Q27. - Algebra 2

Question

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

$a-2b+c=7$

$6a+2b-2c=4$

$4a+6b+4c=14$

Step-by-Step Solution

Answer

The required solution is $\left(2,-1,3\right)$.

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.

$\begin{array}{c}a=\frac{\left|\begin{array}{ccc}7& -2& 1\\ 4& 2& -2\\ 14& 6& 4\end{array}\right|}{\left|\begin{array}{ccc}1& -2& 1\\ 6& 2& -2\\ 4& 6& 4\end{array}\right|}\\ =\frac{224}{112}\\ =2\end{array}$

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.

$\begin{array}{c}b=\frac{\left|\begin{array}{ccc}1& 7& 1\\ 6& 4& -2\\ 4& 14& 4\end{array}\right|}{\left|\begin{array}{ccc}1& -2& 1\\ 6& 2& -2\\ 4& 6& 4\end{array}\right|}\\ =\frac{-112}{112}\\ =-1\end{array}$

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.

$\begin{array}{c}c=\frac{\left|\begin{array}{ccc}1& -2& 7\\ 6& 2& 4\\ 4& 6& 14\end{array}\right|}{\left|\begin{array}{ccc}1& -2& 1\\ 6& 2& -2\\ 4& 6& 4\end{array}\right|}\\ =\frac{336}{112}\\ =3\end{array}$

Step 4- Write the solution

The obtained value of a is 2, b is $-1$ and c is 3 so the solution is $\left(2,-1,3\right)$.