Determinants

Recognise the concept and properties of determinants of order 2 and order 3

  • Theory
    [TheoryMissing]
  • Examples
    Consider the system of equations \((*)\;\left\{ \begin{array}{l}(k - 1)x + y + z = 1\\x + (k - 1)y + z = k\\x + y + (k - 1)z = {k^2}\end{array} \right.\), where k is a real number.
    (a) Find the range of values of k such that (*) has a unique solution.
    (b) Solve (*) when it has a unique solution.
    (c) Suppose that \(k = 2\). Solve (*).
  • Solutions
    (a) ∵ The system of equations has a unique solution.
    ∴ \(\begin{array}{1}\left| {\,\begin{array}{*{20}{c}}{k - 1}&1&1\\1&{k - 1}&1\\1&1&{k - 1}\end{array}\,} \right| \ne 0\\
    \left| {\,\begin{array}{*{20}{c}}{k + 1}&1&1\\{k + 1}&{k - 1}&1\\{k + 1}&1&{k - 1}\end{array}\,} \right| \ne {0^{^{^{^{^{^{^{^{^{}}}}}}}}}}\\
    (k + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{k - 1}&1\\1&1&{k - 1}\end{array}\,} \right| \ne {0^{^{^{^{^{^{^{^{^{}}}}}}}}}}\\
    (k + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\0&{k - 2}&0\\0&0&{k - 2}\end{array}\,} \right| \ne {0^{^{^{^{^{^{^{^{^{}}}}}}}}}}\\
    (k + 1)\,\left| {\,\begin{array}{*{20}{c}}{k - 2}&0\\0&{k - 2}\end{array}\,} \right| \ne {0^{^{^{^{^{^{}}}}}}}\\(k + 1){(k - 2)^2} \ne {0^{^{^{}}}}\end{array}\)
    \(\underline{\underline {k \ne - 1}} \) and \(\underline{\underline {k \ne 2}} \)

    (b) \(\begin{array}{1}D = \left| {\,\begin{array}{*{20}{c}}{k - 1}&1&1\\1&{k - 1}&1\\1&1&{k - 1}\end{array}\,} \right|\\
    = (k + 1){(k - 2)^2}^{^{}}\end{array}\)
    \(\begin{array}{1}{D_x} = \left| {\,\begin{array}{*{20}{c}}1&1&1\\k&{k - 1}&1\\{{k^2}}&1&{k - 1}\end{array}\,} \right|\\
    = \left| {\,\begin{array}{*{20}{c}}1&0&0\\k&{ - {\rm{ }}1}&{1 - k}\\{{k^2}}&{1 - {k^2}}&{k - 1 - {k^2}}\end{array}\,} \right|\\
    = \left| {\,\begin{array}{*{20}{c}}{ - {\rm{ }}1}&{1 - k}\\{1 - {k^2}}&{k - 1 - {k^2}}\end{array}\,} \right|\\
    = {k^2}(2 - k)\end{array}\)
    \(\begin{array}{1}{D_y} = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&1\\1&k&1\\1&{{k^2}}&{k - 1}\end{array}\,} \right|\\
    = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&1\\1&k&1\\0&{{k^2} - k}&{k - 2}\end{array}\,} \right|\\
    = (k - 1)\,\left| {\,\begin{array}{*{20}{1}}k&1\\{{k^2} - k}&{k - 2}\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{1}}1&1\\{{k^2} - k}&{k - 2}\end{array}\,} \right|\\
    = - k + 2\end{array}\)
    \(\begin{array}{1}{D_z} = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&1\\1&{k - 1}&k\\1&1&{{k^2}}\end{array}\,} \right|\\
    = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&0\\1&{k - 1}&1\\1&1&{{k^2} - 1}\end{array}\,} \right|\\
    = - {\rm{ }}\left| {\,\begin{array}{*{20}{1}}{k - 1}&1\\1&1\end{array}\,} \right| + ({k^2} - 1)\,\left| {\,\begin{array}{*{20}{1}}{k - 1}&1\\1&{k - 1}\end{array}\,} \right|\\
    = (k - 2)({k^3} - k - 1)\end{array}\)
    \(\frac{{{D_x}}}{D} = \frac{{{k^2}(2 - k)}}{{(k + 1){{(k - 2)}^2}}} = \frac{{ - {\rm{ }}{k^2}}}{{(k + 1)(k - 2)}}\),
    \(\frac{{{D_y}}}{D} = \frac{{ - {\rm{ }}k + 2}}{{(k + 1){{(k - 2)}^2}}} = \frac{{ - {\rm{ }}1}}{{(k + 1)(k - 2)}}\),
    \(\frac{{{D_z}}}{D} = \frac{{(k - 2)({k^3} - k - 1)}}{{(k + 1){{(k - 2)}^2}}} = \frac{{{k^3} - k - 1}}{{(k + 1)(k - 2)}}\)
    ∴ The solution of the system is \(x = \frac{{ - {\rm{ }}{k^2}}}{{(k + 1)(k - 2)}},{\rm{ }}y = \frac{{ - {\rm{ }}1}}{{(k + 1)(k - 2)}},{\rm{ }}z = \frac{{{k^3} - k - 1}}{{(k + 1)(k - 2)}}\).

    (c) When \(k = 2\),
    \(\left\{ \begin{array}{l}x + y + z = 1\;...............\;(1)\\x + y + z = 2\;..............\;(2)\\x + y + z = 4\;..............\;(3)\end{array} \right.\)
    (1), (2) and (3) cannot hold simultaneously.
    ∴ The system (*) has no solutions.