# Cramer’s Rule Example

### 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.