# CRAMER’S RULE

## CRAMER’S RULE

Let a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃ system linear equation. If  $$\Delta =\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|\ne 0$$ and $$d=\left[ \begin{matrix}{{d}_{1}} \\{{d}_{2}} \\{{d}_{3}} \\\end{matrix} \right]$$ then $${{\Delta }_{1}}=\left| \begin{matrix}{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|$$, $${{\Delta }_{2}}=\left| \begin{matrix}{{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\\end{matrix} \right|$$, $${{\Delta }_{3}}=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\\end{matrix} \right|$$ is the solution of above three equation.

$$x=\frac{{{\Delta }_{1}}}{\Delta }$$, $$y=\frac{{{\Delta }_{2}}}{\Delta }$$ and $$z=\frac{{{\Delta }_{3}}}{\Delta }$$.

Example: solve the equation x + y + z = 9, 2x + 5y + 7z = 52, and 2x + y – z = 0 by CRAMER’S Method.

Solution: Given that x + y + z = 9, 2x + 5y + 7z = 52, and 2x + y – z = 0

$$\Delta =\left| \begin{matrix}1 & 1 & 1 \\2 & 5 & 7 \\2 & 1 & -1 \\\end{matrix} \right|$$ and $$d=\left[ \begin{matrix}9 \\52 \\0 \\\end{matrix} \right]$$.

We can find the determinant of given matrix

$$\Delta =\left| \begin{matrix}1 & 1 & 1 \\2 & 5 & 7 \\2 & 1 & -1 \\\end{matrix} \right|$$ ,

= 1(-5 – 7) -1 (-2 – 14) + 1(2 – 10) = -12 + 16 – 8 = -4

Now we can find $${{\Delta }_{1}}=\left| \begin{matrix}9 & 1 & 1 \\52 & 5 & 7 \\0 & 1 & -1 \\\end{matrix} \right|$$,

= 9(-5 – 7) -1 (-52 – 0) + 1(52 – 0) = -108 + 52 + 52 = -4

Now we can find $${{\Delta }_{2}}=\left| \begin{matrix}1 & 9 & 1 \\2 & 52 & 7 \\2 & 0 & -1 \\\end{matrix} \right|$$,

= 1(-52 – 0) -9 (-2 – 14) + 1(0 – 104) = -52 + 144 – 104 = -12

Now we can find $${{\Delta }_{3}}=\left| \begin{matrix}1 & 1 & 9 \\2 & 5 & 52 \\2 & 1 & 0 \\\end{matrix} \right|$$,

= 1(0 – 52) – 1(0 – 104) + 9(2 – 10) = -52 + 104 – 72 = -20

By CRAMER’S RULE

$$x=\frac{{{\Delta }_{1}}}{\Delta }=\frac{-4}{-4}=1$$,

$$y=\frac{{{\Delta }_{2}}}{\Delta }=\frac{-12}{-4}=3$$,

$$z=\frac{{{\Delta }_{3}}}{\Delta }=\frac{-20}{-4}=5$$.

Therefore x = 1, y = 3 and z = 5 is the solution.