Answer
Cramer’s rule is a faster method for solving a determinant.
Work Step by Step
A linear system in three variables is:
$\begin{align}
& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\
& {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\
& {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\
\end{align}$
Where ${{a}_{1}},{{a}_{2}},{{a}_{3,}}{{b}_{1}}\text{,}{{\text{b}}_{2}}\text{,}{{\text{b}}_{3}}\text{,}{{\text{c}}_{1}}\text{,}{{\text{c}}_{2}}\text{and }{{\text{c}}_{3}}\text{ are coffecients and }{{\text{d}}_{1}},{{d}_{2}}\And {{c}_{3}}\text{ are constants}\text{.}$
Then,
$\begin{align}
& x=\frac{{{D}_{x}}}{D},y=\frac{{{D}_{y}}}{D},z=\frac{{{D}_{z}}}{D},\text{ where D}\ne \text{0} \\
& \\
\end{align}$
Where,
$D=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\text{ }$
These are the coefficients of variables x, y, z.
${{D}_{x}}=\left| \begin{matrix}
{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$
Replace $x$ -coefficients in $D$ with the constants on the right of the three equations.
${{D}_{y}}=\left| \begin{matrix}
{{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$ Replace $y$ -coefficients in $D$ with the constants on the right of the three equations.
${{D}_{z}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right|$ Replace $z$ -coefficients in $D$ with the constants on the right of the three equations.
Example, consider the given linear system
$\begin{align}
& 2x+1y+1z=3 \\
& 1x-1y-1z=0 \\
& 1x+2y+1z=0 \\
\end{align}$
So,
$D=\left| \begin{matrix}
2 & 1 & 1 \\
1 & -1 & -1 \\
1 & 2 & 1 \\
\end{matrix} \right|$
Now, we will find the variable x
${{D}_{x}}=\left| \begin{matrix}
3 & 1 & 1 \\
0 & -1 & -1 \\
0 & 2 & 1 \\
\end{matrix} \right|$
Solve for $D$ along row 1,
$\begin{align}
& \text{D=2}\left| \begin{matrix}
-1 & -1 \\
2 & 1 \\
\end{matrix} \right|-1\left| \begin{matrix}
1 & -1 \\
1 & 1 \\
\end{matrix} \right|+1\left| \begin{matrix}
1 & -1 \\
1 & 2 \\
\end{matrix} \right| \\
& D=2\left\{ -1\cdot 1-(-1)2 \right\}-1\left\{ 1\cdot 1-(-1)1 \right\}+1\left\{ 1\cdot 2-(-1)1 \right\} \\
& D=2(-1+2)-1(1+1)+1(2+1) \\
& D=2-2+3=3
\end{align}$
Similarly,
${{D}_{x}}=\left| \begin{matrix}
3 & 1 & 1 \\
0 & -1 & -1 \\
0 & 2 & 1 \\
\end{matrix} \right|$
Expand along column 1,
$\begin{align}
& {{D}_{x}}=3\left| \begin{matrix}
-1 & -1 \\
2 & 1 \\
\end{matrix} \right|-0\left| \begin{matrix}
1 & 1 \\
-1 & 1 \\
\end{matrix} \right|+0\left| \begin{matrix}
1 & 1 \\
-1 & -1 \\
\end{matrix} \right| \\
& {{D}_{x}}=3\left\{ -1\cdot 1-(-1)(2) \right\}-0+0 \\
& {{D}_{x}}=3(-1+2)=3
\end{align}$
Therefore,
$\begin{align}
& x=\frac{{{D}_{x}}}{D} \\
& x=\frac{3}{3}=1 \\
\end{align}$
By this method, one can easily evaluate the determinant in a faster way.
Cramer’s rule is a faster method for solve a determinant.
