Question

Solve the system of equations using Cramer's Rule: \[ 2x+3y=8, \qquad x-y=-1. \]

Hint:

In Cramer's Rule: \[ x=\frac{D_x}{D}, \quad y=\frac{D_y}{D} \] First compute \(D\). If \(D=0\), the method cannot give a unique solution.

Answer 1

Concept: Cramer's Rule is used to solve a system of linear equations when the determinant of the coefficient matrix is non-zero. For the system \[ a_1x+b_1y=c_1, \qquad a_2x+b_2y=c_2 \] the solution is \[ x=\frac{D_x}{D}, \qquad y=\frac{D_y}{D} \] where \[ D= \begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}, \quad D_x= \begin{vmatrix} c_1 & b_1\\ c_2 & b_2 \end{vmatrix}, \quad D_y= \begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix} \]
Step 1: Form the determinant \(D\). \[ D= \begin{vmatrix} 2 & 3\\ 1 & -1 \end{vmatrix} \] \[ =2(-1)-3(1)=-2-3=-5 \] Since \(D \neq 0\), the system has a unique solution.
Step 2: Find \(D_x\). \[ D_x= \begin{vmatrix} 8 & 3\\ -1 & -1 \end{vmatrix} \] \[ =8(-1)-3(-1)=-8+3=-5 \]
Step 3: Find \(D_y\). \[ D_y= \begin{vmatrix} 2 & 8\\ 1 & -1 \end{vmatrix} \] \[ =2(-1)-8(1)=-2-8=-10 \]
Step 4: Calculate the values of \(x\) and \(y\). \[ x=\frac{D_x}{D}=\frac{-5}{-5}=1 \] \[ y=\frac{D_y}{D}=\frac{-10}{-5}=2 \]
Step 5: Final Solution. \[ \boxed{x=1, \qquad y=2} \]