(a) Express \(\frac{5 + \sqrt{2}}{3 - \sqrt{2}} - \frac{5 - \sqrt{2}}{3 + \sqrt{2}}\) in the form \(a + b\sqrt{2}\).
(b) Solve the following equations simultaneously using the determinant method.
\(3x - y - z = -2\)
\(x + 5y + 2z = 5 \)
\(2x + 3y + z = 0\)
(a) \(\frac{5 + \sqrt{2}}{3 - \sqrt{2}} - \frac{5 - \sqrt{2}}{3 + \sqrt{2}}\)
= \(\frac{(5 + \sqrt{2})(3 + \sqrt{2}) - (5 - \sqrt{2})(3 - \sqrt{2})}{3^{2} - (\sqrt{2})^{2}}\)
= \(\frac{(15 + 5\sqrt{2} + 3\sqrt{2} + 2) - (15 - 5\sqrt{2} - 3\sqrt{2} + 2)}{9 - 2}\)
= \(\frac{17 + 8\sqrt{2} - 17 + 8\sqrt{2}}{7}\)
= \(\frac{0}{7} + \frac{16\sqrt{2}}{7}\)
\(\therefore a = 0 ; b = \frac{16}{7}\)
(b) \(3x - y - z = -2\)
\(x + 5y + 2z = 5\)
\(2x + 3y + z = 0\)
We write the set of equations in matrix form.
\(\begin{pmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2 \\ 5 \\ 0 \end{pmatrix}\)
We find the inverse of the matrix, say A,
\(|A| = \begin{vmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{vmatrix} \)
= \(3(5 - 6) + 1(1 - 4) - 1(3 - 10) = 1\)
Cofactors, C
\(C = \begin{pmatrix} -1 & 3 & -7 \\ -2 & 5 & -11 \\ 3 & 7 & 16 \end{pmatrix}\)
\(C^{T} = \begin{pmatrix} -1 & -2 & 3 \\ 3 & 5 & 7 \\ -7 & -11 & 16 \end{pmatrix}\)
Inverse \(\frac{C^{T}}{|A|} = \begin{pmatrix} -1 & -2 & -3 \\ 3 & 5 & 7 \\ -7 & -11 & -16 \end{pmatrix}\)
\(x = \begin{pmatrix} -1 & -2 & -3 \\ 3 & 5 & 7 \\ -7 & -11 & -16 \end{pmatrix} \begin{pmatrix} -2 \\ 5 \\ 0 \end{pmatrix}\)
= \(\begin{pmatrix} 2 - 10 \\ -6 + 25 \\14 - 55 \end{pmatrix}\)
= \(\begin{pmatrix} -8 \\ 19 \\ -41 \end{pmatrix}\)
\(\therefore x = -8 ; y = 19 ; z = -41\)
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