Cramer’s Rule Example
1) Solve the system of equation X – 2Y + 3Z = 2, 2X + 4Y + 2Z = -1, X + 2Y – 2Z = 5.
Solution: Given that
X – 2Y + 3Z = 2,
2X + 4Y + 2Z = -1,
X + 2Y – 2Z = 5
We can written as matrix form
\(\Delta =\left( \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & 2 \\ 1 & 2 & -2 \\\end{matrix} \right)\) ,
= 1(-8 – 4) + 2(-4 – 2) + 3(4 – 4)
= -12-12 + 0
= -24 ≠ 0
\({{\Delta }_{1}}=\left( \begin{matrix} 2 & -2 & 3 \\ -1 & 4 & 2 \\ 5 & 2 & -2 \\\end{matrix} \right)\),
= 2(-8 – 4) + 2(2 – 10) + 3(-2 – 20)
= -24 – 16 – 66
= -106
\({{\Delta }_{2}}=\left( \begin{matrix} 1 & 2 & 3 \\ 2 & -1 & 2 \\ 1 & 5 & -2 \\\end{matrix} \right)\),
= 1(2 – 10) – 2(-4 – 2) + 3(10 + 1)
= -8 + 12 + 33
= 37
\({{\Delta }_{3}}=\left( \begin{matrix} 1 & -2 & 2 \\ 2 & 4 & -1 \\ 1 & 2 & 5 \\\end{matrix} \right)\),
= 1(20 + 2) + 2(10 + 1) + 2(4 – 4)
= 22 + 22 + 0
= 44
∴ \(X=\frac{{{\Delta }_{1}}}{\Delta }\,=\,\frac{-106}{-24}\,=\,\frac{53}{12}\).
\(Y\,=\,\frac{{{\Delta }_{2}}}{\Delta }\,=\,\frac{37}{-24}\,=\,\frac{-37}{24}\).
\(Z=\frac{{{\Delta }_{3}}}{\Delta }\,=\,\frac{44}{-24}\,=\,\frac{-11}{6}\).
Solution is x = \(\frac{53}{12}\), y = \(\frac{-37}{12}\) and z = \(\frac{-11}{6}\) unique solution.