Systems of linear equations

Solve the systems of linear equations of order 2 and order 3 by Cramer’s rule, inverse matrices and Gaussian elimination

  • Theory
  • Examples
    Consider the following linear systems (where a, b and c are distinct real numbers and \(ab + bc + ac \ne 0\)):
    \((*)\,\left\{ \begin{array}{l}x + {a^2}y + {a^3}z = 0\\x + {b^2}y + {b^3}z = 0\\x + {c^2}y + {c^3}z = 0\end{array} \right.\) and \((**)\,\left\{ \begin{array}{l}x + {a^2}y + {a^3}z = 1 + {a^2} + {a^3}\\x + {b^2}y + {b^3}z = 1 + {b^2} + {b^3}\\x + {c^2}y + {c^3}z = 1 + {c^2} + {c^3}\end{array} \right.\)
    (a) Prove that (*) has no non-trivial solutions.
    (b) Find the solution of (**).
  • Solutions
    (a)
    \(\begin{array}{1}D = \left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}\,} \right|\\
    = {\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\0&{{b^2} - {a^2}}&{{b^3} - {a^3}}\\0&{{c^2} - {a^2}}&{{c^3} - {a^3}}\end{array}\,} \right|^{}}\\ = {\left| {\,\begin{array}{*{20}{c}}{{b^2} - {a^2}}&{{b^3} - {a^3}}\\{{c^2} - {a^2}}&{{c^3} - {a^3}}\end{array}\,} \right|^{}}\\ = (b - a)(c - a)\,{\left| {\,\begin{array}{*{20}{c}}{b + a}&{{b^2} + ab + {a^2}}\\{c + a}&{{c^2} + ac + {a^2}}\end{array}\,} \right|^{}}\\ = (b - a)(c - a){(b{c^2} + abc + {a^2}b + a{c^2} + {a^2}c + {a^3} - {b^2}c - abc - {a^2}c - a{b^2} - {a^2}b - {a^3})^{^{^{}}}}\\ = (b - a)(c - a){(b{c^2} + a{c^2} - {b^2}c - a{b^2})^{^{^{}}}}\\ = (b - a)(c - a){[bc(c - b) + a({c^2} - {b^2})]^{^{^{}}}}\\ = (b - a)(c - a)(c - b){[bc + a(c + b)]^{^{}}}\\ = (b - a)(c - a)(c - b){(ab + bc + ac)^{^{}}}\end{array}\)
    ∵ a, b and c are distinct real numbers and \(ab + bc + ac \ne 0\)
    ∴ \(D \ne 0\) i.e. (*) has no non-trivial solutions.

    (b)
    \(\begin{array}{1}{D_x} = \left| {\,\begin{array}{*{20}{c}}{1 + {a^2} + {a^3}}&{{a^2}}&{{a^3}}\\{1 + {b^2} + {b^3}}&{{b^2}}&{{b^3}}\\{1 + {c^2} + {c^3}}&{{c^2}}&{{c^3}}\end{array}\,} \right|\\ = {\left| {\,\begin{array}{*{20}{c}}{1 + {a^3}}&{{a^2}}&{{a^3}}\\{1 + {b^3}}&{{b^2}}&{{b^3}}\\{1 + {c^3}}&{{c^2}}&{{c^3}}\end{array}\,} \right|^{}}\\ = {\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}\,} \right|^{}}\\ = (b - a)(c - a)(c - b){(ab + bc + ac)^{}}\end{array}\)
    \(\begin{array}{1}{D_y} = \left| {\,\begin{array}{*{20}{c}}1&{1 + {a^2} + {a^3}}&{{a^3}}\\1&{1 + {b^2} + {b^3}}&{{b^3}}\\1&{1 + {c^2} + {c^3}}&{{c^3}}\end{array}\,} \right|\\ = {\left| {\,\begin{array}{*{20}{c}}1&{{a^2} + {a^3}}&{{a^3}}\\1&{{b^2} + {b^3}}&{{b^3}}\\1&{{c^2} + {c^3}}&{{c^3}}\end{array}\,} \right|^{}}\\ = {\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}\,} \right|^{}}\\ = (b - a)(c - a)(c - b){(ab + bc + ac)^{}}\end{array}\)
    \(\begin{array}{1}{D_z} = \left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{1 + {a^2} + {a^3}}\\1&{{b^2}}&{1 + {b^2} + {b^3}}\\1&{{c^2}}&{1 + {c^2} + {c^3}}\end{array}\,} \right|\\ = {\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^2} + {a^3}}\\1&{{b^2}}&{{b^2} + {b^3}}\\1&{{c^2}}&{{c^2} + {c^3}}\end{array}\,} \right|^{}}\\ = {\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}\,} \right|^{}}\\ = (b - a)(c - a)(c - b){(ab + bc + ac)^{}}\end{array}\)
    \(\begin{array}{1}\frac{{{D_x}}}{D} = \frac{{(b - a)(c - a)(c - b)(ab + bc + ac)}}{{(b - a)(c - a)(c - b)(ab + bc + ac)}}\\ = 1\end{array}\)
    \(\begin{array}{1}\frac{{{D_y}}}{D} = \frac{{(b - a)(c - a)(c - b)(ab + bc + ac)}}{{(b - a)(c - a)(c - b)(ab + bc + ac)}}\\ = 1\end{array}\)
    \(\begin{array}{1}\frac{{{D_z}}}{D} = \frac{{(b - a)(c - a)(c - b)(ab + bc + ac)}}{{(b - a)(c - a)(c - b)(ab + bc + ac)}}\\ = 1\end{array}\)
    ∴ The solution of the system is \(x = 1{\rm{ }},{\rm{ }}y = 1{\rm{ }},{\rm{ }}z = 1\).