(a) Express (frac{5 + sqrt{2}}{3 – sqrt{2}} – frac{5 – sqrt{2}}{3 + sqrt{2}}) in the form (a + bsqrt{2}).
(b) Solve the following equations simultaneously using the determinant method.
(3x – y – z = -2)
(x + 5y + 2z = 5 )
(2x + 3y + z = 0)
Explanation
(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 + 5sqrt{2} + 3sqrt{2} + 2) – (15 – 5sqrt{2} – 3sqrt{2} + 2)}{9 – 2})
= (frac{17 + 8sqrt{2} – 17 + 8sqrt{2}}{7})
= (frac{0}{7} + frac{16sqrt{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)