\right)$.