### Determinants

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

• [TheoryMissing]
• 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 (*).
• (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.