Question

Solve the equations:
$x+y+z+u=0$ ,
$ax+by+cz+du=0$ ,
${{a}^{2}}x+{{b}^{2}}y+{{c}^{2}}z+{{d}^{2}}u=0$ ,
${{a}^{3}}x+{{b}^{3}}y+{{c}^{3}}z+{{d}^{3}}u=k$

Answer 1

First let us try to find a general solution for x, y, z and u. First we try to develop a solution for a general form of a simultaneous equation.

& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}u={{e}_{1}} \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}u={{e}_{2}} \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z+{{d}_{3}}u={{e}_{3}} \\\ & {{a}_{4}}x+{{b}_{4}}y+{{c}_{4}}z+{{d}_{4}}u={{e}_{4}} \\\ \end{aligned}$$ Solve these 4 equations using Cramer’s formula to get the required values. **Complete step-by-step answer:** We can take two approaches to solve this problem. First, what we can do is to create a matrix A, X and D such that $A=\left( \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\\ {{a}_{4}} & {{b}_{4}} & {{c}_{4}} & {{d}_{4}} \\\ \end{matrix} \right)\ ,X=\left( \begin{matrix} x \\\ y \\\ z \\\ u \\\ \end{matrix} \right)\ ,\ D=\left( \begin{matrix} {{e}_{1}} \\\ {{e}_{2}} \\\ {{e}_{3}} \\\ {{e}_{4}} \\\ \end{matrix} \right)$ Thus, the above equation can be written in the form $AX\ =\ D\ \Rightarrow \ X\ =\ {{A}^{-1}}D$ But then calculating A-1 can be very cumbersome since A is $4\times 4$matrix. Another approach is to solve using Cramer’s rule, for that we will have five determinants ${{\Delta }_{1}},\ {{\Delta }_{2}},\ {{\Delta }_{3}},\ {{\Delta }_{4}}$and $\Delta $ such that $\begin{aligned} & \Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\\ {{a}_{4}} & {{b}_{4}} & {{c}_{4}} & {{d}_{4}} \\\ \end{matrix} \right| \\\ & {{\Delta }_{1}}=\ \left| \begin{matrix} {{e}_{1}} & {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\\ {{e}_{2}} & {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\\ {{e}_{3}} & {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\\ {{e}_{4}} & {{b}_{4}} & {{c}_{4}} & {{d}_{4}} \\\ \end{matrix} \right|\ ,\ {{\Delta }_{2}}=\left| \begin{matrix} {{a}_{1}} & {{e}_{1}} & {{c}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{e}_{2}} & {{c}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{e}_{3}} & {{c}_{3}} & {{d}_{3}} \\\ {{a}_{4}} & {{e}_{4}} & {{c}_{4}} & {{d}_{4}} \\\ \end{matrix} \right|\ ,\ \\\ & {{\Delta }_{3}}=\ \left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{e}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{e}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{e}_{3}} & {{d}_{3}} \\\ {{a}_{4}} & {{b}_{4}} & {{e}_{4}} & {{d}_{4}} \\\ \end{matrix} \right|\ ,\ {{\Delta }_{4}}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} & {{e}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} & {{e}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} & {{e}_{3}} \\\ {{a}_{4}} & {{b}_{4}} & {{c}_{4}} & {{e}_{4}} \\\ \end{matrix} \right| \\\ \end{aligned}$ Then $x,y,z$and $u$can be calculated by Cramer’s formula given as: $x=\dfrac{{{\Delta }_{1}}}{\Delta },\ \ y=\dfrac{{{\Delta }_{2}}}{\Delta },\ \ z=\dfrac{{{\Delta }_{3}}}{\Delta },\ \ u=\dfrac{{{\Delta }_{4}}}{\Delta }$ Here $\Delta \ =\ \left| \begin{matrix} 1 & 1 & 1 & 1 \\\ a & b & c & d \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} & {{d}^{2}} \\\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} & {{d}^{3}} \\\ \end{matrix} \right|$ At first it might seem very complicated but we have formula to find determinant of $4\times 4$ which is If we have matrix of $4\times 4$, say $\Delta \ =\ \left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}} \\\ {{a}_{41}} & {{a}_{42}} & {{a}_{43}} & {{a}_{44}} \\\ \end{matrix} \right|$, then it’s determinant can be evaluated by duplicating the first three columns to form three extra columns. then for each of the first four terms in the top row, $${{a}_{11}}$$, $${{a}_{12}}$$, $${{a}_{13}}$$ $${{a}_{13}}$$multiply by the determinant of the $3\times 3$matrix immediately below and to the right. Add these together to get the determinant of $4\times 4$ matrix. So, determinant of $4\times 4$ matrix will be, $$\Delta \ =\ \left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}} \\\ {{a}_{41}} & {{a}_{42}} & {{a}_{43}} & {{a}_{44}} \\\ \end{matrix} \right|={{a}_{11}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\\ {{a}_{32}} & {{a}_{33}} & {{a}_{34}} \\\ {{a}_{42}} & {{a}_{43}} & {{a}_{44}} \\\ \end{matrix} \right|+{{a}_{12}}\left| \begin{matrix} {{a}_{23}} & {{a}_{24}} & {{a}_{21}} \\\ {{a}_{33}} & {{a}_{34}} & {{a}_{31}} \\\ {{a}_{43}} & {{a}_{44}} & {{a}_{41}} \\\ \end{matrix} \right|+{{a}_{13}}\left| \begin{matrix} {{a}_{24}} & {{a}_{21}} & {{a}_{22}} \\\ {{a}_{34}} & {{a}_{31}} & {{a}_{32}} \\\ {{a}_{44}} & {{a}_{41}} & {{a}_{42}} \\\ \end{matrix} \right|+{{a}_{14}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ {{a}_{41}} & {{a}_{42}} & {{a}_{43}} \\\ \end{matrix} \right|$$ So, for $\Delta \ =\ \left| \begin{matrix} 1 & 1 & 1 & 1 \\\ a & b & c & d \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} & {{d}^{2}} \\\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} & {{d}^{3}} \\\ \end{matrix} \right|$, $$\Delta \ =\ \left| \begin{matrix} 1 & 1 & 1 & 1 \\\ a & b & c & d \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} & {{d}^{2}} \\\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} & {{d}^{3}} \\\ \end{matrix} \right|=1\left| \begin{matrix} b & c & d \\\ {{b}^{2}} & {{c}^{2}} & {{d}^{2}} \\\ {{b}^{3}} & {{c}^{3}} & {{d}^{3}} \\\ \end{matrix} \right|+1\left| \begin{matrix} c & d & a \\\ {{c}^{2}} & {{d}^{2}} & {{a}^{2}} \\\ {{c}^{3}} & {{d}^{3}} & {{a}^{3}} \\\ \end{matrix} \right|+1\left| \begin{matrix} d & a & b \\\ {{d}^{2}} & {{a}^{2}} & {{b}^{2}} \\\ {{d}^{3}} & {{a}^{3}} & {{b}^{3}} \\\ \end{matrix} \right|+1\left| \begin{matrix} a & b & c \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ {{a}^{3}} & {{a}^{3}} & {{a}^{3}} \\\ \end{matrix} \right|$$ We know that, if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as, $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$ Then, $\Delta $ can be simplified similarly and then upon simplification we will get $\Delta =(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)$ Now, ${{\Delta }_{1}}=\left| \begin{matrix} 0 & 1 & 1 & 1 \\\ 0 & b & c & d \\\ 0 & {{b}^{2}} & {{c}^{2}} & {{d}^{2}} \\\ k & {{b}^{3}} & {{c}^{3}} & {{d}^{3}} \\\ \end{matrix} \right|=-k\left| \begin{matrix} 1 & 1 & 1 \\\ b & c & d \\\ {{b}^{2}} & {{c}^{2}} & {{d}^{2}} \\\ \end{matrix} \right|$ This is exactly the form of determinant we simplified earlier. Therefore, ${{\Delta }_{1}}=k(b-c)(c-d)(d-b)$ Similarly, $\begin{aligned} & {{\Delta }_{2}}=k(a-c)(c-d)(d-a) \\\ & {{\Delta }_{3}}=-k(a-b)(b-d)(d-a) \\\ & {{\Delta }_{4}}=k(a-b)(b-c)(c-a) \\\ \end{aligned}$ $\begin{aligned} & x=\dfrac{{{\Delta }_{1}}}{\Delta }=\dfrac{-k}{(b-a)(d-a)(c-a)} \\\ & y=\dfrac{{{\Delta }_{2}}}{\Delta }=\dfrac{-k}{\left( b-a \right)(b-c)(b-d)} \\\ & z=\dfrac{{{\Delta }_{3}}}{\Delta }=\dfrac{-k}{(c-a)(c-b)(c-d)} \\\ & u=\dfrac{{{\Delta }_{4}}}{\Delta }=\dfrac{-k}{(d-a)(d-b)(d-c)} \\\ \end{aligned}$ Hence, the question is solved. **Note:** The first method discussed above, i.e. to solve by solving the equation $AX=D$might be very lengthy. Therefore, it is advisable to solve using Cramer’s rule. While solving the determinants we should be very careful with the signs of cofactors. Always remember that, determinant of $4\times 4$ matrix will be,$$\Delta \ =\ \left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}} \\\ {{a}_{41}} & {{a}_{42}} & {{a}_{43}} & {{a}_{44}} \\\ \end{matrix} \right|={{a}_{11}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\\ {{a}_{32}} & {{a}_{33}} & {{a}_{34}} \\\ {{a}_{42}} & {{a}_{43}} & {{a}_{44}} \\\ \end{matrix} \right|+{{a}_{12}}\left| \begin{matrix} {{a}_{23}} & {{a}_{24}} & {{a}_{21}} \\\ {{a}_{33}} & {{a}_{34}} & {{a}_{31}} \\\ {{a}_{43}} & {{a}_{44}} & {{a}_{41}} \\\ \end{matrix} \right|+{{a}_{13}}\left| \begin{matrix} {{a}_{24}} & {{a}_{21}} & {{a}_{22}} \\\ {{a}_{34}} & {{a}_{31}} & {{a}_{32}} \\\ {{a}_{44}} & {{a}_{41}} & {{a}_{42}} \\\ \end{matrix} \right|+{{a}_{14}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ {{a}_{41}} & {{a}_{42}} & {{a}_{43}} \\\ \end{matrix} \right|$$ This will be helpful in solving questions based on the similar pattern. While calculating $\Delta $, we developed the analogue from the result above but one must be very cautious in doing so as we can see that the analogue of the above form in a $4\times 4$determinant form we get 6 terms and not four terms. If not careful one might be tempted to just write $\Delta =(a-b)(b-c)(c-d)(d-a)$ which will be a blunder mistake. Therefore, one should solve the determinant and not try to find any shortcut. However, the method to solve would be similar as discussed in the solution.