Q30. - Algebra 2

Question

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

$4x+2y-3z=-32$

$-x-3y+z=54$

$2y+8z=78$

Step-by-Step Solution

Answer

The required solution is $\left(11,-17,14\right)$.

Step 1- Determine the value of x

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

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

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

$\begin{array}{c}x=\frac{\left|\begin{array}{ccc}-32& 2& -3\\ 54& -3& 1\\ 78& 2& 8\end{array}\right|}{\left|\begin{array}{ccc}4& 2& -3\\ -1& -3& 1\\ 0& 2& 8\end{array}\right|}\\ =\frac{-902}{-82}\\ =11\end{array}$

Step 2 – Determine the value of y

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

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

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

$\begin{array}{c}y=\frac{\left|\begin{array}{ccc}4& -32& -3\\ -1& 54& 1\\ 0& 78& 8\end{array}\right|}{\left|\begin{array}{ccc}4& 2& -3\\ -1& -3& 1\\ 0& 2& 8\end{array}\right|}\\ =\frac{1394}{-82}\\ =-17\end{array}$

$\begin{array}{c}y=\frac{\left|\begin{array}{ccc}4& -32& -3\\ -1& 54& 1\\ 0& 78& 8\end{array}\right|}{\left|\begin{array}{ccc}4& 2& -3\\ -1& -3& 1\\ 0& 2& 8\end{array}\right|}\\ =\frac{1394}{-82}\\ =-17\end{array}$

Step 3 – Determine the value of z

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

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

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

$\begin{array}{c}y=\frac{\left|\begin{array}{ccc}4& 2& -32\\ -1& -3& 54\\ 0& 2& 78\end{array}\right|}{\left|\begin{array}{ccc}4& 2& -3\\ -1& -3& 1\\ 0& 2& 8\end{array}\right|}\\ =\frac{-1148}{-82}\\ =14\end{array}$

Step 4- Write the solution

The obtained value of x is 11, y is $-17$ and z is 14 so the solution is $\left(11,-17,14\right)$.