(a) If (frac{sqrt{5} + 4}{3 – 2sqrt{5}} – frac{2 + sqrt{5}}{4 – 2sqrt{5}} = a + bsqrt{5}), find the values of a and b.
(b)(i) Evaluate : (begin{vmatrix} 2 & -1 & 2 \ 1 & 3 & 4 \ 1 & 2 & 1 end{vmatrix})
(ii) Using the result in b(i), find, correct to two decimal places, the value of x in the system of equations.
(2x – y + 2z + 5 = 0)
(x + 3y + 4z – 1 = 0)
(x + 2y + z + 2 = 0)
Explanation
(a) (frac{sqrt{5} + 4}{3 – 2sqrt{5}} – frac{2 + sqrt{5}}{4 – 2sqrt{5}})
= (frac{(sqrt{5} + 4)(4 – 2sqrt{5}) – (2 + sqrt{5})(3 – 2sqrt{5})}{(3 – 2sqrt{5})(4 – 2sqrt{5})})
= (frac{4sqrt{5} – 10 + 16 – 8sqrt{5} – 6 + 4sqrt{5} – 3sqrt{5} + 10}{12 – 6sqrt{5} – 8sqrt{5} + 4(5)})
= (frac{10 – 3sqrt{5}}{32 – 14sqrt{5}})
Rationalizing,
((frac{10 – 3sqrt{5}}{32 – 14sqrt{5}}) (frac{32 + 14sqrt{5}}{32 + 14sqrt{5}}))
= (frac{320 + 140sqrt{5} – 96sqrt{5} – 42(5)}{1024 + 448sqrt{5} – 448sqrt{5} – 196(5)})
= (frac{110 + 44sqrt{5}}{44})
= (frac{5}{2} + sqrt{5}).
You can do the rationalizing before solving if you wish to. I just like mine a little complex.
(b)(i) (begin{pmatrix} 2 & -1 & 2 \ 1 & 3 & 4 \ 1 & 2 & 1 end{vmatrix})
= (2(3 – 8) – (-1)(1 – 4) + 2(2 – 3))
= (2(-5) + 1(-3) + 2(-1))
= (-10 – 3 – 2)
= (-15)
(ii) (2x – y + 2z = -5 … (1))
(x + 3y + 4z = 1 ….. (2))
(x + 2y + z = -2 ….. (3))
(x = frac{begin{vmatrix} -5 & -1 & 2 \ 1 & 3 & 4 \ -2 & 2 & 1 end{vmatrix}}{-15})
= (frac{-5(3 – 8) + 1(1 + 8) + 2(2 + 6)}{-15})
= (frac{25 + 9 + 16}{-15})
= (frac{50}{-15} = -3.33) (2 d.p)