Question

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

$3a+c=23$

$4a+7b-2c=-22$

$8a-b-c=34$

Short answer

The required solution is$\left(\frac{141}{29},-\frac{102}{29},\frac{244}{29}\right)$

Step-by-step solution

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.

$\begin{array}{c}a=\frac{\left|\begin{array}{ccc}23& 0& 1\\ -22& 7& -2\\ 34& -1& -1\end{array}\right|}{\left|\begin{array}{ccc}3& 0& 1\\ 4& 7& -2\\ 8& -1& -1\end{array}\right|}\\ =\frac{-423}{-87}\\ =\frac{141}{29}\end{array}$

– Determine the value of b 

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

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

– 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.

$\begin{array}{c}c=\frac{\left|\begin{array}{ccc}3& 0& 23\\ 4& 7& -22\\ 8& -1& 34\end{array}\right|}{\left|\begin{array}{ccc}3& 0& 1\\ 4& 7& -2\\ 8& -1& -1\end{array}\right|}\\ =\frac{-732}{-87}\\ =\frac{244}{29}\end{array}$

Step 4- Write the solution

The obtained value of a is$\frac{141}{29}$ , b is$-\frac{102}{29}$ and c is $\frac{244}{29}$so the solution is .$\left(\frac{141}{29},-\frac{102}{29},\frac{244}{29}\right)$