Solution: Assume there be a system of n simultaneous linear equations and with ‘n’ unknown given by
${{a}_{11}}{{X}_{1}}+{{a}_{12}}{{X}_{2}}+…+{{a}_{1n}}{{X}_{n}}={{b}_{1}}$
${{a}_{21}}{{X}_{1}}+{{a}_{22}}{{X}_{2}}+…+{{a}_{2n}}{{X}_{n}}={{b}_{2}}$
${{a}_{n1}}{{X}_{1}}+{{a}_{n2}}{{X}_{2}}+…+{{a}_{nn}}{{X}_{n}}={{b}_{n}}$
$LetD=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & \cdots & {{a}_{1n}} \\
{{a}_{21}} & {{a}_{22}} & \cdots & {{a}_{2n}} \\
\vdots & \vdots & {} & \vdots \\
{{a}_{n1}} & {{a}_{n1}} & \cdots & {{a}_{nn}} \\
\end{matrix} \right|$Let ${{D}_{1}}$ be the determinant obtained from D after replacing the ${{j}^{th}}$column by
$\left| \begin{matrix}
{{b}_{1}} \\
{{b}_{2}} \\
\vdots \\
{{b}_{n}} \\
\end{matrix} \right|$Then,
${{X}_{1}}=\frac{{{D}_{1}}}{D},{{X}_{2}}=\frac{{{D}_{2}}}{D},…{{X}_{n}}=\frac{{{D}_{n}}}{D}$provided that $D\ne 0$
$3x+y+z=2$
$2x-4y+3z=-1$
$4x+y-3z=-11$
So by comparing with the theorem, let’s find $D,{{D}_{1}}And{{D}_{2}}$
$\Rightarrow D=\left| \begin{matrix}
3 & 1 & 1 \\
2 & -4 & 3 \\
4 & 1 & -3 \\
\end{matrix} \right|$
Solving for the determinant, expanding along 1st row
$\Rightarrow D=3\left[ (-4)(-3)-(3)(1) \right]-1\left[ (2)(-3)-12 \right]+1\left[ 2-4(-4) \right]$
$\Rightarrow D=3\left[ 12-3 \right]-\left[ -6-12 \right]+\left[ 2+16 \right]$
$\Rightarrow D=27+18+18$
$\Rightarrow D=63$
Again,
$\Rightarrow {{D}_{1}}=\left| \begin{matrix}
2 & 1 & 1 \\
-1 & -4 & 3 \\
-11 & 1 & -3 \\
\end{matrix} \right|$
Solving for the determinant, expanding along 1st row
$\Rightarrow {{D}_{1}}=2\left[ (-4)(-3)-(3)(1) \right]-1\left[ (-1)(-3)-(-11)(3) \right]+1\left[ (-1)-(-4)(-11) \right]$
$\Rightarrow {{D}_{1}}=2\left[ 12-3 \right]-1\left[ 3+33 \right]+1\left[ -1-44 \right]$
$\Rightarrow {{D}_{1}}=2\left[ 9 \right]-36-45$
$\Rightarrow {{D}_{1}}=18-36-45$
$\Rightarrow {{D}_{1}}=-63$
Now,
$\Rightarrow {{D}_{2}}=\left| \begin{matrix}
3 & 2 & 1 \\
2 & -1 & 3 \\
4 & -11 & -3 \\
\end{matrix} \right|$
Solving for the determinant, expanding along 1st row
$\Rightarrow {{D}_{2}}=3\left[ 3+33 \right]-2\left[ -6-12 \right]+1\left[ -22+4 \right]$
⇒ ${{D}_{2}}=3\left[ 36 \right]-2(-18)-18$
⇒ ${{D}_{2}}=126$
$\Rightarrow {{D}_{3}}=\left| \begin{matrix}
3 & 1 & 2 \\
2 & -4 & -1 \\
4 & -1 & -11 \\
\end{matrix} \right|$
Solving for the determinant, expanding along 1st row
$\Rightarrow {{D}_{3}}=3\left[ 44+1 \right]-1\left[ -22+4 \right]+2\left[ 2+16 \right]$
⇒ ${{D}_{3}}=3\left[ 45 \right]-1(-18)-2(18)$
$\Rightarrow {{D}_{3}}=135+18+36$
⇒ ${{D}_{3}}=189$
Therefore, by Cramer’s Rule, we will get
$\Rightarrow X=\frac{{{D}_{1}}}{D}$
$\Rightarrow X=\frac{-63}{63}$
$\Rightarrow X=-1$
$\Rightarrow Y=\frac{{{D}_{2}}}{D}$
$\Rightarrow Y=\frac{126}{63}$
$\Rightarrow Y=2$
$\Rightarrow Z=\frac{{{D}_{3}}}{D}$
$\Rightarrow Z=\frac{189}{63}$
$\Rightarrow Z=3$