Question

In the following set of equations (from a quantum mechanics problem), Aand Bare the unknowns, kand Kare given, and$\mathit{i}\mathbf{=}\sqrt{\mathbf{-}\mathbf{1}}$.Use Cramer's rule to find Aand show that$\mathbf{|}\mathit{A}{\mathbf{|}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$

$\{\begin{array}{l}A-B=-1\\ ikA-KB=ik\end{array}$

Short answer

By using Cramer’s rule $A=\frac{K+ik}{-K+ik}$.

It is proved that $|A{|}^2=1$.

Step-by-step solution

Formula and concept used

Using the formula, the determinant of $\mathbf{2}\mathbf{\times }\mathbf{2}$matrixes$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\mathbf{=}\mathit{a}\mathit{d}\mathbf{-}\mathit{c}\mathit{b}$.

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 $a_1{b}_2-a_2{b}_1\ne 0$),

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

Solving for role="math" localid="1664284685454" $\mathit{y}$in a similar way,role="math" localid="1664284668716" $\mathit{y}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{2}}{\mathbf{a}}_{\mathbf{1}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{a}}_{\mathbf{2}}}{{\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_{\mathbf{3}}\mathbf{-}{\mathbf{a}}_{\mathbf{3}}{\mathbf{b}}_{\mathbf{1}}}$.

Using the definition of second order determinant, write the solution x and y in the form

role="math" localid="1664284649178" $\mathit{x}\mathbf{=}\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{x}}}{\mathbf{D}}\mathbf{,}\mathbf{}\mathit{y}\mathbf{=}\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{y}}}{\mathbf{D}}$

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 D by 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.

Step 2: By using Cramer’s rule find A

Identifying$D,D_x$ and $D_y$of the given system of equation$\left\{\begin{array}{l}A-B=-1\\ ikA-KB=ik\end{array}\right.$.

$$\begin{gathered} D=\left|\begin{array}{cc}1& -1\\ ik& -K\end{array}\right| \\ {D}_A=\left|\begin{array}{cc}-1& -1\\ ik& -K\end{array}\right| \\ {D}_B=\left|\begin{array}{cc}1& -1\\ ik& ik\end{array}\right| \end{gathered}$$

Finding an expression for $D$using the formula to find the determinant of$2\times 2$ matrixes $D=\left(1\right)(-K)-\left(ik\right)(-1)=-K+ik$.

Finding an expression for$D_x$using the formula to find the determinant of$2\times 2$matrixes$D_A=(-1)(-K)-\left(ik\right)(-1)=K+ik$.

Finding an expression for$D_y$using the formula to find the determinant of$2\times 2$matrixes,

$$\begin{gathered} D_B=\left(1\right)\left(ik\right)-\left(ik\right)(-1) \\ =ik+ik \\ =2ik \end{gathered}$$

Finding$A$ using Cramer's rule$A=\frac{D_A}{D}$.

$A=\frac{K+ik}{-K+ik}$

To show that |A|2=1

Finding the value $\left|A\right|$of using the rule $|a+ib|=\sqrt{a^2+b^2}$

$$\begin{gathered} \left|\frac{K+ik}{-K+ik}\right|=\frac{|K+ik|}{|-K+ik|} \\ =\frac{\sqrt{K^2+k^2}}{\sqrt{{(-K)}^2+k^2}} \\ =\frac{\sqrt{K^2+k^2}}{\sqrt{K^2+k^2}} \\ =1 \end{gathered}$$

The value of $|A{|}^2$

role="math" localid="1664284551229" $$\begin{gathered} {\left|A\right|}^2=1^2 \\ =1 \end{gathered}$$

Hence proved.