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The standard form for linear equations in two variables is $\mathit{A}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{C}$.

The determinant of$\mathbf{2}\mathit{x}\mathbf{2}$matrixes$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\mathbf{=}\mathbf{}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathbf{-}\mathbf{}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{\$}$.

Consider the set of equations$\left\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}\right.$ If multiplying the first equation by $b_2$, the second by $b_1,$, and then subtract the results and solve for $x$,

(if $a_1{b}_2-a_2{b}_1≠0$),

$x=\frac{c_1{b}_2-c_2{b}_1}{a_1{b}_3-a_3{b}_1}.$

Solving for role="math" localid="1664282277055" $\mathit{y}$in a similar way,role="math" localid="1664282264503" $\mathit{y}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{2}}{\mathbf{a}}_{\mathbf{1}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{a}}_{\mathbf{2}}}{{\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_{\mathbf{3}}\mathbf{-}{\mathbf{a}}_{\mathbf{3}}{\mathbf{b}}_{\mathbf{1}}}$.

Using the definition of second order determinant, write the solutionandin the form,

role="math" localid="1664282245159" $\mathit{x}\mathbf{=}\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{x}}}{\mathbf{D}}\mathbf{,}\mathbf{}\mathit{y}\mathbf{=}\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{y}}}{\mathbf{D}}$

This determinant is called the determinant of the coefficients. To find the numerator determinant for $x$, start with $D$, erase the $x$coefficients $a_1$and $a_2$, and replace them by the constants $c_1$and $c_2$from the right-hand sides of the equations. Similarly, replace the y coefficients in by$D$ the constant terms to find the numerator determinant in y .This method of solution of a set of linear equations is called Cramer's rule.

Rewriting the first equation in standard form,

$$\begin{gathered} n{x}^{\text{'}}=\mathrm{γ}(x-vt)n \\ {x}^{\text{'}}=\mathrm{γ}x-\mathrm{γ}vt \\ \mathrm{γ}x-\mathrm{γ}vt=x^{\text{'}} \end{gathered}$$

Rewriting the second equation in standard form,

$$\begin{gathered} t^{\text{'}}=\mathrm{γ}\left(t-vx/c^2\right) \\ {t}^{\text{'}}=\mathrm{γ}t-\mathrm{γ}v/c^{2}x \\ \mathrm{γ}t-\mathrm{γ}v/c^{2}x=t^{\text{'}} \\ -\mathrm{γ}v/c^{2}x+\mathrm{γ}t=t^{\text{'}} \end{gathered}$$

Identifying $D,D_x$and $D_t$of the given system of equation $\left\{\begin{array}{l}\mathrm{γ}x-\mathrm{γ}vt=x^{\text{'}}\\ -\mathrm{γ}v/c^{2}x+\mathrm{γ}t=t^{\text{'}}\end{array}\right.$

$$\begin{gathered} D=\left|\begin{array}{cc}\mathrm{γ}& -\mathrm{γ}v\\ -\mathrm{γ}v/c^2& \mathrm{γ}\end{array}\right| \\ {D}_x=\left|\begin{array}{cc}x\text{'}& -\mathrm{γ}v\\ t\text{'}& \mathrm{γ}\end{array}\right| \\ {D}_t=\left|\begin{array}{cc}\mathrm{γ}& x\text{'}\\ -\mathrm{γ}v/c^2& t\text{'}\end{array}\right| \end{gathered}$$

Finding an expression $D$for using the formula to find the determinant of $2×2$matrices,

$$\begin{gathered} D=\left(\mathrm{γ}\right)\left(\mathrm{γ}\right)-\left(-\mathrm{γ}v/c^2\right)(-\mathrm{γ}v) \\      ={\mathrm{γ}}^2-{\mathrm{γ}}^2{v}^2/c^2 \\      ={\mathrm{γ}}^2\left(1-v^2/c^2\right) \\      =1 \end{gathered}$$

Finding an expression for role="math" localid="1664281815770" $D_x$using the formula to find the determinant of $2×2$matrices,

role="math" localid="1664281957930" $$\begin{gathered} D_x=\left(x^{\text{'}}\right)\left(\mathrm{γ}\right)-\left(t^{\text{'}}\right)(-\mathrm{γ}v) \\         =\mathrm{γ}{x}^{\text{'}}+\mathrm{γ}v{t}^{\text{'}} \\         =\mathrm{γ}\left(x^{\text{'}}+v{t}^{\text{'}}\right) \end{gathered}$$

Finding an expression for role="math" localid="1664281783782" $D_y$using the formula to find the determinant of $2×2$matrices,

role="math" localid="1664281914908" $$\begin{gathered} D_t=\left(\mathrm{γ}\right)\left(t^{\text{'}}\right)-\left(-\mathrm{γ}v/c^2\right)\left(x^{\text{'}}\right) \\        =\mathrm{γ}{t}^{\text{'}}+\mathrm{γ}v{x}^{\text{'}}/c^2 \\        =\mathrm{γ}\left(t^{\text{'}}+v{x}^{\text{'}}/c^2\right) \end{gathered}$$

Finding$x$ using Cramer's rule $x=\frac{D_x}{D}$.

$$\begin{gathered} x=\frac{\mathrm{γ}\left(x^{\text{'}}+v{t}^{\text{'}}\right)}{1} \\      =\mathrm{γ}\left(x^{\text{'}}+v{t}^{\text{'}}\right) \end{gathered}$$

Finding t using Cramer's rule $t=\frac{D_t}{D}$.

$$\begin{gathered} t=\frac{\mathrm{γ}\left(t^{\text{'}}+v{x}^{\text{'}}/c^2\right)}{1} \\     =\mathrm{γ}\left(t^{\text{'}}+v{x}^{\text{'}}/c^2\right) \end{gathered}$$