Use Cramer’s rule to solve each of the following systems of linear equations over the field of rational numbers.
\(x + y + z = 11\) \(\phantom{(a)} 2x - 6y - z = 0\) \(\phantom{(a)} 3x + 4y + 2z = 0\).
\(3x - 2y = 7\) \(\phantom{(b)} 3y - 2z = 6\) \(\phantom{(b)} 3z - 2x = -1\).
Solution
For (a), the coefficient determinant is \[ \Delta=\det\begin{bmatrix}1&1&1\\2&-6&-1\\3&4&2\end{bmatrix}=-9. \] Replacing the first, second, and third columns by \((11,0,0)^t\) gives respectively \[ \Delta_x=72,\qquad \Delta_y=63,\qquad \Delta_z=-234. \] Cramer’s rule therefore gives \[ (x,y,z)=(-8,-7,26). \]
For (b), \[ \Delta=\det\begin{bmatrix}3&-2&0\\0&3&-2\\-2&0&3\end{bmatrix}=19. \] The three replaced-column determinants are \(95,76,57\), so \[ (x,y,z)=\left(\frac{95}{19},\frac{76}{19},\frac{57}{19}\right)=(5,4,3). \]