Question

Use Cramer's rule to solve the systems in Exercises 22 through 24.

24. $\left|\begin{array}{cc}2x+3y& =8\\ 4y+5z& =3\\ 6x+7z& =-1\end{array}\right|$

Short answer

Therefore

$x=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$

Step-by-step solution

Definition. 

In matrices, Cramer's rule expresses the solution in terms of the determinants of the coefficient matrix (i.e., for a square matrix) and of matrices obtained from it by replacing one column by the column vector of right-hand-sides of the equations.

To solve. 

Using Cramer's rule, we solve:

$$\begin{gathered} \det \mathrm{A}=\left|\begin{array}{ccc}2& 3& 0\\ 0& 4& 5\\ 6& 0& 7\end{array}\right| \\ \det \mathrm{A}=146 \\ {\mathrm{x}}_1=\frac{\left|\begin{array}{ccc}8& 3& 0\\ 3& 4& 5\\ -1& 0& 7\end{array}\right|}{146} \\ {\mathrm{x}}_1=\frac{146}{146} \\ {\mathrm{x}}_1=1 \end{gathered}$$

$$\begin{gathered} x_2=\frac{\left|\begin{array}{ccc}2& 8& 0\\ 0& 3& 5\\ 6& -1& 7\end{array}\right|}{146} \\ {x}_2=2 \\ {x}_3=\frac{\left|\begin{array}{ccc}2& 3& 8\\ 0& 4& 3\\ 6& 0& -1\end{array}\right|}{146} \\ {x}_3=\frac{-146}{146} \\ {x}_3=-1 \end{gathered}$$

Therefore,

$x=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$