Question

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

22.$\left|\begin{array}{ccc}3x+& 7y=& 1\\ 4x+& 11y=& 3\end{array}\right|$

Short answer

Therefore,

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

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:

$$\begin{gathered} detA=\left|\begin{array}{cc}3& 7\\ 4& 11\end{array}\right| \\ detA=5 \\ {x}_1=\frac{\left|\begin{array}{cc}1& 7\\ 3& 11\end{array}\right|}{5} \\ {x}_1=\frac{-10}{5} \\ {x}_1=-2 \\ {x}_2=\frac{\left[\begin{array}{cc}3& 1\\ 4& 3\end{array}\right]}{5} \\ {x}_2=\frac{5}{5} \\ {x}_2=1 \end{gathered}$$

Therefore,

$x=\left|\begin{array}{c}-2\\ 1\end{array}\right|$.